DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS

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1 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS DONALD M. DAVIS Abstract. The partal Strlng numbers T n (k used here are defned as k. Ther 2-exponents ν(t n (k are mportant n odd ( n algebrac topology. We provde many specfc results, applyng to all values of n, statng that, for all k n a certan congruence class mod 2 t, ν(t n (k = ν(k k 0 + c 0, where k 0 s a 2-adc nteger and c 0 a postve nteger. Our analyss nvolves several new general results for ν( ( n 2+1, the proofs of whch nvolve a new famly of polynomals. Followng Clarke ([3], we nterpret T n as a functon on the 2-adc ntegers, and the 2-adc ntegers k 0 descrbed above as the zeros of these functons. 1. Man results The partal Strlng numbers T n (k used here are defned, for ntegers n and k wth n postve, by T n (k = odd ( n k. Other versons can be defned localzed at other prmes and summed over restrcted congruences. Let ν( denote the exponent of 2 n an nteger. The numbers ν(t n (k are mportant n algebrac topology ([1], [4], [6], [8], [9], [12], and work on evaluatng these numbers has appeared n the above papers as well as [3], [5], [11], [14], and [15]. In ths paper, we gve complete results for n 36 and also for n = 2 e + 1 and 2 e + 2, and we gve two famles of results applyng to all values of n but wth k restrcted to certan congruence classes. In [7], some of these results wll be appled to obtan new results for v 1 -perodc homotopy groups of the specal untary groups. We also present n Secton 2 some new results about ν( ( n 2+1 k. The proofs of these, n Secton 3, ntroduce a new famly of polynomals q m (x, whch mght be of Date: September 22, Key words and phrases. Strlng number, dvsblty, Hensel s Lemma Mathematcs Subect Classfcaton: 11B73, 05A99. 1

2 2 DONALD M. DAVIS ndependent nterest. Fnally, n Secton 4 we dscuss our results n the context of analytc functons on the 2-adc ntegers, and Hensel s Lemma. We begn wth the result whch s easest to state, and hence best llustrates the nature of our results. Theorem 1.1. Let e 2, n = 2 e + 1 or 2 e + 2, and 1 2 e 1. (1 There s a 2-adc nteger x,n such that for all ntegers x Moreover x,2 e +1 ν(t n (2 e 1 x + = ν(x x,n + n 2. { (mod 2 +1 f = 2 e 2 or 2 e 1 1 (mod 2 +1 otherwse and (mod 2 f 1 2 e 2 x,2 e (mod 2 +1 f 2 e 2 < < 2 e 1 1 (mod 2 +1 f = 2 e 1. (2 Let g(x = ν(t n (2 e 1 x+ (n 2. Then x,n = 2 t 0 +2 t 1 +, where t 0 = g(0 and t +1 = g(2 t t. If e = 1, the result s dfferent. See Table 1.3. For example, the last 24 dgts of the bnary expanson of x 4,9 are , and so we can make the followng more explct statement. 7 x 0 (2 8 x 3 (4 ν(t 9 (4x + 4 = 9 x 5 (8 10 x 9 (16, and contnue ndefntely, ust notng the last poston n whch the bnary expansons of x and x 4,9 dffer. Our next result utlzes Maple calculatons n ts proof. Although each case apples to nfntely many values of x, we wll explan n the proof how each case can be reduced to a small number of verfcatons.

3 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 3 Theorem 1.2. For each n 36, there s a partton of Z nto fntely many congruence classes C = [ mod 2 m ] such that, for each, ether (a there exsts a 2-adc nteger x 0 and a postve nteger c 0 such that ν(t n (2 m x + = ν(x x 0 + c 0 for all ntegers x, or (b there exsts a postve nteger y 0 such that ν(t n (k = y 0 for all k n C. The congruence classes C and ntegers c 0 and y 0 are as n Tables 1.3 and 1.4. Let g(x = ν(t n (2 m x + c 0. Then x 0 = 2 t t 1 +, wth t 0 = g(0 and t +1 = g(2 t t. We conecture that the general form of the theorem can be extended to all ntegers n;.e., that for each n there s a partton of Z nto fntely many congruence classes on each of whch ether ν(t n (k = ν(k k 0 + c 0 for some k 0 and c 0 or else ν(t n (k s constant on C. In the tables, the letter refers to any nteger.

4 4 DONALD M. DAVIS Table 1.3. Values of C, c 0, and y 0 n Theorem 1.2 n C c 0 y 0 n C c 0 y ( (2 3 1 (2 1 1 ( , 1 ( , 1 ( ( (2 7 1 (2 4 1 (2 9 9 ( ( , 2 ( , 2 (4 10 1, 3 (4 8 1, 3 ( , 2 ( , 3 (4 11 0, 4 (8 12 0, 1 ( (8 11 4, 5 ( ( ( ( ( ( (4 18 1, 5 (8 13 2, 6 (8 17 2, 6 (8 16 3, 7 ( ( ( , 2, 4, 6 ( , 2, 4, 6 (8 18 1, 3, 5, 7 (8 16 1, 3, 5, 7 ( , 2, 5, 6 ( , 1 (8 20 0, 3 (8 19 2, 3, 6, 7 (8 19 7, 15 ( , 12 ( ( , 21 ( ( ( ( ( ( (256 29

5 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 5 Table 1.4. More values of C, c 0, and y 0 n Theorem 1.2 n C c 0 y 0 n C c 0 y , 4 ( , 4 (8 22 3, 7 (8 19 1, 5 ( ( ( ( (8 25 6, 22 ( , 22 ( , 21 ( , 23 ( ( ( ( ( , 2, 3, 4 ( , 3, 4, 5 (8 23 5, 6, 7, 8 ( , 1, 7, 8, 9, 15 ( , 13, 14, 15 ( , 22 ( ( , 5 ( , 6 (8 25 4, 6 (8 24 5, 7 (8 26 1, 7, 9, 15 ( , 2, 8, 10 ( , 2, 8, 10 ( , 3, 9, 11 ( , 6 ( , 7 ( (8 26 2, 3, 10, 11 ( , 2, 9, 10 ( , 1, 8, 9 ( , 4, 8, 12 ( , 5, 12, 13 ( ( ( ( ( ( ( (8 35 3, 11 ( , 12 ( , 5, 9, 13 ( , 6, 10, 14 ( , 12 ( , 13 ( , 6, 10, 14 ( , 7, 11, 15 ( ( ( ( ( ( (16 35 One can notce a lot of nce patterns n these tables, and formulate (and sometmes prove conectures about ther extenson to all values of n. One nterestng dea, followng Clarke ([3], s to note that snce T n (k mod 2 m only depends on k mod 2 m 1, T n ( extends to a functon T n : Z 2 Z 2, where Z 2 denotes the 2-adc ntegers. Here the metrc on Z 2 s gven, as usual, by d(x, y = x y, where z := 1/2 ν(z.

