with an even sum and for k 1mod4 1, 2,, n with an odd sum. ,, n of Pascal s triangle count the subsets of 1, 2,, n

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1 Some remars o Rogers-Szegö olyomials ad Losaitsch s triagle Joha Cigler Faultät für Mathemati Uiversität Wie johacigler@uivieacat Abstract I this exository aer we collect some simle facts about aalogues of Pascal s triagle where the etries cout subsets of the itegers with a eve or odd sum ad show that they are related to q Newto ad Rogers-Szegö olyomials I articular we cosider a iterestig triagle due to Losaitsch from this oit of view We also setch some extesios of these results to sets whose sums have fixed residues modulo a rime 1 Itroductio This exository aer has bee isired by a Pascal-lie triagle which has bee obtaied by the chemist S Losaitsch (Sima Lozaic) [6] Some of its roerties ca be foud i [1] ad [7] A We show that the elemets L( ) of Losaitsch s triagle have several iterestig iterretatios: a) For 1mod4 L( ) is the umber of subsets of 1 with a eve sum ad for 1mod4 1 with a odd sum b) Let the umber of subsets of W be the set of all tules w with ad i i1 The L( ) is the umber of those tules which have a eve umber of iversios c) For w W let w 1 1 be the reversal of w The L( ) also couts the set of all equivalece classes ww d) Let be a q biomial coefficiet The L ( ) L ( ) q modq 1 For helful remars I wat to tha J Schoissegeier ad V Losert 11 As is well ow the etries of Pascal s triagle cout the subsets of 1 with elemets which will be called sets for short A set S will be called eve if the sum ( S) of its elemets is eve ad odd if this sum is odd By covetio the emty set is eve i

2 E be the set of all eve subsets of 1 Let elemets ad let umber of its elemets For examle (53) 6 O be the set of all odd subsets of 1 ad e ( ) E the umber of its ad o ( ) O the e because E 53 {13}{15}{134}{145}{35}{345} Let us ote the trivial fact e ( ) o ( ) (11) Lemma 11 e ( ) e ( ) o ( ) 1 o ( ) o ( ) e ( ) 1 (1) This is a refiemet of the formula 1 Proof Let S be a subset of 1 To rove (1) cosider 3 ossibilities: a) S 1 b) S cotais recisely oe of the umbers 1 ad The remaiig ( 1) set ca be a arbitrary subset of 1 because oe of these umbers is eve ad the other oe is odd There are such subsets 1 c) S cotais both 1 ad Sice ( 1) is odd the remaiig ( ) subset must have the oosite arity of the give subset 1 To each set S we associate its idicator fuctio c S defied by cs () i 1if i S ad cs () i 0else or equivaletly the tule ws ( ) 1 with i cs() i Let W be the set of all tules w with 01 1 let iv( w) 1 i j : i 1 j 0 ad i i For each w be the umber of iversios of w We call a tule eve if iv( w ) is eve ad odd else i1

3 For examle there are 4 eve ad odd quadrules with iv iv iv 0110 iv 1001 because iv iv Istead of W we could also cosider the set of all roducts w CC 1 C of elemets A ad elemets B which satisfy BA qab A iversio is the a air iv(w) C C ( B A) with i j which gives CC C q A B i j Let us associate to each S i1 i i S 1 i 1 i 1 i 1 with 1 i1 i i the set 1 1 Lemma 1 For each set S 1 Proof 1 ( ) iv we have S ws This holds for 1 ad all because iv ws i 1 11 Thus i i1 S i imlies S 1 i Let us assume that it is true for 1 ad all ad cosider a set S It is trivially true for Let us suose that it is true for 1 Let S 0 be the restrictio of S to {1 1} S c ( ) S 1 1 If S ad ( S ) i ad the If S the ( S) ( S0) ivws0 iv ws If S the 1 1 ( S) S0 iv ws0 iv w S0 iv w S because there are w S elemets 0 i 0 We will ow cosider matrices whose colums are either e or o where e is the colum with etries e ( ) ad o the colum with etries o ( ) for 3

