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a(n) = 1/((d-1)*n + 1)*Sum_{i=0..n} binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i), with d = 6.
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%I #72 Jan 10 2024 16:33:08

%S 1,2,22,342,6202,122762,2571326,56031470,1257199154,28849835538,

%T 673953255142,15973925161030,383186776643946,9285457458463770,

%U 226959074854361742,5588974707042304222,138529985051020001634,3453373395317346136610,86526667346028323084726,2177844556015530807952438

%N a(n) = 1/((d-1)*n + 1)*Sum_{i=0..n} binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i), with d = 6.

%C See Yang-Jiang paper, related to large Schröder numbers, which correspond to the formula in the Name, instead with d=2.

%H Seiichi Manyama, <a href="/A363006/b363006.txt">Table of n, a(n) for n = 0..500</a>

%H Lin Yang, Yu-Yuan Zhang, and Sheng-Liang Yang, <a href="https://doi.org/10.1016/j.laa.2023.12.021">The halves of Delannoy matrix and Chung-Feller properties of the m-Schröder paths</a>, Linear Alg. Appl. (2024).

%H Sheng-liang Yang and Mei-yang Jiang, <a href="https://journal.lut.edu.cn/EN/abstract/abstract528.shtml">Pattern avoiding problems on the hybrid d-trees</a>, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. See p. 145. (in Mandarin.)

%F G.f. satisfies A(x) = 1 + x * A(x)^5 * (1 + A(x)). - _Seiichi Manyama_, May 29 2023

%F From _Seiichi Manyama_, Aug 09 2023: (Start)

%F a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(6*n-k,n-1-k) for n > 0.

%F a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(5*n,k-1) for n > 0. (End)

%t With[{d = 6}, Table[(1/((d - 1) n + 1)) Sum[Binomial[(d - 1) n + 1, n - i] Binomial[(d - 1) n + i, i], {i, 0, n}], {n, 0, 12}] ]

%o (PARI) a(n) = my(d=6); sum(i=0, n, binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i))/((d-1)*n + 1); \\ _Michel Marcus_, May 16 2023

%Y Cf. A006318 (d=2), A027307 (d=3), A144097 (d=4), A260332 (d=5).

%Y Cf. A217364, A363305, A364195, A364196.

%K nonn

%O 0,2

%A _Michael De Vlieger_, May 16 2023