6 6 DONALD M. DAVIS The 2-adc ntegers 2 m x 0 + whch occur n Theorem 1.2 are ust the zeros of the functon T n. We can count the number of zeros to be gven as n Table 1.5, and mght try to formulate a guess about the general formula for ths number of zeros. Table 1.5. Number of zeros of T n n Number of 0 s of T n Our second general result establshes for all n, except those 1 less than a 2-power, the values of ν(t n (k for k n the congruence class contanng 0. We could almost certanly nclude n = 2 e 1 nto ths theorem, but the detals of provng that case are so detaled as to be perhaps not worthwhle here. Theorem 1.6. Let n 5 and ( 2, 1 f n = 2 e (a, b = ( 1, 2 f 2 e < n 3 2 e 1 (0, 1 f 3 2 e 1 < n 2 e+1 2. Then there exsts a 2-adc nteger x n such that for all ntegers x ν(t n (2 e+a x = ν(x x n + n b. The cases n = 2 e + 1 and 2 e + 2 of ths theorem overlap wth Theorem 1.1. For these n, we have x n = 1 + x 2 e 1,n. For all n n Theorem 1.6, there s an algorthm for x n totally analogous to that of Theorem 1.1. Our next result s of a smlar nature, but apples to many more congruence classes. The cases to whch t apples are those n whch the 2-exponent of a certan sum (see (2.34 and (2.35 s determned by exactly one of ts summands, and for whch the mod 4 result 2.6 suffces to prove t. The algorthm for computng x 0 s lke that of

7 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 7 Theorem 1.2. Here and throughout, α(n denotes the number of 1 s n the bnary expanson of n. Theorem 1.7. Suppose 2 e + 2 t n < 2 e + 2 t+1 wth 0 t e 1. Let S n = {p : max(0, n 2 e 2 e 1 p < 2 e 1 and ( n 1 p p 1 (2}. If p S n, say that an nteger q < 2 e 1 s assocated to p f q = p or q = p + 2 w wth w = ν(n 1 or w > t. If q s assocated to an nteger p of S n, then there exsts a 2-adc nteger x 0 such that for all ntegers x ν(t n (2 e 1 x + q = ν(x x 0 + n 2 α(p 0, where p 0 s the resdue of p mod 2 t. A bt of work s requred to get any sort of feel for the complcated condton n ths theorem. In Table 1.8, we lst for n from 17 to 31, the values of p n S n, then the addtonal values of q covered by the theorem, and fnally the values of for whch Theorem 1.2, as depcted n Tables 1.3 and 1.4, gves a value for the congruence mod 8 whch s not covered by Theorem 1.7. The strength of the theorem s, of course, that t apples to all values of n (except 2-powers. Table 1.8. Comparson of Theorems 1.7 and 1.2 n p S n addtonal q mod 8 results mssed 17 0, 1, 2, 4 3, 5, , 2, 4 1, 3, 5, , 1, 3, 4, 5 7 2, , 4 2, 6 1, 3, 5, , 1, 2, 5, , 2, 6 1, 3, , 1, 3, 7 2, , 2, 3, , 2, , 4 3, , 4, , ,

8 8 DONALD M. DAVIS 2. Proofs of man theorems In ths secton, we prove the four man theorems lsted n Secton 1. A central ngredent n the proofs s results about ν ( ( n 2+1 k. We begn by provdng sx results about ths, of whch all but the frst are new. The proofs of most of these appear n Secton 3. The frst result was proved n [9, 3.4]. Proposton 2.1. ([9, 3.4] For any nonnegatve ntegers n and k, ν ( ( n 2+1 k ν([n/2]!. In usng ths, and many tmes throughout the paper, we use (2.2 ν(n! = n α(n. The next result s a refnement of Proposton 2.1. Here and throughout, S(n, k denote Strlng numbers of the second knd. Proposton 2.3. Mod 4 S(k, n + 2nS(k, n 1 ɛ = 0, b = 0 ( 1 2n+ɛ n! 2+b k (2n + 1S(k, n + 2(n + 1S(k, n 1 ɛ = 1, b = 0 2nS(k, n 1 ɛ = 0, b = 1 S(k, n + 2(n + 1S(k, n 1 ɛ = 1, b = 1. The proofs of the last three propostons all nvolve new polynomals q m (x, whch mght be of ndependent nterest. See Defnton 3.1 for the defnton, whch pervades Secton 3. Proposton 2.4. For any nonnegatve ntegers n and k, ν ( ( n 2+1 k n k α(n. Proposton 2.5. For any nonnegatve ntegers n and k wth n > k, ν ( ( n 2+1 k n 1 k α(k wth equalty ff ( n 1 k k s odd. The fnal proposton s a refnement of Proposton 2.5.

9 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 9 Proposton 2.6. If n and k are nonnegatve ntegers wth n > k, then, mod 4, ( n 2+1 k /(2 n 1 2k k! ( n 1 k {2 ( n 1 k f n 1 and k are even k + k 2 0 otherwse. The followng corollary wll also be useful. Corollary 2.7. For n 3, > 0, and p Z, ν( ( n 2+1 (2 + 1 p max(ν([ n ]!, n α(n 2 wth equalty f n {2 e + 1, 2 e + 2} and = 2 e 1. Proof. The sum equals k 0 T k, where T k = 2 k( ( p n k 2+1 +k. By Proposton 2.1, ν(t k ν([ n 2 ]!, whle by Proposton 2.4, ν(t k n α(n, mplyng the desred nequalty. If n = 2 e + 1 and = 2 e 1, then ν(t 0 = 2 e 1 1 by Proposton 2.5, whle for k > 0, ν(t k 2 e 1 by 2.1. If n = 2 e + 2 and = 2 e 1, ν(t 0 = 2 e 1 by 2.5, ν(t 1 > 2 e 1 by 2.3, and ν(t k > 2 e 1 for k > 1 by 2.1. Our proofs of the theorems of Secton 1 wll make essental use of the followng result of [5]. Here and throughout, we wll employ the useful notaton { mn mn(m, n f m n (m, n = a number > m f m = n. Note that mn (m, m s not a well-defned number, and that ν(m+n = mn (ν(m, ν(n. Lemma 2.8. ([5] Let N denote the set of nonegatve ntegers. A functon f : Z N { } s of the form f(n = ν(n E for some 2-adc nteger E ff t satsfes f(n + 2 d = mn (f(n, d for all d N and all n Z. In ths case, E = 0 2e, where e 0 = f(0 and e k+1 = f(2 e e k. We begn the proofs of the theorems of Secton 1 by dscussng the proof of Theorem 1.2. The way that the cases of Theorem 1.2 are dscovered s by havng Maple compute values of ν(t n (k for ranges of values of k. For example, seeng that ν ( k takes on the values 17, 25, 17, 18, 17, 19, 17, 18, 17, 20, 17, 18 as k goes 10, 18, odd ( 19

10 10 DONALD M. DAVIS 26,..., 98 makes one pretty sure that for all ntegers x we have ν(t 19 (8x + 2 = ν(x x for some 2-adc nteger x 0, and you could even guess that the last 9 dgts n the bnary expanson of x 0 are But to prove t, more s requred. Ths s a case not covered by any of our three general theorems, but the proofs of all four of our theorems have smlar structure. Let f(x = ν(t 19 (8x Then f(x + 2 d = 17 + ν ( ( ( x+2 + ( 19 ( ( x+2 (( d 1 = mn ( f(x, ν ( ( ( x+2 (( d Thus the clam that ν(t 19 (8x + 2 = ν(x x for some 2-adc nteger x 0 wll follow from Lemma 2.8 once we show that (2.10 ν ( ( ( x+2 (( d 1 = d + 17 for all x and d 0. We expand the two powers of (2 + 1, obtanng terms, for k 0 and > 0, wth 2-exponent (2.11 ν ( ( 8x+2 k + ν 2 3+d ( ( + k + + ν k+. Let ψ(s = s + ν( ( s. Snce ν ( 2 3+d = 3 + d ν(, t wll suffce to show that the mnmum value of ψ(k + ν( + ν ( 8x+2 k s 14, and that ths value occurs for an odd number of pars (k,. Maple computes that the mnmum value of ψ(s s 16, whch occurs when s = 3, 5, 7, or 9, and that ψ(s = 17 for s = 1, 4, 6, 8, and 10. For s 11, ψ(s 7 + s by 2.1. Ths nformaton makes t easy to check that the mnmum value of ψ(k + ν( + ν ( 8x+2 k s ndeed 14, and ths value occurs exactly when (k, = (0, 8, (2, 8, or (1, 8. Ths completes the proof that for all ntegers x we have ν(t 19 (8x + 2 = ν(x x for some 2-adc nteger x 0. Each of the cases of Theorem 1.2 can be establshed n ths manner, although many of the cases are covered by our general theorems 1.1, 1.6, and 1.7. The cases n whch T n (k s constant on a congruence class are proved smlarly, although Lemma 2.8 need not be used. For example, to show ν(t 13 (8x + 7 = 11 for

