cp's OEIS Frontend

Also products of terms in rows of A047996.

A necklace of sets of beads is a cycle where each element of the cycle is itself a set of beads, the total size being the total number of beads.

Equivalently, a(n) is the number of cyclic compositions of n. These could also be loosely described as cyclic partitions.

Inverse Mobius transform of A059966. - Jianing Song, Nov 13 2021

This sequence seems to represent the number of modal families (i.e., sets of scales who are each other's modes) in a musical tuning with n notes per octave. - Luke Knotts, May 28 2025

Also number of necklaces with 2*n beads, n white and n black (to get the correspondence, start at root, walk around outside of tree, use white if move away from the root, black if towards root).

Also number of terms in polynomial expression for permanent of generic circulant matrix of order n.

a(n) is the number of equivalence classes of n-compositions of n under cyclic rotation. (Given a necklace, split it into runs of white followed by a black bead and record the lengths of the white runs. This gives an n-composition of n.) a(n) is the number of n-multisets in Z mod n whose sum is 0. - David Callan, Nov 05 2003

a(n) is the number of cyclic equivalence classes of triangulations of a once-punctured n-gon. - Esther Banaian, May 06 2025

Equivalently, T(n,k) = number of necklaces with k black beads and n-k white beads (binary necklaces of weight k).

The same sequence arises if we take the table U(n,k) = number of necklaces with n black beads and k white beads and read it by antidiagonals (cf. A241926). - Franklin T. Adams-Watters, May 02 2014

U(n,k) is also equal to the number of ways to express 0 as a sum of k elements in Z/nZ. - Jens Voß, Franklin T. Adams-Watters, N. J. A. Sloane, Apr 30 2014 - May 05 2014. See link ("A Note on Modular Partitions and Necklaces") for proof.

The generating function for column k is given by the substitution x_j -> x^j/(1-x^j) in the cycle index of the Symmetric Group of order k. - R. J. Mathar, Nov 15 2018

From Petros Hadjicostas, Jul 12 2019: (Start)

Regarding the comments above by Voss, Adams-Watters, and Sloane, note that Fredman (1975) proved that the number S(n, k, v) of vectors (a_0, ..., a_{n-1}) of nonnegative integer components that satisfy a_0 + ... + a_{n-1} = k and Sum_{i=0..n-1} i*a_i = v (mod n) is given by S(n, k, v) = (1/(n + k)) * Sum_{d | gcd(n, k)} A054535(d, v) * binomial((n + k)/d, k/d) = S(k, n, v).

This result was also proved by Elashvili et al. (1999), who also proved that S(n, k, v) = Sum_{d | gcd(n, k, v)} S(n/d, k/d, 1). Here, S(n, k, 0) = A241926(n, k) = U(n, k) = T(n + k, k) (where T(n, k) is the current array). Also, S(n, k, 1) = A245558(n, k). See also Panyushev (2011) for more general results and for generating functions.

Finally, note that A054535(d, v) = c_d(v) = Sum_{s | gcd(d,v)} s*Moebius(d/s). These are the Ramanujan sums, which equal also the von Sterneck function c_d(v) = phi(d) * Moebius(d/gcd(d, v))/phi(d/gcd(d, v)). We have A054535(d, v) = A054534(v, d).

It would be interesting to see whether there is a proof of the results by Fredman (1975), Elashvili et al. (1999), and Panyushev (2011) for a general v using Molien series as it is done by Sloane (2014) for the case v = 0 (in which case, A054535(d, 0) = phi(d)). (Even though the columns of array A054535(d, v) start at v = 1, we may start the array at column v = 0 as well.)

(End)

U(n, k) is the number of equivalence classes of k-tuples of residues modulo n, identifying those that differ componentwise by a constant and those that differ by a permutation. - Álvar Ibeas, Sep 21 2021

a(n) is the number of necklaces with 4 black beads and n white beads.

Also nonnegative integer 2 X 2 matrices with sum of elements equal to n, up to rotational symmetry.