4 Sets with a eve sum 1 Let us first cosider the matrix e ( ) e0 e1 e e3 whose etries are the umber of eve sets (cf OEIS [7] A8011) The first terms are Closely related is the matrix ( ) sets (cf OEIS [7] A ) o o o o o whose etries are the umber of odd Sice i both triagles the colums c ad c have the same arity by Lemma 11 the etries a ( ) of these matrices satisfy a ( ) a ( ) a ( ) ad thus 1 1 a ( ) a ( ) a ( ) (1) 1 Therefore the olyomials a ( x) a( ) x satisfy the recursio 0 a ( x) 1 x a ( x) x(1 x) () 1 By alyig this to ad 1 we get the homogeeous recursio a ( x) (1 x) a ( x) 1 x a ( x) (1 x) 1 x a ( x) (3) 1 3 Sice z 3 (1 x) z 1 x z(1 x) 1 x z(1 x) z 1 x z 1 x we see that a ( x) c0(1 x) c1 1x c 1 x for some coefficiets c i Let ow e ( x) e( ) x ad 0 o ( x) o( ) x 0 4

5 From the iitial values e0( x) 1 e1( x) 1 e( x) 1 x we comute 1 1x 1x 1x 1x c0 c1 c 4 1x 4 1x This gives e (1 x) 1x ( x) (1 x) (1 x) 1 x e 1( x) ad aalogous formulae for o ( x ) 1 (4) From (4) we see that e ( x ) is alidromic ie 1 x e e( x) x if 03mod 4 The geeratig fuctios are 1xz 1x z (1 x) z ad 1 (1 ) 1 1 e ( x) z 0 xz x z (5) 1 (1 ) 11 xz x z xz o ( x) z 0 xz x z (6) 1 (1 ) (1 x) z 1 (1 ) 11 xz x z From (5) we coclude that the geeratig fuctio of 1 e( x) e( ) x x e 0 x is 1 (1 ) 1 1 1z x 1 z ( x 1) z e ( x) z 0 xz x z (7) 1 (1 ) 1 1 xz x z It is ow easy to derive some exlicit formulae From (4) we see that (1 x) (1 x) j e ( x) (1 x) (1 x) x (8) j j 5

6 This imlies 1 e ( ) (9) j j j As secial case we get the well-ow formula e( ) (10) If we write (5) i the form zxz exz ( ) 0 1 z xz 1 1z x z 1 1 z 1z we see by comarig coefficiets of x that z 1 ( 1) (1 z) (1 ) 1 z e ( ) z (11) z z 1 ( 1) (1 ) 1 z 1 1 e ( 1) z (1) 1 0 z Remar It is erhas iterestig that these geeratig fuctios ca also be writte i the followig way: x e4 ( x) 4 41 e ( 4 x ) (1 x) (1 x) 4 1 x o4 1( x) 41 4 e ( 41) x (1 x) (1 x) 4 x o4 ( x) 4 43 e ( 4) x (1 x) (1 x) 4 3 x e4 3( x) e ( 43) x (1 x) (1 x) (13) 6

7 3 Losaitsch s triagle 31 Let us ow cosider aother class of triagles where the colums c ad c have oosite arity By Lemma 11 the etries of these matrices satisfy b ( ) b ( ) b ( ) 1 (31) Therefore the olyomials b( x) bx ( ) satisfy the recursio 0 b ( x) 1 x b ( x) x(1 x) (3) By alyig this to ad 1 we get b ( x) (1 x) b ( x) 1 x b ( x) (1 x) 1 x b ( x) (33) 1 3 The best ow secial case of a matrix satisfyig (31) is Losaitsch s triagle L ( ) (cf [1] [7] A034851) which is defied by the recursio L ( ) L ( ) L ( ) 1 with iitial values L(0 ) [ 0] ad L(1 ) [ 1] ad L ( ) 0 for 0 (34) The first terms are This matrix has bee obtaied by the chemist SM Losaitsch (or Lozaic) [6] i his ivestigatio of araffi Therefore we call the umbers L ( ) Losaitsch umbers The same triagle has also bee cosidered i [1] i the study of some sort of eclaces where these umbers have bee called eclace umbers Further iformatio ca be foud i OEIS [7] A Let us first observe that L e o o e e o ( ) (35) For the right-had side satisfies (34) with the same iitial ad boudary values as Losaitsch s triagle 7

