Math 774 Homework 3. Austin Mohr. October 11, Let µ be the usual Möbius function, namely 0, if n has a repeated prime factor; µ(d) = 0,

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1 Math 774 Homework 3 Austin Mohr October 11, 010 Problem 1 Let µ be the usual Möbius function, namely 0, if n has a repeate prime factor; µ(n = 1, if n = 1; ( 1 k, if n is a prouct of k istinct primes. Proposition 1. For all n N, µ( = { 1, if n = 1; 0, if n > 1. Proof. The conclusion is clear for n = 1. Let now n an consier some iviing n. If has a repeate prime factor, then µ( contributes nothing to the sum. Thus, letting A = { : ivies n an has at most one copy of each istinct prime factor of n}, we have µ( = µ(. A Let now A. If is the prouct of an even number of istinct primes, then µ( = 1. Similarly, if is the prouct of an o number of istinct primes, then µ( = 1. Eviently, the number elements of A that are the prouct of an even number of prime factors is the same as the number of elements of A that are the prouct of an o number of prime factors (these can be viewe as the number of bitstrings of a fixe length. Therefore, µ( = 0, as esire. A 1

2 Proposition. For functions f, g : Z R an for all n N, f(n = g( if an only if g(n = ( n µ f(. Proof. ( We have, for all n N, ( n µ f( = ( n µ g(m m = µ m n {:m,} = g(m m n = m n g(m l n m {:m,} ( n g(m µ (l. ( n µ By 1, we see that the sum on the right is equal to zero except in the case when m = n. Thus, we have ( n µ f( = g(nµ(1 as esire. ( We have, for all n N, g( = m = m n = g(n, ( µ f(m m {:m,} = m n f(m = m n f(m l n m ( µ f(m m ( µ m {:m,} µ (l.

3 By 1, we see that the sum on the right is equal to zero except in the case when m = n. Thus, we have g( = f(nµ(1 as esire. Problem = f(n, Let S n consist of two a 1 s, two a s,..., two a n s. Proposition 3. The number of sequences obtaine from S n such that, for all i, the a i s are not ajacent is given by n ( n (n k! ( 1 k k n k. k=0 Proof. Let A i enote the set of all sequences in which the two a i s are together. Without loss of generality, we compute the size of A 1, as each A i has the same carinality. We may treat a 1 a 1 as single element a an etermine the number of permutations on S n \ {a 1, a 1 } {a}. Using multinomial coefficients, we see that (n 1! A 1 = n 1. In a similar fashion, we see that A 1 A k = (n k! n k, for any k. Letting S be the set of sequences with the esire properties, the inclusion-exclusion formula gives S = n ( n (n k! ( 1 k k n k. k=0 For n = 3, the above formula gives 30 sequences of the esire type. Below we explicitly list all 30 such sequences (using a, b, an c, for simplicity. 3

4 abcabc bcabca cabcab abcacb bcabca cabcba abcbac bcabac cabacb abcbca bcacba cababc abacbc bcacab cacbab acbabc bcbaca cbacab acbacb bacbac cbacba acbcab bacabc cbabca acbcba bacacb cbabac acabcb babcac cbcaba Proposition 4. The number of arrangements of S n on a circle is given by (n 1! n + (n 1!. Proof. Since we consier rotations of a circle to be equivalent, we may insist that the first position be a 1. As before, we fill the remaining n 1 positions in (n 1! n 1 ways. Now, rotating the secon a 1 into the first position gives (except in a special case an arrangement that has alreay been counte, so we upate our count to (n 1! n. The case in which this rotation oes not give another arrangement is when the first n characters are in precisely the same orer as the last n characters. Since we insist on having a 1 in the first position, there are (n 1! such arragements (arranging a,..., a n forces the remaining placements. Thus, our final count becomes (n 1! n + (n 1!. Proposition 5. Let A n enote the number of sequences obtaine from S n such that, for all i, the a i s are not ajacent The number of arrangements of S n on a circle subject to the same constraints is given by (n 3A n 1 + (n 1!. Proof. As before, we use rotation to insist that a 1 be in the first position. Next, arrange the elements of the multiset {a, a,..., a n, a n } as a sequence, which can be accomplishe in A n 1 ways. Now, we may freely appen a 1 to the beginning of this sequence an then associate the en of the sequence the the element just prior to a 1 in a circle. It remains to insert the other a 1. Since it cannot be ajacent to the first a 1, there are n 3 possible placements. As before, however, rotating the secon a 1 into the first position may prouce another circle that has alreay been enumerate. We upate our count in the same way as before, thus arriving at (n 3A n 1+(n 1! istinct circles. 4