The g.f. is Z(C_4,x), the 4-variate cycle index polynomial for the cyclic group C_4, with substitution x[i]->1/(1-x^i), i=1,...,4. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 4-necklaces whose 4 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_4,x). - Wolfdieter Lang, Feb 15 2005

Turning over the necklace is not allowed (the group is cyclic not dihedral). T(n,k) = T(k,n) follows immediately from the formula. - N. J. A. Sloane, May 03 2014

T(n, k) is the number of equivalence classes of k-tuples of residues modulo n, identifying those that differ componentwise by a constant and those that differ by a permutation. - Álvar Ibeas, Sep 21 2021

Number of periodic palindromes of length n using a maximum of k different symbols.

From Petros Hadjicostas, Sep 02 2018: (Start)

According to Christian Bower's theory of transforms, we have boxes of different sizes and different colors. The size of a box is determined by the number of balls it can hold. In this case, we assume all balls are the same and are unlabeled. Assume also that the number of possible colors a box with m balls can have is given by c(m). We place the boxes on a circle at equal distances from each other. Two configurations of boxes on the circle are considered equivalent if one can be obtained from the other by rotation. We are interested about circular configurations of boxes that are circular palindromes (i.e., necklaces with boxes that are the same when turned over). Let b(n) be the number of circularly palindromic configurations of boxes on a circle when the total number of balls in the boxes is n (and each box contains at least one ball).

Bower calls the sequence (b(n): n >= 1), the CPAL ("circular palindrome") transform of the input sequence (c(m): m >= 1). If the g.f. of the input sequence (c(m): m >= 1) is C(x) = Sum_{m>=1} c(m)*x^m, then the g.f. of the output sequence (b(n): n >= 1) is B(x) = Sum_{n >= 1} b(n)*x^n = (1 + C(x))^2/(2*(1 - C(x^2))) - 1/2.

In the current sequence, each box holds only one ball but can have one of k colors. Hence, c(1) = k but c(m) = 0 for m >= 2. Thus, C(x) = k*x. Then, for fixed k, the output sequence is (b(n): n >= 1) = (T(n, k): n >= 1), where T(n, k) = number of necklaces with n beads and k colors that are the same when turned over. If we let B_k(x) = Sum_{n>=1} T(n, k)*x^n, then B_k(x) = (1 + k*x)^2/(2*(1 - k*x^2)) - 1/2. From this, we can easily prove the formulae below.

Note that T(n, k=2) - 1 is the total number of Sommerville symmetric cyclic compositions of n. See pp. 301-304 in his paper in the links below. To see why this is the case, we use MacMahon's method of representing a cyclic composition of n with a necklace of 2 colors (see p. 273 in Sommerville's paper where the two "colors" are an x and a dot . rather than B and W). Given a Sommerville symmetrical composition b_1 + ... + b_r of n (with b_i >= 1 for all i and 1 <= r <= n), create the following circularly palindromic necklace with n beads of 2 colors: Start with a B bead somewhere on the circle and place b_1 - 1 W beads to the right of it; place a B bead to the right of the W beads (if any) followed by b_2 - 1 W beads; and so on. At the end, place a B bead followed with b_r - 1 W beads. (If b_i = 1 for some i, then a B bead follows a B bead since there are 0 W beads between them.) We thus get a circularly palindromic necklace with n beads of two colors. (The only necklace we cannot get with this method is the one than has all n beads colored W.)

It is interesting that the representation of a necklace of length n, say s_1, s_2, ..., s_n, as a periodic sequence (..., s_{-2}, s_{-1}, s_0, s_1, s_2, ...) with the property s_i = s_{i+n} for all i, as was done by Marks R. Nester in Chapter 2 of his 1999 PhD thesis, was considered by Sommerville in his 1909 paper (in the very first paragraph of his paper). (End)

The array is symmetric; for the entries on or below the diagonal see A245559.

If the congruence in the definition is changed from Sum_{j=0..n-1} j*u_j == 1 mod n to Sum_{j=0..n-1} j*u_j == 0 mod n we get the array shown in A241926, A047996, and A037306.