8 By (4) we see that e ( ) 1 if 03mod 4 ad o ( ) 1 if 1 mod 4 Therefore we get Proositio 31 Losaitsch s triagle is the uiquely determied matrix where for each the th colum is either e or o ad all elemets of the mai diagoal are 1 Let us give aother characterizatio of Losaitsch s triagle which aears as secial case i [5] Theorem 3 (Stehe G Harte ad AJ Radcliffe) The Losaitsch umber L ( ) is the umber of all tules L W with a eve umber of iversios Proof from all subsets of {1 } to W is 1 a bijectio betwee the sets for which ( S) mod ad L 1 For 1mod4 we have 0mod ad therefore L e( ) ad for 1 1 mod 4 we have 1mod ad therefore L o ( ) Thus L L( ) By Lemma 1 we see that the ma S ws 3 The oosite matrix L ( ) o e e o o e also A The first terms are is OEIS [7] A03485 ad essetially The Losaitsch olyomials recursio (33) L ( x) L( ) x ad 0 L ( x) L( ) x satisfy the 0 I the same way as above we see that 8

9 (1 x) 1x 1x 1x 1x L ( x) 1x 1 x 4 1x 4 1x This ca be simlified to L L (1 x) 1x ( x) 1 1 (1 x) (1 x) 1x ( x) (36) Note that ad or L 1( x) (1 x) L( x) (37) 1 x L L( x) x (38) L ( ) L ( ) (39) Thus we get (cf [1] Theorem 8) by observig (11) 1 L( 1) L( 1) L ( ) ad L ( ) else (310) I the same way as above we get the geeratig fuctio 1 (1 ) xx z (1 x) z L ( x) z 0 xz x z (311) 1 (1 ) 11 xz x z Some other roerties of the Losaitsch olyomials ca be foud i [1] Comarig with (13) we get Proositio 33 xe( x) Lx ( ) L ( x ) 1 (1 x) (1 x) (31) 9

10 There exists also aother iterestig relatio betwee the umbers e ( ) ad L ( ) Proositio 34 Proof z e ( z ) L ( z) (1 z) 1z (313) It suffices to comute the coefficiet of x i (7) This gives 1 1 (1 ) 1 z z (1 z) z z z e ( z ) (1 ) z 1 z (1 z) 1 z z L 1 ( z) 1 (1 z) 1z ad (1 ) 1 (1 ) 1 z z z z z z e ( 1) z (1 ) z 1 z (1 z) 1 z 1 z L 1 3( z) 3 (1 z) 1z 33 Let us ow derive aother iterestig combiatorial iterretatio of the Losaitsch umbers For w1 W let w 1 be the reversal of w Defie a equivalece relatio o W by w w If w w we call w alidromic Theorem 35 ww ad let P be the subset of all alidromic classes Let r ( ) R ad ( ) P deote the umber of their elemets The r ( ) L ( ) ( ) L ( ) L ( ) R be the set of all equivalece classes Let (314) 10

11 Proof By (310) we have L ( 1) L ( 1) 0 ad L ( ) L ( ) else For ( ) we get the same result: ( ) because each alidromic set is uiquely determied by its restrictio to 1 ad each subset of 1 has a uique extesio to a alidromic set of 1 It is clear that ( 1) 0 (11) alidromic set of 1 1 Thus ( ) L ( ) L ( ) Sice there are 1 ( ) because each subset of 1 1 has a uique extesio to a o-alidromic classes i W we get 1 1 r ( ) ( ) ( ) ( ) L ( ) Therefore 1 L( ) ( ) is the umber of o-alidromic equivalece classes Examle For examle W 5 has 6 equivalece classes Thus L(5 ) 6 ad L(5) 4 ad (5) L(5] L(5) Remar Some secial cases of Theorem 35 are metioed i OEIS A0060 A A ad A i the followig formulatio: Let B( ) be the umber of bracelets ie eclaces allowig turig over with white 1 blue ad red beads the B( ) L ( ) To verify this it suffices to choose a reresetative of the bracelet whose first term is the blue bead If we relace the first 1 letters 1 by 10 we see that Theorem 35 imlies [1]Theorem 4 11