11 all x, we frst defne DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 11 θ(k = 2 k ( k. Maple and 2.1 show = 10 k = 5, 6 ν(θ(k = 11 k = 1, 2, 7 > 11 other k. Snce T 13 (8x + 7 = ( ( 8x+7 k θ(k, and 8x+7 k s odd for k {5, 6, 1, 2, 7}, we obtan, mod 2 12, T 13 (8x + 7 ( ( 8x+7 5 θ(5 + 8x+7 6 θ( Maple shows θ(5 θ( mod Snce ( 8x+7 ( 5 + 8x+7 ( 6 = 8x+7 5 (1 + 8x+2 0 (mod 4, 6 we obtan ( ( 8x+7 5 θ(5 + 8x+7 6 θ(6 0 mod 2 12, from whch our desred concluson follows. Ths concludes our comments regardng the proof of Theorem 1.2. Now we work toward proofs of the more general results, Theorems 1.1, 1.6, and 1.7. Frst we recall some background nformaton. We wll often use that (2.12 ( 1!S(k, = ( 2 (2 k T (k. Sometmes we have k <, n whch case S(k, = 0, and so T (k = ( 2 (2 k when k <. Other tmes we use (2.12 to say that T (k ±!S(k, mod 2 k. Many tmes we wll use wthout comment the fact, related to (2.2, that ν (( m n = α(n + α(m n α(m. Closely related s the fact that ( m n s odd ff each dgt n the bnary expanson of m s at least as large as the correspondng dgt of n. We wll sometmes say that ( m n s even due to the 2 t -poston, meanng that n ths poston m has a 0 and n has a 1. Other basc formulas that we use wthout comment are and, f 0 < < 2 t, then α(n 1 = α(n 1 + ν(n α(2 t+1 A + 2 t + = α(a + t α( 1.

12 12 DONALD M. DAVIS We also use the well-known formula (2.13 S(k +, k ( k+2 1 k 1 mod 2. A generalzaton to mod 4 values was gven n [2] and wll be used several tmes. We wll not bother to state all eght cases of that theorem ust those that we need. The proof of Theorem 1.1 utlzes the followng two lemmas. Lemma Let 1 2 e 1 wth e 2. Then { ν(t 2 e +1(2 e 1 = 2 e 1 + {2 e 2, 2 e 1 } + 2 e + otherwse, whle = 2 e e 2 ν(t 2 e +2(2 e 1 + = 2 e + 2 e 2 < < 2 e 1 > 2 e + = 2 e 1, Proof. For the frst part, by the remarks followng (2.12, we must show ν( { ( 2 e e 1 + = 2 e 1 1 {2 e 2, 2 e 1 } 2 e 1 otherwse, or equvalently 1 (2 e 1! ( 2 e e 1 + { 1 mod 2 {2 e 2, 2 e 1 } 0 mod 2 otherwse. By Proposton 2.3, the LHS s congruent mod 2 to S(2 e 1 +, 2 e 1, and by (2.13 ths s ( 2 e e 1 1, whch s as requred. The second part of the lemma reduces smlarly to showng ( 1 ( mod e 2 e +2 (2 e 1 +1! 2 2 e mod 4 2 e 2 < < 2 e 1 0 mod 4 = 2 e 1. By 2.3, the LHS s congruent mod 4 to (2.16 S(2 e 1 +, 2 e S(2 e 1 +, 2 e 1. Mod 2, ths s ( 2 e e 1 whch s odd f 1 2 e 2. Now assume 2 e 2 < < 2 e 1. The second term of (2.16 s easly seen to be 0 mod 4 usng (2.13. For the frst term of (2.16, we use part of [2, Thm 3.3], whch relates mod 4 values of S(n, k to bnomal coeffcents. It mples that, f e 3, the mod 4 value of the frst term s ( 2 e 2 +k 2 e 3,

13 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 13 where 0 k < 2 e 3. The 2-exponent n ths number s 1+α(2 e 3 +k α(2 e 2 +k = 1, as desred. If = 2 e 1, both terms of (2.16 are 2 mod 4, by a smlar analyss. Lemma If p Z, δ = 1 or 2, and ν(n = e + wth 1, then ν ( ( 2 e +δ 2+1 (2 + 1 p ((2 + 1 n 1 = 2 e + + δ 1. Proof. The sum equals >0 T, where T := 2 ( n ( 2 e +δ 2+1 (2 + 1 p. For evaluaton of the 2-exponent of the -sum here, we use Corollary 2.7. We obtan that f 2 e+, then ν(t + e + ν( + { 2 e + δ e 1 2 e 1 1 > 2 e 1, wth equalty f = 2 e 1. Ths s 2 e + + δ 1 wth equalty ff = 2 e 1. If > 2 e+, then ν(t > 2 e + snce 2 e+ + 2 e 1 > 2 e + for 1. Now we easly prove Theorem 1.1. Proof of Theorem 1.1. Let δ {1, 2}, 1 2 e 1, and let g(x = ν(t 2 e +δ(2 e 1 x + 2 e e + 2 δ. Note that the expresson that we wsh to evaluate for Theorem 1.1 s g(x 1+2 e 2+δ. For d 0, wrtng T n ( as a sum of two parts as we dd n (2.9, g(x+2 d = mn ( g(x, 2 e +2 δ+ν( ( 2 e +δ 2+1 (2+1 p ((2+1 2d+e 1 1, where p = 2 e 1 x + 2 e 1 +. By Lemma 2.17, the RHS equals mn (g(x, d, and so g(x = ν(x E δ for some E δ by Lemma 2.8. By Lemma 2.14 = δ = 1, {2 e 2, 2 e 1 } > δ = 1, {2 e 2, 2 e 1 } ν(e δ = g(0 = 1 δ = 2, 1 2 e 2 = δ = 2, 2 e 2 < < 2 e 1 > δ = 2, = 2 e 1. Our desred g(x 1+2 e 2+δ equals ν(x 1 E δ +2 e 2+δ, and x,2 e +δ := 1+E δ s as clamed.

