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 A245558 Square array read by antidiagonals: T(n,k) = number of n-tuples of nonnegative integers (u_0,...,u_{n-1}) satisfying Sum_{j=0..n-1} j*u_j == 1 mod n and Sum_{j=0..n-1} u_j = m. 8
 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 7, 8, 7, 3, 1, 1, 4, 9, 14, 14, 9, 4, 1, 1, 4, 12, 20, 25, 20, 12, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 18, 40, 66, 75, 66, 40, 18, 5, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,8 COMMENTS 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 LINKS Taylor Brysiewicz, Necklaces count polynomial parametric osculants, arXiv:1807.03408 [math.AG], 2018. A. Elashvili, M. Jibladze, Hermite reciprocity for the regular representations of cyclic groups, Indag. Math. (N.S.) 9 (1998), no. 2, 233-238. MR1691428 (2000c:13006). A. Elashvili, M. Jibladze, D. Pataraia, Combinatorics of necklaces and "Hermite reciprocity", J. Algebraic Combin. 10 (1999), no. 2, 173-188. MR1719140 (2000j:05009). See p. 174. M. L. Fredman, A symmetry relationship for a class of partitions, J. Combinatorial Theory Ser. A 18 (1975), 199-202. I. M. Gessel and C. Reutenauer, Counting permutations with given cycle structure and descent set, J. Combin. Theory, Ser. A, 64, 1993, 189-215, Theorem 9.4. EXAMPLE Square array begins:   1, 1,  1,  1,   1,   1,    1,    1,    1,    1, ...   1, 1,  2,  2,   3,   3,    4,    4,    5,    5, ...   1, 2,  3,  5,   7,   9,   12,   15,   18,   22, ...   1, 2,  5,  8,  14,  20,   30,   40,   55,   70, ...   1, 3,  7, 14,  25,  42,   66,   99,  143,  200, ...   1, 3,  9, 20,  42,  75,  132,  212,  333,  497, ...   1, 4, 12, 30,  66, 132,  245,  429,  715, 1144, ...   1, 4, 15, 40,  99, 212,  429,  800, 1430, 2424, ...   1, 5, 18, 55, 143, 333,  715, 1430, 2700, 4862, ...   1, 5, 22, 70, 200, 497, 1144, 2424, 4862, 9225, ...   ... Reading by antidiagonals, we get:   1;   1, 1;   1, 1,  1;   1, 2,  2,  1;   1, 2,  3,  2,  1;   1, 3,  5,  5,  3,   1;   1, 3,  7,  8,  7,   3,   1;   1, 4,  9, 14, 14,   9,   4,  1;   1, 4, 12, 20, 25,  20,  12,  4,  1;   1, 5, 15, 30, 42,  42,  30, 15,  5,  1;   1, 5, 18, 40, 66,  75,  66, 40, 18,  5, 1;   1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1;   ... MAPLE # To produce the first 10 rows and columns (as on page 174 of the Elashvili et al. 1999 reference): with(numtheory): cnk:=(n, k) -> add(mobius(n/d)*d, d in divisors(gcd(n, k))); anmk:=(n, m, k)->(1/(n+m))*add( cnk(d, k)*binomial((n+m)/d, n/d), d in divisors(gcd(n, m))); # anmk(n, m, k) is the value of a_k(n, m) as in Theorem 1, Equation (4), of the Elashvili et al. 1999 reference. r2:=(n, k)->[seq(anmk(n, m, k), m=1..10)]; for n from 1 to 10 do lprint(r2(n, 1)); od: MATHEMATICA rows = 12; cnk[n_, k_] := Sum[MoebiusMu[n/d] d, {d , Divisors[GCD[n, k]]}]; anmk[n_, m_, k_] := (1/(n+m)) Sum[cnk[d, k] Binomial[(n+m)/d, n/d], {d, Divisors[GCD[n, m]]}]; r2[n_, k_] := Table[anmk[n, m, k], {m, 1, rows}]; T = Table[r2[n, 1], {n, 1, rows}]; Table[T[[n-k+1, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 05 2018, from Maple *) CROSSREFS This array is very similar to but different from A011847. Cf. A051168, A092964, A241926, A047996, A037306, A245559. Rows include A001840, A006918, A051170, A011796, A011797, A031164. Main diagonal is A022553. Sequence in context: A176298 A259575 A169623 * A011847 A091325 A193596 Adjacent sequences:  A245555 A245556 A245557 * A245559 A245560 A245561 KEYWORD nonn,tabl AUTHOR N. J. A. Sloane, Aug 07 2014 STATUS approved

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Last modified August 9 16:51 EDT 2022. Contains 356026 sequences. (Running on oeis4.)