12 4 Rogers-Szegö olyomials ad Losaitsch s triagle 41 There are close coectios with some q biomial theorems For the relevat facts about q calculus I refer to [] Let be a q biomial coefficiet It turs out that both e ( ) ad L( ) are residues modulo q q 1 of q biomial coefficiets Theorem 41 1 e ( ) oq ( ) q mod q 1 L ( ) Lq ( ) modq 1 (41) Proof Recall that 1 j 1 qxq x j1 0 (4) j1 j j Sice 1 qx j q x e ( ) oq ( ) x modq 1 j1 we get the first lie of (41) j1 j Now observe that mod ad mod Therefore 1 mod q 1q modq 1 e ( ) oq ( ) ad 1 mod q 1qq mod q 1 o ( ) qe ( ) Comarig with (35) we get the secod lie of (41) for 03mod 4 for 1mod 4 1

13 Corollary 4 The Losaitsch olyomials are related to the Rogers-Szegö olyomials by ( ) ( ) ( )mod 1 r ( x q) 0 x L x ql x r x q q (43) ad the olyomials e ( x ) are related to the q Newto olyomials 1 j ( x q) 1q x q x j1 0 by e( x) qo( x) ( x q)mod q 1 (44) From (43) it is obvious that L ( x ) is alidromic We already ow that e ( x ) is alidromic if 03mod 4 This ca also be deduced from (41) For For 3mod4 this follows from 0mod4 we have 44 1 mod 41 mod This imlies that q q o(4 ) qe(4 )modq 1 4 we have e(4 ) o(4 ) e ( x) o ( x) 1 x sice by (4) if 1mod But i this case 4 Several theorems about q biomial coefficiets give results for Losaitsch umbers or for the umbers of eve or odd sets if we relace the olyomials e ( x ) ad L ( x) by the equivalece classes ( x) e ( x) qo ( x) ad ( x) L ( x) ql ( x) modulo q 1 The first terms of the matrix e ( ) qo ( ) are 13

14 Let us ote that 1 (1 q) ( x) (1 q) 1x 1x mod q 1 (45) This holds for 0 ad 1 By iductio we get j (1 q) ( x) (1 q) 1q x 1 q x (1 q) ( x) 1 j1 1 1 x(1 q) 1 x 1 x (1 q) 1 x 1 x ad 1 j 1 (1 q) ( x) (1 q) 1q x 1 q x (1 q) ( x) 1 j1 1 qx q x x q x x x q x x q Here we have used that 1 qx(1 q) 1 qx q x (1 x)(1 q)modq 1 1 (1 ) 1 1 (1 )(1 ) 1 1 (1 ) 1 1 mod 1 The first terms of L ( ) ql ( ) are I this case we get 1 (1 q) ( x) (1 q) 1x mod q 1 ( x) (1 x) ( x) (46) The recurrece 1 1 r ( x q) ( x1) r ( x q) q 1 xr ( x q) (47) for the Rogers-Szegö olyomials (cf eg []) imlies 1( x) (1 x) ( x) ad 1 Sice 1 ( x) ( x 1) ( x) q 1 x ( x) 1 (1 q) q 1 (1 q) (1 q) mod q 1 we get 1 q ( x) ( x 1) 1 q ( x) 1 q x ( x) 1 ( x1) 1 q ( x) 1 q x ( x) 1q x 1 ( x) which imlies the first lie of (46) by iductio 14

15 Now it is well ow that q iv( w q ) (48) ww This follows for examle from the q biomial theorem AB A B 0 for q commutig variables BA qab (cf eg []) For if we write ( AB) C C with Ci A B we get iv( w) ( A B) A B q ww i1 i If we reduce (48) modulo q 1 we get agai Theorem 3 43 Sice each equivalece class modulo q 1 ( q1)( q 1) is also a equivalece class modulo q 1 ad modulo q 1 we see that L ( ) L ( ) ad L ( ) L ( ) q1 q1 Thus 1 L ( ) q1 q1 (49) By (310) we get the well-ow result 1 The first terms of the table q1 q1 0 q1 0 are else (410) 15