14 14 DONALD M. DAVIS The proof of Theorem 1.6 s smlar n nature, but longer. Proof of Theorem 1.6. Usng Lemma 2.8 and argung as n (2.10, t suffces to prove that for d 0 and any nteger x (2.18 ν ( ( n 2+1 ( e+ax (( e+a+d 1 = n b + d. Indeed, f g(x = ν ( ( n 2+1 ( e+a x n + b, then (2.18 mples g(x + 2 d = mn (g(x, d and Theorem 1.6 then follows from Lemma 2.8. We wrte the sum n (2.18 as T wth (2.19 ν(t = + e + a + d ν( + ν ( ( n 2+1 ( e+ax. We wll show that n all cases ν(t s mnmzed for a unque value of. The second case of the theorem wll follow from provng that f 2 e < n 3 2 e 1 and ν(p e 1, then + e 1 ν( + ν ( ( n 2+1 (2 + 1 p n 2 wth equalty ff = 2 e 1. Expandng (2 + 1 p as k 0 2k( p k k leads us to needng that for > 0 ( e ν( + ν ( ( n 2+1 n 1 wth equalty ff = 2 e 1, and ( k + 2e ν( ν(k + ν ( ( n 2+1 +k > n for, k > 0. The equalty n (2.20 when = 2 e 1 follows easly from Proposton 2.5. Also by 2.5, the dfference n (2.20 becomes ( e + 1 ν( + ν ( ( n 2+1 n e α( ν( = e 1 α( 1. Ths s > 0 f 2 e 1 and < 3 2 e 2, whle f = 3 2 e 2, then ( n 1 = 0 and so (2.22 s > 0 by 2.5.

15 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 15 Now suppose > 3 2 e 2. Then ν( > 3 2 e 2, and snce n 3 2 e 1, (2.23 n ν([ n 2 ]! 3 2e 2 + e 1. Thus, usng Proposton 2.1, we obtan + e 1 + ν( + ν ( ( n 2+1 > 3 2 e 2 + e ν([ n ]! n, 2 establshng strct nequalty n (2.20. Now we verfy (2.21. By 2.5, (2.21 s satsfed f (2.24 ν( + ν(k + α( + k 2e 2 or f (2.25 ν( + ν(k + α( + k = 2e 1 and ( n 1 k +k By 2.1 and (2.23, (2.21 s also satsfed f ( k ν( ν(k 3 2 e 2 e. 0 (mod 2. If + k > 3 2 e 2, then (2.26 s satsfed. If {, k} = {2 e 1, 2 e 2 }, then (2.25 s satsfed. Assume WLOG that ν( ν(k. Then (2.24 s mpled by ν(+1+α( + k 1 2e 2 and ths s satsfed whenever + k 3 2 e 2 and (, k (2 e 1, 2 e 2. The thrd case of the theorem wll follow from provng that, referrng to (2.19, f 2 e+1 2 t+1 < n 2 e+1 2 t wth 1 t < e 1 and ν(p e, then + e + d ν( + ν ( ( n 2+1 (2 + 1 p n 1 + d wth equalty ff = 2 e 2 t. Expandng (2 + 1 p, ths reduces to showng f > 0 then ( e + 1 ν( + ν ( ( n 2+1 n wth equalty ff = 2 e 2 t, and f, k > 0, then ( k + 2e ν( ν(k + ν ( ( n 2+1 +k n. If > 2 e, snce n 2 e+1 2, + e + 1 ν( 2 e + e + 1 > n ν([ n 2 ]!, and so strct nequalty holds n (2.27 by 2.1. By Theorem 2.5, (2.27 s satsfed f (2.29 e ν( + α( = α( 1 + 1,

16 16 DONALD M. DAVIS and equalty holds n (2.27 ff equalty holds n (2.29 and ( n 1 s odd. If < 2 e, then α( 1 e 1 wth equalty ff = 2 e 2 r for some r. Thus (2.29 holds wth equalty ff = 2 e 2 t by Lemma If = 2 e, by Proposton 2.6 the LHS of (2.27 s n + 1. Thus (2.27, ncludng consderaton of equalty, has been establshed for all. By 2.1, (2.28 s satsfed f + k + 2e ν( ν(k n ν([ n 2 ]!, and hence, snce n 2 e 2, t s satsfed f Ths s satsfed f + k > 2 e. + k + 2e ν( ν(k 2 e + e 1. By 2.5, (2.28 s also satsfed f ν( + ν(k + α( + k 2e 1. Ths s satsfed f = k = 2 e 1 and f ν( + α( + k 1 2e 2, whch s true for all other (, k wth + k 2 e. The frst case, n = 2 e, wll follow smlarly from ( e 1 ν( + ν ( ( 2 e 2 e 2+1 for > 0 wth equalty ff = 2 e 2, whle f, k > 0, then ( k + 2e 2 ν( ν(k + ν ( ( 2 e 2+1 +k 2 e + 2. Equalty n (2.30 wth = 2 e 2 follows from Proposton 2.6 snce ( 2 e 1 2 e 2 2 e 2 2 mod 4. If > 2 e 1, then strct nequalty n (2.30 s mpled by 2.1. If = 2 e 1, t s mpled by Proposton 2.3. It s mpled by Theorem 2.4 f ν( e 3, whch s true for < 2 e 1 provded 2 e 2. Smlarly, (2.31 s mpled by 2.1 f + k 2 e 1 unless = k = 2 e 2, n whch case t s mpled by 2.3. If + k < 2 e 1, then ν( + ν(k 2e 5, and so the clam follows from Proposton 2.4. The followng lemma was used n the above proof. Lemma If 2 e+1 2 t+1 m < 2 e+1 2 t wth 0 t < e and = 2 e 2 r wth 0 r < e, then ( m s odd ff r = t.

17 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 17 ( = 0. If r = t, then m ( = 2 e 2 t +d 2 e 2 t Proof. If r < t, then 0 m <, so ( m wth 0 d < 2 t and hence s odd. If r > t, then the bnary expanson of m has a 0 n the 2 r poston, whle has a 1 there. The followng lemma wll be useful n the proof of Theorem 1.7. Lemma In the notaton of Theorem 1.7, f p S n, then α(p p 0 1 and ( and n 2 e 1 p e 1 +p 0 are odd. ( n p0 1 p 0 Proof. Let p = A2 t + p 0 and n = 2 e + 2 t + wth both p 0 and nonnegatve and less than 2 t. If p 0 1 < 0, then ( 2 e A2 t A2 s odd, whch easly mples α(a 1, t whle f p 0 1 0, then ( 2 e +2 t A2 t A2 s odd. Ths mples that A s 0 or an even t 2-power. Now, f p p 0, we can wrte p = p t+r wth r 0. If r = 0, then ( 2 e + 1 p 0 odd mples 1 p 0 < 0 and ( 2 t + 1 p 0 ( p 0 odd, whch mples n p0 1 p 0 odd. If r > 0, then the odd bnomal coeffcent can be consdered mod 2 as ( ( 2 e 2 t+r 2 t + 1 p 0 2 t+r p 0, whch mples ( n p 0 1 p 0 s odd. Now we may assume ( n p 0 ( 1 p 0 s odd. Thus 2 t + 1 p 0 p 0 s odd and hence so s ( 2 e 1 +2 t + 1 p 0 2 e 1 +p 0. The proof of Theorem 1.7 s smlar to the others, but longer yet. 2 t +p 0 Proof of Theorem 1.7. Smlarly to the proofs of the other three theorems, t suffces to prove for d 0 and any nteger x (2.34 ν ( ( n 2+1 (2+1 2 e 1 x+q ((2+1 2e 1+d 1 = d+n 2 α(p 0. Here, and for the remander of ths secton, n, e, t, q, p, and p 0 are as n Theorem 1.7. To prove (2.34, t suffces to show for k 0 and > 0 (2.35 ν ( ( ( 2 e 1 x+q k +e 1 ν(++k+ν n 2+1 +k n 2 α(p 0 wth equalty ff = 2 e 1 and k = p 0. We frst prove the equalty. Note that f q s assocated to p S n, then ( 2 e 1 x+q p 0 s odd. We must show that ν ( ( n e 1 +p 0 = n 2 p0 2 e 1 α(p 0. Ths follows from Proposton 2.5 snce ( n 2 e 1 1 p 0 2 e 1 +p 0 s odd by Lemma 2.33.