5 For n =, the above formula gives 1 circles of the esire type an, for n = 3, 4 circles. Below we explicitly list all such sequences (using a, b, an c, for simplicity. Problem 3 n = abab n = 3 abcabc abcacb abcbac abacbc Let I n be the ientity matrix of orer n, let J n be the all ones matrix of orer n, an let A n = J n I n. Proposition 6. The number of erangements on n istinct elements is equal to per A n. Proof. A typical term of the expansion of per A n is of the form a 1σ(1 a nσ(n, for some permutation σ. From the structure of A n, we see that { 0 if σ(i = i a iσ(i = 1 otherwise. Thus, a term in the expansion is equal to 1 precisely when it is a erangement, an so summing over all permutations counts the number of erangements. Proposition 7. For all n N, et A n = ( 1 n 1 (n 1. Proof. Our goal is to perform row operations so that A n is in upper-triangular form, an so has eterminant equal to the prouct of its iagonal. For convenience, we emonstrate the operations on A 4. The operations reaily generalize to A n. 5

6 Row = Row - Row 3 Row 3 = Row 3 - Row 4 Row 4 = Row 4 - Row Transpose Row 1 an Row Row 4 = Row 4 + Row + Row Thus, accounting for the change in sign inuce by the transposition of rows, we have et A n = ( 1( 1 n (n 1 = ( 1 n 1 (n 1. Proposition 8. The expansion of et A n has n! D n terms of 0, Dn+( 1n 1 (n 1 terms of 1, an Dn+( 1n (n 1 terms of -1. 6

7 Proof. Due to the structure of A n, we get term of 0 precisely when the acting permutation has a fixe point. Hence, there are n! D n terms of 0. By a previous observation, we know that et A n = ( 1 n 1 (n 1. Letting a be the number of terms of 1 an b be the number of terms of -1, we have the system which yiels the solution a + b = D n a b = ( 1 n 1 (n 1, a = D n + ( 1 n 1 (n 1 b = D n + ( 1 n (n 1. Proposition 9. Let U n be the n th ménage number an C n be the circulant of size n. For all n N, per(j n I n C n = U n. Proof. For convenience, we consier the the case where n = 4. The result reaily generalizes. The matrix J 4 I 4 C 4 is of the form As previously observe, a term in the expansion of the permanent is equal to 1 precisely when a iσ(i = 1 for all i an is equal to 0, otherwise. We see that a 1σ(1 = 1 whenever σ(1 / {1, 4}, a σ( = 1 whenever σ( / {1, }, a 3σ(3 = 1 whenever σ(3 / {, 3}, an a 4σ(4 = 1 whenever σ(4 / {3, 4}. These are precisely the conitions on the women in the ménage problem once the men have been seate (uner the stanar seating arrangement. Thus, summing over all permutations gives the ménage number U 4. Similarly, per(j n I n C n = U n. 7