16 Remar If we comare (410) with Theorem 35 we see that the umber of alidromic tules i W coicides with ( ) (411) (47) imlies q1 1 1 r ( x 1) ( x1) r ( x 1) ( 1) 1 xr ( x 1) with iitial values r 0 ( x 1) 1 ad r( x 1) 1 x 1 This gives by iductio r x x 1 ( 1) 1 ( 1) (1 ) 1 r x x x (41) Of course this also follows from (36) I the geeral case we get from the defiig relatio Pascal matrix 1 1 q 1 of the q Proositio 43 The equivalece classes ( ) e ( ) qo ( ) satisfy ( ) q ( 1 ) ( 1 1) (413) with ( 0) 1for all ad (0 ) 0 for 0 ad the equivalece classes ( ) L ( ) ql ( ) satisfy ( ) q( 1 ) ( 1 1) (414) with ( 0) 1 for all ad (0 ) 0 for 0 Remar It is clear that ( ) ad ( ) are uiquely determied by these rules Thus we get a very simle aroach to the matrices L( ) ad e ( ) ad their roerties From ( ) q ( 1 ) q ( 1 1) ( ) ( 1) q( 1) ( ) ( ) q( ) 1 we get e ( ) e ( ) e ( ) 1 ad thus agai (1) I the same way we get (34) 16

17 Formula (414) imlies that L ( ) L ( 1 ) L ( 11) L ( 1) L ( 11) L ( 1 ) (415) This is obvious for the iterretatio of L( ) as the set of w W with a eve umber of iversios if we cosider the last etry of w 44 Let us give some more examles: a) The recurrece (47) imlies L 1( x) ql 1( x) ( x 1) L( x) ql( x) which agai gives (37) ad L( x) ql( x) ( x 1) L 1( x) ql 1( x) ( q 1) x L( x) ql( x) which imlies L x x L x xl x xl x x L x xl x x x ( ) ( 1) 1( ) ( ) ( ) ( 1) 1( ) ( ) (1 ) b) The well-ow idetity q q 0 1 q 0 imlies L ( ) ql ( ) q L ( 1) ql ( 1) L( ) ql( ) which gives L( ) L( ) L( 1) L( 1) L( ) (416) c) A more trasaret roof of (31) rus as follows: The well-ow idetity x x 0 imlies (1 x)(1 qx) 1q x (1 qx) 1 q x(1 x)(1 qx) 1 q x x (1 x) 1 q x 1 q x x x q x 0 If we reduce this idetity modulo q 1 we get (31) 4 j The same roof also gives agai (13) We have oly to observe the arity of 17

18 d) The q Fiboacci olyomials 1 1 ( 1) F ( s q) q s 0 satisfy F ( s q) F ( s q) q sf ( s q) (cf [3]) 3 1 For q 1 we get the Fiboacci olyomials Note that (1) F () s s 0 F gives the Fiboacci umbers For q 1 we get Therefore the olyomials F s F s ( 1) F ( s 1) F s sf s 1 1 (417) 1 ( sq) ( 1 s ) F( sq)modq 1 0 satisfy ( sq) ( sq) q s ( sq)mod q The first terms are 0111 s1 s sq1 s s sq1 s s s s q Let ow ( sq) f() s qf() s The We get 1 f() s L( 1 ) s (418) 0 f () s f () s sf () s 1 1 f () s f () s sf () s 1 This gives by iductio or by cosiderig (417) f f F () s F s () s 1 F 1() s F 1 s sf s () s (419) 18

19 For s 1 we get a aalogue of the Fiboacci umbers The first terms are f (1) f(1) L( 1 ) (40) 0 This sequece also occurs i OEIS[7] A1056 F F F 1F From (419) we get f(1) f 1(1) e) As is well ow the sequece 1 1 F ( 1) ( 1) is eriodic with 0 eriod 6 Cosider the q aalogue 1 1 f( ) ( 1) q 0 The first terms are Note that these are artial sums of Euler s etagoal umber series q q q q q q q q q As show i [4] it satisfies f( ) f( 1) q f( 3) q f( 4) Let ( ) f( )mod( q 1) The it is easy to verify that ( 1) ( ) with iitial values 0111 q q q 0 q q q1 1 1 Therefore we get Proositio 44 1 The sequece ( 1) e ( 1 ) is eriodic with eriod 1 with iitial values

20 5 A more geeral case 51 Let e ( j ) be the umber of subsets of 1 whose sums are cogruet to j modulo The 1 j j1 j j qxx q xe jq q j1 1 ( ) mod 1 j1j j j0 O the other had we ow that j1 1 j 1 qxq x 0 Therefore we get q Let ow 1 j e ( j ) q q modq 1 q (41) 1 1 j ( q ) e ( j q ) q modq 1 j0 (4) The equivalece classes ( q ) satisfy ( q ) q ( 1 q ) ( 1 1 q ) mod q 1 (43) with ( 0 q) 1for all ad (0 q) 0 for 0 q For examle the first terms of ( 3 q) are 0