18 18 DONALD M. DAVIS Strct nequalty n (2.35 when = 2 e 1 and k p 0 follows from Lemma 2.36 usng Propostons 2.1 and 2.6. Here we also use that f p S n and k satsfes (2.37 then k < 2 e 1 and hence the x n ( 2 e 1 x+q k does not play an essental role. Lemma If n, e, t, q, p, and p 0 are as n Theorem 1.7 and ( k n ν([ n 2 ]! 2 2e 1 α(p 0, then (2.38 α(q k + α(p 0 α(q 1. If the LHS of (2.38 equals 1, then ether n s even and ( n 1 2 e 1 k 2 e 1 +k 0 mod 4 or ( ( n 1 2 e 1 k 2 e 1 +k = 0 = n 1 2 e 1 k 2 e 1 +k 2. If the LHS of (2.38 equals 0, then ether 0 mod 2 or k = p0. ( n 1 2 e 1 k 2 e 1 +k Proof. We begn by provng (2.38. Usng Lemma 2.33 and that q s assocated to p, we have (2.39 q = p 0 + δ2 t+r + ɛ2 w wth δ and ɛ equal to 0 or 1, r 0, and w = ν(n 1 or w > t. The only way that (2.38 could fal s f k = q = p t+r + 2 w. But (2.37 mples k 2 t + t 2, whch s nconsstent wth w > t and wth r > 0. Thus n s even and w = ν(n 1 and r = 0. Let n = 2 e + 2 t + c2 t 1 + 2b wth c {0, 1} and b 2 t 2 1. If c = 0, then (2.37 reduces to p 0 + α(p t w t 3, whch s false. If, on the other hand, c = 1, the assumpton that ( n p 1 p s odd and p 2 t mples that p 0 2 t 1 + 2b, so k > 2 t + 2 t 1 contradcts k 2 t + t 2. There are three concevable ways n whch equalty could hold n (2.38. One s ɛ = 0, δ = 1, and k = q;.e., k = q = p 2 t. But k 2 t mples 2 e 1 + k > n 1 2 e 1 k and hence ( ( n 1 2 e 1 k 2 e 1 +k = 0. We also have n 1 2 e 1 k 2 e 1 +k 2 = 0; the only way ths could fal s f n = 2 e + 2 t+1 1 and k = p = 2 t, but then ( n p 1 p s not odd. Another s ɛ = δ = 1 and α(q k = 1. In ths case, the only way to have k < 2 t s f w = ν(n 1 and k = q 2 t+r, where r s as n (2.39. In ths case, ( n 1 2 e 1 k 0 2 e 1 +k s dvsble by 2 t f there are at least t carres n mod 4, usng the result that ( a+b b the bnary addton of a and b. In ths case, ether the bnomal coeffcent equals 0, or else, f v = ν(n, there wll be carres n the 2 v 1 and 2 v postons n the relevant

19 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 19 bnary addton. The thrd possblty, ɛ = 1, δ = 0, and k = q, mples that n s even and ν(q = ν(n 1 and leads to ( n 1 2 e 1 k 2 e 1 +k 0 mod 4, exactly as above. Fnally we show that f ( n 1 2 e 1 k 2 e 1 +k s odd and the LHS of (2.38 equals 0, then k = p 0. It s not dffcult to see that f 0 k < n 2 e 1 and ( n 1 2 e 1 k 2 e 1 +k s odd, then k < 2 t and ( n 1 k k s odd. Frst suppose α(q k = 2 and α(q α(p 0 = 2. If q = p t+r + 2 t+s wth 0 r < s, then to keep k < 2 t, we must have k = p 0. If q = p t+r + 2 ν(n 1 wth r 0, then we must have k = p ν(n 1 2 u for some u. If u ν(n 1, then s even due to the 2 mn(u,ν(n 1 -poston. ( n 1 2 e 1 k 2 e 1 +k Now suppose α(q k = 1 and α(q α(p 0 = 1. If q = p t+r wth r 0, then k = p 0 s the only way to have k < 2 t. If q = p ν(n 1, then the argument at the end of the precedng paragraph apples. Ths completes the proof of Lemma 2.36, and hence the proof that when = 2 e 1, (2.35 holds wth equalty exactly as clamed there. We contnue the proof of Theorem 1.7 by establshng strct nequalty n (2.35 when 0 < < 2 e 1 and 0 k < 2 e 1. The followng elementary lemma wll be useful. Lemma Suppose 0 < < 2 e 1 and 0 k < 2 e 1. Let φ(, k = α( + k + ν( α(k. Then (1 φ(, k e 1; (2 φ(, k = e 1 ff = 2 e 1 2 h and 0 k < 2 h for some 0 h < e 1; (3 φ(, k = e 2 ff ether = 2 e 1 2 h and 2 e 2 k < 2 e h for some 0 h < e 1, or = 2 e 1 2 l 2 h, 0 h l < e 1, and 0 k < 2 h or 2 l k < 2 l + 2 h. Proof. Let h = ν(, and let k 0 = k (k mod 2 h. Then φ(, k = h+α(+k 0 α(k 0. The only way to get α( + k 0 α(k 0 = e h 1 s f + k 0 = 2 e 2 h and α(k 0 = 1, or +k 0 = 2 e 1 2 h and k 0 = 0. But the frst s mpossble snce k 0 < 2 e 1. Smlarly the only ways to get α( +k 0 α(k 0 = e h 2 s f (α( +k 0, α(k 0 = (e h 1, 1 or (e h, 2, and these can only be accomplshed n the ways lsted n part (3.

20 20 DONALD M. DAVIS Let ( m s s = ( m s s {2 ( m s f m and s are even + s 2 0 otherwse and ( (2.41 ν m s ( 2 s = mn(2, ν( m s. s Let φ(, k be as n Lemma The desred strct nequalty n (2.35 when 0 < < 2 e 1 and 0 k < 2 e 1 follows from the followng result usng Proposton 2.6. Theorem If n, e, t, q, p, and p 0 are as n Theorem 1.7, 0 < < 2 e 1, and 0 k < 2 e 1, then (2.43 α(q k + e 1 + ν 2 ( n 1 k +k α(q α(p0 + φ(, k. Proof. By Lemma 2.33, p = p 0 or p t+s wth s 0. Hence α(q α(p 0 2. Also ν(p ν(n, a consequence of the oddness of ( n 1 p p, wll be used often wthout comment. The theorem wll follow from showng: f φ(, k = e 1, then 2 f α(q α(p ( ν n 1 k 0 = 2 and k = q 2 +k 1 f α(q α(p 0 = 2 and α(q k = 1 1 f α(q α(p 0 = 1 and k = q, and f φ(, k = e 2, then ν ( n 1 k +k 1 f α(q α(p0 = 2 and k = q. We call these cases 1 through 4. Let n = 2 e +2 t + wth 0 < 2 t. Our hypothess s that ( 2 e +2 t ɛ2 t+s + p 0 1 ɛ2 t+s +p 0 s odd. Case 3: We have q = p r wth r t or r = ν(n 1, n whch latter case and p 0 are dvsble by 2 r+1. We must show that ( 2 e 1 +2 t +2 h + 1 p 0 2 r 2 e 1 2 h +p 0 +2 s even. r Here 2 h > p r. If r t, then the bnomal coeffcent s even due to the 2 r - or 2 h -poston, whle f r = ν(n 1, t s even due to the 2 ν(n 1 -poston. Case 2: Here q = p 0 +2 s +2 r wth s t and r = ν(n 1 or r > s. Also k = q 2 v and 2 h > k. The bnomal coeffcent whch we must show s even s ( 2 e h + 2 t + 1 p 0 2 s 2 r + 2 v C :=. 2 e 1 2 h + p s + 2 r 2 v