Differs from A011847 from row n = 9, k = 4 on; if the rows are surrounded by 0's, this yields A051168 without its rows 0 and 1, i.e., a(1) is A051168(2,1). - M. F. Hasler, Sep 29 2018

This array was first studied by Fredman (1975). - Petros Hadjicostas, Jul 10 2019

Row less the last element is palindrome for n=odd or n=power of 2 where n is the row number (observation-conjecture).

From Petros Hadjicostas, Jul 13 2019: (Start)

By reading carefully the proof of Lemma 5.1 (pp. 65-66) in Barnes (1959), we see that he actually proved a general result (even though he does not state it in the lemma).

According to the definition of this sequence, for 1 <= k <= n, T(n, k) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = 0 (mod n). The proof of Lemma 5.1 in Barnes (1959) implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s) for 1 <= k <= n.

For fixed k >= 1, the g.f. of the column (T(n, k): n >= 1) (with T(n, k) = 0 for 1 <= n < k) is (x^k/k) * Sum_{s|k} phi(s) * (-1)^(k - (k/s)) / (1 - x^s)^(k/s), which generalizes Herbert Kociemba's formula from A032801.

Barnes' (1959) formula is a special case of Theorem 4 (p. 66) in Ramanathan (1944). If R(n, k, v) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = v (mod n), then he proved that R(n, k, v) = (1/n) * Sum_{s | gcd(n,k)} (-1)^(k - (k/s)) * binomial(n/s, k/s) * C_s(v), where C_s(v) = A054535(s, v) = Sum_{d | gcd(s,v)} d * Moebius(s/d) is Ramanujan's sum (even though it was first discovered around 1900 by the Austrian mathematician R. D. von Sterneck).

Because C_s(v = 0) = phi(s), we get Barnes' (implicit) result; i.e., R(n, k, v=0) = T(n, k) for 1 <= k <= n.

For k=2, we have R(n, k=2, v=0) = T(n, k=2) = A004526(n-1) for n >= 1. For k=3, we have R(n, k=3, v=0) = T(n, k=3) = A058212(n) for n >= 1. For k=4, we have R(n, k=4, v=0) = A032801(n) for n >= 1. For k=5, we have R(n, k=5, v=0) = T(n, k=5) = A008646(n-5) for n >= 5.

The reason we have T(2*m+1, k) = A037306(2*m+1, k) = A047996(2*m+1, k) for m >= 0 and k >= 1 is the following. When n = 2*m + 1, all divisors s of gcd(n, k) are odd. In such a case, k - (k/s) is even for all k >= 1, and thus (-1)^(k - (k/s)) = 1, and thus T(n = 2*m+1, k) = (1/n) * Sum_{s | gcd(n, k)} phi(s) * binomial(n/s, k/s) = A037306(2*m+1, k) = A047996(2*m+1, k).

By summing the products of the g.f. of column k times y^k from k = 1 to k = infinity, we get the bivariate g.f. for the array: Sum_{n, k >= 1} T(n, k)*x^n*y^k = Sum_{s >= 1} (phi(s)/s) * log((1 - x^s + (-x*y)^s)/(1 - x^s)) = -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x*y)^s).

Letting y = 1 in the above bivariate g.f., we get the g.f. of the sequence (Sum_{1 <= k <= n} T(n, k): n >= 1) is -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x)^s) = -x/(1 - x) + Sum_{m >= 0} (phi(2*m + 1)/(2*m + 1)) * log(1 - 2*x^(2*m+1)), which is the g.f. of sequence A082550. Hence, sequence A082550 consists of the row sums.

There is another important interpretation of this array T(n, k) which is related to some of the references for sequence A047996, but because the discussion is too lengthy, we omit the details.

(End)

Numbers with ordered partitions that have periods of length 6.

From each ordered partition of the numbers (15+j) with 0

The a(n) sequence begins with 16 and each member has 1 period; the b(n) sequence begins with 17 and each member has 2 periods; the c(n) sequence begins with 18 and each member has 3 periods; the d(n) sequence begins with 19 and each member has 2 periods; the e(n) sequence begins with 20 and each member has 1 period of length 6