8 Problem 4 The number of ways to choose a 0,..., a 9 as a permutation of {0,..., 9} so that no column of the array a 0 a 1 a a 3 a 4 a 5 a 6 a 7 a 8 a 9 repeats any number is given by 10 j=0 (10 j! j g(10, kg(10, j k, k=0 where g(n, k enotes the number of ways to choose k nonajacent elements from n elements arrange on a cycle. Proof. Let Pj i enote the property that a given permutation of the a s has a j equal to the value in row i (i = 0 or 1 an column j of the matrix (an so the permutation oes not meet the esire constraints. Let the properties be arrange on two cycles. The first cycle will be P0 0, P 0 1, P 1 0, P 1 1,..., P 4 0, P 4 1, P 0 0. The secon cycle will be P5 0, P 5 1, P 6 0, P 6 1,..., P 9 0, P 9 1, P 5 0. Observe that every permutation failing to meet the constraint of inucing no repeate element in the matrix correspons to a subset of {Pj i 0 i 1, 0 j 10} having size at most 10 an containing no pair of ajacent P s (where ajacency is efine by the cycles. Here, the size of the subset specifies the number of entries in which the permutation fails an the chosen P s specify for what reason the permutation fails. Notice that any two ajacent P s cannot occur at the same time, as they either specify a failure in the same column for two ifferent reasons (for example, P0 0 an P 0 1, or they specify that two ifferent a s are assigne the same element of {0,..., 9} (for example, P0 1 an P 1 0. Now, we compute the esire quantity in the following way. Let A j be the number of permutations failing the matrix conition in at least j columns. We may view this as summing over all possible j 0 + j 1 = j, where we select j 0 properties from the first cycle an j 1 properties from the secon cycle while maintaining that no pair of selecte properties are ajacent. We then permute the remaining elements freely. All tol, we have A j = (10 j! j g(kg(j k. k=0 8

9 Thus, by inclusion-exclusion, we obtain that the number of permutations satisfying the matrix constraint is given by as esire. Problem 5 10 j=0 (10 j! j g(10, kg(10, j k, k=0 We now etermine the number of wheels of length n in r. Let A = [r] be the set of colors. Let S n,r be the set of n-samples of A, {(a 1, a n a i A}. Define the shift φ : S n,r S n,r by φ(a 1,..., a n := (a,..., a n, a 1. We say two samples α, β S n,r are circularly equivalent, written α β, if there exists k such that φ k (α = β. A wheel is an equivalence class S n,r uner. That is, two colorings give the same wheel if one is a rotation of the other. A wheel W has perio if, for all α W, φ (α = α an, for all <, φ (α α. Proposition 10. If a wheel has length n an perio, then n. Proof. By the ivision algorithm, there are q an r with 0 r < such that n = q + r. It follows irectly that α = φ n (α = φ q+r (α = φ r (φ q (α = φ r (α. Since 0 r <, it must be that r = 0, an so n = q, as esire. Proposition 11. Let M( enote the number of wheels of length an perio in r colors. Prove that, for all n M( = r n. 9

10 Proof. Observe first that S n,r = r n. By the previous proposition, we know that summing over all iviing n is equivalent to summing over all possible perios for wheels of S n,r (one can easily construct a wheel having perio for any iviing n. Thus, it suffices to show that, for each iviing n, M( counts all elements of S n,r belonging to wheels with perio. Let S be such an element. Now, S can be viewe as the concatenation of n copies of a sample of length an perio. (For example, when n =, S = (a 1,..., a, a 1,..., a. Thus, efining S amounts to efining the first entries of S. To o so, we first choose any of the M( wheels, an then choose one of the circularly equivalent samples represente by the chosen wheel. Therefore, there are M( possibilities for S. As previous observe, summing over all iviing n gives the esire result. Proposition 1. For all n N, M(n = 1 µ(r n. n Proof. Let f(n = r n an g(n = nm(n. Applying the Möbius inversion formula to the result of the previous proposition, we have nm(n = ( n µ r, from which it follows that M(n = 1 n ( n µ r = 1 µ ( r n. n Proposition 13. The total number of wheels of length n in r colors is given by T (n := M(. Proof. As previously observe, the collection of possible perios for wheels of length n are precisely the values of iviing n. Now, a wheel of length n can be viewe as the concatenation of n copies of a wheel of length an 10

11 perio. Thus, efining the wheel amounts to efining its first entries. For each value of, this can be accomplishe in M( ways. Thus, summing over all iviing n (that is, over all possible perios, we enumerate all wheels of length n. As an example, the number of wheels of length 3 on r colors is given by T (3 = 3 M( = M(1 + M(3 = µ(r µ(r = µ(1r + 1 ( µ(1r 3 + µ(3r 3 = r + 1 ( r 3 r 3 = r3 + r. 3 11