21 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 21 If v = r or s, t reduces to Case 3, ust consdered. If r = ν(n 1, then C s even due to the 2 mn(v,ν(n 1 -poston. Otherwse C s even due to the 2 h -poston, snce 2 t + 1 p 0 2 s 2 r + 2 v s negatve. Case 1: Now q s as n Case 2, but k = q. We must show that there are at least two carres n the bnary addton of 2 e 1 2 h + p s + 2 r and 2 h t + 1 2p 0 2 s+1 2 r+1. If r = ν(n 1, carres occur n postons 2 r and 2 r+1. If r > s, carres occur n 2 r and 2 h. The second term n the defnton of ( m s s s easly seen to be nconsequental here. Case 4: Agan q s as n Case 2, k = q, and (, k s one of the two types n Theorem For the frst type of (, k, f r > s, then + k > n 1 k, so ( n 1 k +k = 0, whle f r = ν(n 1, the bnomal coeffcent s even due to the 2 ν(n 1 -poston. If (, k s of the second type and k = p s + 2 ν(n 1, then ( n 1 k +k s even due to the 2 ν(n 1 -poston, snce p 0, n, and are all dvsble by 2 ν(n. If = 2 e 1 2 l 2 h and 2 l k < 2 l + 2 h wth k = p s + 2 r wth r > s, we clam that ( n 1 k +k s even due to the 2 e 2 -poston. Indeed, 2 e 1 2 h + k < 2 e 1, so + k has a 1 n the 2 e 2 -poston, whle 2 e 1 n k 1 < 2 e t+1 2 h < 2 e e 2 snce t < e 3. If = 2 e 1 2 e 2 2 h and 2 t+1 < k < 2 h, one easly verfes that ( n 1 k +k s even due to the 2 e 3 -poston. Fnally, f = 2 e 1 2 l 2 h wth h < l < e 2 and 2 t+1 < k < 2 h, then ( n 1 k +k s even due to the 2 e 2 -poston, as s easly proved. Our fnal step n the proof of Theorem 1.7 s to prove strct nequalty n (2.35 when > 2 e 1. Proposton 2.1 mples the result f k 2 t or f > 2 e. Thus, by Proposton 2.6, t suffces to prove (2.43 when 2 e 1 < 2 e and 0 k < 2 t. Recall that q s as n (2.39. Because k < 2 t, t must be the case that f δ = 1, then 2 t+r appears n q k, and smlarly 2 w f ɛ = 1 and w > t. These wll contrbute to α(q k. Thus the only ways to have D k := α(q k (α(q α(p 0 0 are (a k = p 0 and D k = 0; (b k = p ν(n 1 and D k = 1; and (c k = p ν(n 1 2 v and D k = 0.

22 22 DONALD M. DAVIS Smlarly to Lemma 2.40, we have for 2 e 1 2 e and 0 k < 2 e 1, φ(, k e wth equalty ff = 2 e 2 h and 0 k < 2 h for some 0 h < e, or = 2 e. We wll be done once we prove the followng lemma. Lemma If p S n and 2 e 1 < 2 e and 0 k < 2 t, then (1 f k = p 0 or p ν(n 1 2 v for some v, and φ(, k = e, then ( n 1 k +k s even. ( (2 f k = p ν(n 1 and φ(, k = e, then ν n 1 k 2 +k = 2. (3 f φ(, k = e 1 and k = p ν(n 1, then ( n 1 k +k s even. ( = 0 = ν n 1 k, so now we may assume < 2 e. Proof. If = 2 e, then ν ( n 1 k +k +k If k = p 0 or p ν(n 1 and φ(, k = e, then + k > n 1 k (and hence ( n 1 k +k = 0 unless h = e 2 and t = e 1. But part of the defnton of Sn sad that f t = e 1, then p 0, and hence + k > n 1 k n ths case, too. For part (2, we also need that ( n 1 k +k 2 s even, but t wll also be 0, usng that 2 ν(n 2 0. s even due to the 2 mn(v,ν(n 1 -poston f If k = p ν(n 1 2 v, then ( n 1 k +k v ν(n 1, whle f v = ν(n 1, we are n the case k = p 0 already handled. A smlar argument works for part (3, usng the 2 mn(ν(,ν(k -poston, provded ν( ν(k. However, equalty of ν( and ν(k wll not occur, because one can easly prove by nducton on that f 2 e 1 α( + k + ν( α(k < e 1. < 2 e and 0 k < 2 e 1 and ν( = ν(k, then 3. Proofs of results about ν ( ( n 2+1 k In ths secton, we prove four propostons about ν ( ( n 2+1 k whch were stated and used n the prevous secton. The polynomals q m (x whch we ntroduce n Defnton 3.1 mght be of ndependent nterest. Our frst proof utlzes an argument of Sun ([13]. Proof of Proposton 2.3. We mmc the argument n the proof of [13, Thm 1.3]. Let C m,l,b = ( m 2+b( l. Usng an dentty whch relates k to Strlng numbers, we

23 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 23 obtan ( 2n+ɛ 2+b k = ( 2n+ɛ 2+b l ( l l!s(k, l = l C 2n+ɛ,l,b l!s(k, l. Snce C 2n+1,n,0 = 2n + 1, C 2n+1,n 1,0 = 2(2n + 1(n + 1n, C 3 2n,n,0 = 1, C 2n,n 1,0 = 2n 2, C 2n,n,1 = 0, C 2n,n 1,1 = 2n, C 2n+1,n,1 = 1, and C 2n+1,n 1,1 = 2n(n + 1, our result follows from ν(l!c 2n+ɛ,l,b ν((2n + ɛ! l = ν(n! + n l, where we have used [14, Thm 1.1] at the frst step. The remanng proofs utlze a new famly of polynomals q m (x. Defnton 3.1. For m 1, we defne polynomals q m (x nductvely by q 1 (x = x 1, and m (3.2 f (x + 1x(x 1 (x m + 2 = b,m x, =1 (3.3 then (x + 1x(x 1 (x m + 2 = m 2 m b,m q (x. =1 For example, q 2 (x = x 2 x + 2. The relevance of these polynomals s gven by the followng result. Theorem 3.4. For all ntegers x, m( x = 2 x 2m q m (x. Proof. The proof s by nducton on m. Valdty when m = 1 follows from 2 ( x x = (2 + 1 ( ( x = (x + 1 x 2 = (x + 12 x 1.