21 5 Let us first derive a aalogue of (): Proositio 51 Let e ( j x ) e( j ) x For ad each j we get i e( j x ) 1 x e ( j x ) x 1 x i1 i (44) Proof To show this observe first that e ( j ) e ( 0 ) for all j with 0 j if 1 1 This imlies 1 e ( j ) for all j J Schoissegeier has show me a simle roof due to V Losert: Cosider the grou of itegers modulo ad let E( i ) be the set of all subsets S of whose sum of elemets is cogruet to i modulo Let S 1 be the the set of elemets s 1 with s S The ma S S 1 gives a bijectio from E( i ) to E( i ) Iteratig this ma we see that all sets E( i ) have the same size because the multiles of exhaust Now cosider the elemets i with 1 i The umber of sets whose sums are cogruet to j modulo which cotai oe of these umbers is e ( j ) Sice the sum of cosecutive umbers is divisible by the umber of sets which cotai all elemets 1 i is e ( j ) Let ow 1 1 For each ( ) subset of 1 there are recisely 1 subsets of 1 such hat their uio is jmod because cosecutive umbers cotai all residues modulo This gives This gives for 1 sets which are j mod 1 1 e( j ) e( j ) e( j ) 1 which imlies (44) 1

22 By alyig (44) to ad 1 we get the homogeeous recursio e( j x ) (1 x) e 1( j x ) 1 x e( j x ) (1 x) 1 x e 1( j x ) 0 (45) By summig over all j we get 1 1 (46) 1 1 ( q ) ( q ) 1 q q ( q ) For a rimitive th root of uity q (46) reduces to ( ) ( ) ( ) (47) 53 Let us ow derive a aalogue of (4) Let be a rimitive th root of uity The j1 1 j j 1 x x e( j ) Observe that j j j x x e j e x 0 j1 0 j0 1 ( ) ( 0 ) O the other had we have 1 1 j j x x x 1 (1 ) 1 0 j1 1 j1 Sice each roduct of 1 j x over cosecutive values of j equals 1 x we see that 1 i j 1 xbi ( x) 1x 1 j1 for some olyomial bi ( x ) of degree i Therefore the olyomial e ( ) ( 0 ) x e x satisfies e i i (1 x) bi ( x) 1x ( x ) (48)

23 This imlies 1 i ( ) 1 1 bi x z i0 e ( x ) z (49) 0 1 ( x1) z 11x z Observe that this also follows from (45) 54 We defie ow Losaitsch umbers L ( j ) by or equivaletly by j L ( j ) q modq 1 (430) q 1 j ( q ) L ( j q ) modq 1 j0 (431) q The equivalece classes ( q ) satisfy ( ) q q ( 1 q) ( 1 1 q) mod q 1 (43) with ( 0 q) 1for all ad (0 q) 0 for 0 For examle the first terms of ( 3) are The olyomials L ( x j) L( j ) x will be called Losaitsch olyomials 0 Note that all olyomials L ( x j ) are alidromic j Sice e ( j ) [ q] q modq 1 1 we see that 1 L ( j ) e j modq 1 Therefore each colum L ( j ) coicides with a colum ( ai for some i

24 Therefore we get Proositio 5 The matrix L ( 0 ) 0 ei ( for some i ad where each elemet of the mai diagoal is 1 0 From (431) ad (48) we coclude Theorem 53 is the uiquely determied matrix whose colums are L ( 0 ) is the umber of elemets w W such that iv( w ) is a multile of Refereces [1] Tewodros Amdeberha Mahir Bile Ca ad Victor H Moll Broe bracelets Molie series Paraffi wax ad a ellitic curve of coductor 48 SIAM J Discr Math 5 (011) [] Joha Cigler Elemetare q Idetitäte Sémiaire Lotharigie de Combiatoire B05a (1981) [3] Joha Cigler q Fiboacci olyomials Fiboacci Quarterly 41 (003) [4] Joha Cigler A ew class of q Fiboacci olyomials Electr J Comb 10 (003) R19 [5] Stehe G Harte ad AJ Radcliffe Sigatures of strigs AComb 17 (013) [6] SM Losaitsch Die Isomerie- Arte bei de Homologe der Paraffi-Reihe Chem Berichte 30 (1897) [7] OEIS htt://oeisorg/ 4