24 24 DONALD M. DAVIS We show that 2 2m x m( x satsfes the equaton (3.3 whch defnes qm (x. We nsert ths expresson for q (x nto the RHS of (3.3 and obtan 2 m x ( x (2 b,m = 2 ( m x x (2 + 1 (2 m + 2 = 2 m x (x + 1 (x m + 2 ( x m+1 x 2, but ( x m+1 x 2 = 2 x m, snce t s the sum of all ( x m+1 wth n a fxed party. Thus we obtan (x + 1 (x m + 2, as desred. At the second step above, we have used (3.2 wth x = 2. Proposton 2.4 s an mmedate consequence of Theorems 3.4 and 3.5. Theorem 3.5. For all postve ntegers x, ν(q m (x m x + ν((x + 1! = m + 1 α(x + 1. Proof. The proof s by nducton on m. When m = 1, t reduces to α(x+1+ν(x 1 2. For the LHS of (3.3, note that ν((x + 1 (x m + 2 ν((x + 1! (x m, usng (2.2. For the -term ( < m n the sum n (3.3, by nducton on m we have 2-exponent m + ν(b,m + x + ν((x + 1! m x + ν((x + 1!. Thus the nequalty for ν(q m (x follows by nducton. The proof of Proposton 2.5 requres the followng two lemmas, and the result follows easly from the second and Theorem 3.4. Lemma 3.6. If b,m s as n Defnton 3.1, then wth equalty ff ( m s odd. ν(b,m ν(m! ν(! (m

25 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 25 Proof. We have b,m z m m! x 0 m = m z m b m!,m x m 0 =0 = m 0 1 m! (x + 1 mz m = m 0 ( x+1 m z m = (1 + z x+1 (x+1 log(1+z = e = k 0 = k 0 1 k! (log(1 + zk (1 + x k 1 (log(1 + zk k! k =0 ( k x. Here we have ntroduced the notaton (x + 1 m = (x + 1x (x m + 2. Equate coeffcents of x z m, and get 1 b m!,m = 1 1! (k! ([zm ](log(1 + z k. k Here [z m ]p(z denotes the coeffcent of z m n p(z. Let l(z = log(1 + z/z. The clam of the lemma reduces to ν ( 1 (k! ([zm k ]l(z k (m, k or equvalently ν ( 2 k (k! ([zm k ]l(2z k wth equalty ff ( m s odd. k Snce l(2z 1 + z mod 2, and ν((k! k wth equalty ff k =, all terms n the sum have ν( wth equalty ff k = and ( m s odd. Lemma 3.7. Let q m ( be as n Defnton 3.1, and let x be any nteger. ν(q m (x ν(m! wth equalty ff ( x m m s odd. Then Proof. We have q m (x = (x + 1 m m 1 =1 2 m b,m q (x.

26 26 DONALD M. DAVIS Note that ν((x+1 m ν(m! wth equalty ff ( x+1 m s odd. By nducton, the -term T n the sum satsfes ν(t m + ν(m! ν(! (m + ν(! = ν(m! wth equalty ff ( ( m s odd and x s odd. Ths mples the nequalty. Equalty occurs ff m 1 ( (3.8 + x s odd. By Lemma 3.10, s congruent to ( x m m ( x+1 m m =0 ( m ( ( x ( m x+1 m mod 2. Thus the expresson n (3.8 =0, establshng the clam. Proposton 2.6 follows mmedately from Theorem 3.4 and the followng result, whch s a refnement of Lemma 3.7. Theorem 3.9. If m s a postve nteger and x s any nteger, then, mod 4, q m (x/m! ( x m {2 ( x m f x and m are even m + m 2 0 otherwse. The proof of Theorem 3.9 requres several subsdary results. Lemma If m and x are ntegers wth m 0, then m ( ( x ( m x+1 ( m + 2 x+1 (mod 4. m 1 =0 Proof. Ths follows easly from Jensen s Formula (see e.g., [10], whch says that f A, B, and D are ntegers wth D 0, then D D = =0 ( +B D ( A =0 ( 1 ( A+B D Ths mples that the sum n our lemma equals. m =0 ( 1 ( x m. We prove that ths s congruent, mod 4, to the RHS of our lemma when x 0 by nducton on x. The formula s easly seen to be true f x = 0 (note that when x = 0 and m = 1 the LHS

27 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 27 equals 1 whle the RHS equals 3, and the nducton step s by Pascal s formula. For x < 0, let y = x wth y > 0. The equaton to be proved becomes m ( m+y 1 ( m y+m 2 ( m 2 y+m 3 (mod 4. m 1 =0 When y = 1, both sdes equal δ m,0 + 2δ m,1 and the result follows by nducton on y usng Pascal s formula. The next result refnes Lemma 3.6. Lemma If b,m s as n Theorem 3.1, then, mod 4, 2 m!b,m /m! ( {( f s even m + m 1 2c,m, where c,m = ( f s odd. Proof. As n the proof of 3.6, we have ( m!b,m /m! = k m 2 2 k (k! ([zm k ]l(2z k. Snce, mod 4, l(2z 1 z 2z 3, and 2 k /(k! 0 unless k equals 0 or a 2-power, (3.12 equals [z m ](1 z 2z [z m 2e ](1 z 2z 3 +2e e 0 [z m ](1 z 2z ( + 2 e. m 2 e e 0 Replace m by l. We must prove, mod 4, (3.13 A,l + 2B,l C,l + 2D,l, where and A,l = ( l, C,l = [z l ](1 z 2z 3, D,l = ( +2 e l 2, e e 0 B,l = {( ( l 1 l 2 even odd. If = 0, both sdes of (3.13 are congruent to δ l,0 + 2δ l,1. For the RHS, note that f l = 2 f wth f 1, then 2D 0,l 0 as t obtans a 2 from e = f and from e = f 1.

28 28 DONALD M. DAVIS Havng proved the valdty of (3.13 when = 0, we proceed by nducton on. If s even, then, mod 4, A +1,l + 2B +1,l C +1,l 2D +1,l = A,l + A,l 1 + 2(B,l 1 + B,l 2 (C,l C,l 1 2C,l 3 2(D,l + D,l 1 (A,l + 2B,l C,l 2D,l + (A,l 1 + 2B,l 1 C,l 1 2D 1,l 2B,l + 2B,l 2 + 2C,l 1 + 2C,l 3 2 ( ( l ( l ( l l 3 0, and a smlar argument works when s odd. The followng result relates the even parts n 3.9 and Lemma Let p (x = {( x 2 x and even 0 otherwse and c,m = {( ( m 1 m 2 Then, mod 2, f x and m are ntegers wth m 0, m ( (3.15 p (x + c x,m. ( x+1 m 1 =1 (( m Proof. Frst let x be odd. By Lemma 3.10, mod 2, ( x. ( x+1 m 1 ( m 1 Snce p (x = 0 and ( x Now suppose x s even and m odd. We must prove, mod 2, ( x+1 m 1 odd ( m 2 even odd. 0 for odd, ths s equvalent to (3.15 n ths case. ( x (( + even By 3.10, the LHS s congruent to ( ( x m 2 ( x m 1 + ( ( x m 1.. If s odd, ( m 1 and f s even, ( m 0. The desred result s now mmedate. Fnally suppose x and m are both even. Agan usng 3.10, we must show odd ( m 1 ( x even ( m ( x 2 + odd ( m 2 ( x ( m 2,

29 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 29 k even snce ( m 1 0 f s even. The terms on the LHS combne wth the -odd terms on the RHS to yeld ( +1 ( x m 1. Lettng k = + 1, ths becomes odd ( k ( x k+1 ( m k k 1. Snce x and k are even, x k+1 ( k 1 x k k 2, and so all terms cancel. Now we easly prove Theorem 3.9. Proof of Theorem 3.9. The proof s by nducton on m, wth the case m = 1 mmedate. Usng notaton of 3.14, equaton (3.3 yelds, mod 4, q m (x/m! = ( m 1 x+1!2 m b,m q (x m m!! =1 2 ( x+1 m 1 ( x m m + m =0 2 (( x+1 m 1 ( x m m + 2pm (x, ( ( x m m 1 =1 m 1 (( m =1 (( m (( + x 2c,m + 2p (x ( p (x + c x,m as desred. Here we have used 3.10 and 3.11 at the second step and 3.14 at the last step. 4. Relatonshp wth Hensel s Lemma In [5], the author ntroduced Lemma 2.8 and appled t to study ν(t 5 ( and ν(t 6 ( smlarly to what we do here for all T n (. Clarke was quck to observe n [3] that f T n ( s consdered as a functon on Z 2, then our concluson that ν(t n (x = ν(x x 0 +c 0 when x s restrcted to a congruence class C can be nterpreted as sayng that T n (x 0 = 0. He showed that f T n (x 0 = 0 and T n(x 0 0, then T n (x = x x 0 T n(x 0 on a neghborhood of x 0, whch corresponds to our congruence class C. Here agan x = 1/2 ν(x on Z 2, and d(x, y = x y defnes the metrc. Also, T n denotes the dervatve. Moreover, Clarke noted that the teraton toward the root x 0 n our

30 30 DONALD M. DAVIS theorems s a dsgused form of Hensel s Lemma for convergence toward a root of the functon T n. We llustrate by consderng the root of T 13 of the form 4x See Theorem 1.2 and Table 1.3. For our teraton toward x 0, let (4.1 g(x = ν(t 13 (4x Then g(0 = 1, g(2 1 = 5, g( = 6, etc. Thus our early approxmaton to 4x s ( ( , and, contnung, we obtan that the last 18 dgts n the bnary expanson of 4x are ( Note that each 1 n the bnary expanson requres a separate calculaton. Now we descrbe the Hensel pont of vew, followng Clarke ([3]. He showed that where T n(k = ( n 2+1 (2 + 1 k L(2 + 1, L(2 + 1 = ( 1 1 (2 / =1 s the 2-adc logarthm. Hensel s Lemma appled to an analytc functon f nvolves the teraton k n+1 = k n f(k n f (k n, whch, under favorable hypotheses, converges to a root of f. We have f = T 13. Usng Maple, we fnd 8 k 1, 2 (4 (4.4 ν(t 13(k = 9 k 0 (4 11 k 3 (4. To prove ths, whch nvolves an nfnte sum (for L and nfntely many values of k, frst note that our only clam s about the mod 2 11 value of T 13, and so the sum for L may be stopped after = 12. Snce L( mod 4, we are only concerned wth (2 + 1 k mod 2 9. Snce (2 + 1 k mod 2 9 has perod 2 8 n k, performng the computaton for 256 values of k would suffce. Let k 0 = 1. Then Maple computes that k 1 = 1 T 13(1 has bnary expanson endng T 13 ( , and so agrees wth (4.3 mod 64. Next k 2 = k 1 T 13(k 1 has bnary expanson T 13 (k 1

31 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 31 endng , agreeng wth (4.3 mod Fnally k 3 = k 2 T 13(k 2 T 13 (k 2 agrees wth (4.3, and hence s correct at least mod Ths s much faster convergence than ours. Let θ(x = T 13 (4x + 1. Our algorthm essentally apples Hensel s Lemma to θ(x, but ust takes the leadng term each tme. For all x, ν(θ (x = ν(4t 13(4x + 1 = 10, and so our g(x equals ν(θ(x/θ (x. Thus when we let x +1 = x + 2 g(x, we are addng the leadng term of θ(x /θ (x. Once the lmtng value, whch we denote by x 0, s found, the root of T 13 s 4x In [3], Clarke defnes, for an analytc functon f, g(x, h = f(x + h f(x hf (x h 2 and shows that f f(x 0 = 0 and g(x, h 2 r for all relevant x and h, then the desred formula holds for all x satsfyng f(x = x x 0 f (x 0 (4.5 x x 0 < f (x 0 /2 r. He also notes that our T n ( are analytc when restrcted to all 2-adc ntegers of ether party. For f = T 13, Maple suggests that ν(g(x, h 7 for h even, wth equalty ff x + h 0, 3 mod 4. Ths can be easly proved usng Maple calculatons and some elementary arguments. Hence r = 7. Usng (4.4, we obtan the results of Table 1.3 for n = 13 whch are lsted there as C = [1, 2 (4] and [0, 4 (8], snce { T 13(x x 0 1, 2 (4 = x 0 0 (4. Beng less than ths requres x x or 2 3 n (4.5, whose recprocals are the modul of the congruence classes n Table 1.3. Another result of [3] gves a condton, (4.6 f(x < mn ( f (x, f (x 2 2 k 1 2 r (where g(x, h 2 r and f s analytc on c + 2 k Z 2, whch guarantees that teraton of Hensel from x converges to a root of f. For f = T 13 and x 3 (4, T 13(x 2 /2 r ( /2 7 = 2 15 by (4.4, whle by Table 1.3 T 13 (x takes on values 2 11, 2 13,

32 32 DONALD M. DAVIS and Thus the condton (4.6 does not hold, consstent wth our fndng n Table 1.3 that T 13 (x s constant on balls about 7, 3, and 11, so there s no root n these neghborhoods. Clarke s approach s a very attractve alternatve to ours. It converges faster, and t s more closely assocated wth analytc methods, such as the Hensel/Newton convergence algorthm. On the other hand, there s a certan combnatoral smplcty to our approach, especally Lemma 2.8 and ts reducton to consderaton of expressons such as (2.11 and (2.35, and subsequently to (2.38. We fnd t very attractve that for each f = T n, t seems lkely that Z 2 can be parttoned nto fntely many balls B(x 0, ɛ on each of whch f(x s lnear n x x 0 (ncludng the possblty that t s constant. It s not clear whch approach wll be the better way to establsh ths. References [1] M. Bendersky and D. M. Davs, 2-prmary v 1 -perodc homotopy groups of SU(n, Amer. J. Math. 114 ( [2] O-Y. Chan and D. Manna, Dvsblty propertes of Strlng numbers of the second knd, Contemp Math 517 ( [3] F. Clarke, Hensel s Lemma and the dvsblty of Strlng-lke numbers, Jour Number Theory 52 ( [4] M. C. Crabb and K. Knapp, The Hurewcz map on stunted complex proectve spaces, Amer Jour Math 110 ( [5] D. M. Davs, Dvsblty by 2 of Strlng-lke numbers, Proc Amer Math Soc 110 ( [6], Dvsblty by 2 and 3 of certan Strlng numbers, Integers 8 (2008 A56, 25pp. [7], v 1 -perodc 2-exponent of SU(2 e and SU(2 e + 1, submtted, dmd1/su2e.pdf. [8] D. M. Davs and K. Potocka, 2-prmary v 1 -perodc homotopy groups of SU(n revsted, Forum Math 19 ( [9] D. M. Davs and Z. W. Sun, A number-theoretc approach to homotopy exponents of SU(n, Jour Pure Appl Alg 209 ( [10] H. W. Gould, Congruences nvolvng sums of bnomal coeffcents and a formula of Jensen, Amercan Math Monthly 69 ( [11] A. T. Lundell, A dvsblty property for Strlng numbers, Jour Number Theory 10 ( [12], Generalzed e-nvarants and the numbers of James, Quar Jour Math Oxford 25 ( [13] Z. W. Sun, Combnatoral congruences and Strlng numbers, Acta Arthmetca 126 ( [14] Z. W. Sun and D. M. Davs, Combnatoral congruences modulo prme powers, Trans Amer Math Soc 359 (

33 DIVISIBILITY BY 2 OF PARTIAL STIRLING NUMBERS 33 [15] P. T. Young, Congruences for degenerate number sequences, Dscrete Math 270 ( Department of Mathematcs, Lehgh Unversty, Bethlehem, PA 18015, USA E-mal address: dmd1@lehgh.edu