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a(n) = 7*binomial(8*n+7,n)/(8*n+7).
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%I #33 Feb 15 2024 18:08:57

%S 1,7,77,1015,14763,228459,3689595,61474519,1048927880,18236463245,

%T 321899509386,5753527081211,103922382296180,1893943017506925,

%U 34783258504651434,643111366544129175,11960812088346090200,223614812152492437432,4200107505573406222425

%N a(n) = 7*binomial(8*n+7,n)/(8*n+7).

%C Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=8, r=7.

%H Vincenzo Librandi, <a href="/A234466/b234466.txt">Table of n, a(n) for n = 0..200</a>

%H J-C. Aval, <a href="http://arxiv.org/pdf/0711.0906v1.pdf">Multivariate Fuss-Catalan Numbers</a>, arXiv:0711.0906v1, Discrete Math., 308 (2008), 4660-4669.

%H Thomas A. Dowling, <a href="http://www.mhhe.com/math/advmath/rosen/r5/instructor/applications/ch07.pdf">Catalan Numbers Chapter 7</a>.

%H Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.

%H Wojciech Mlotkowski, <a href="http://www.math.uiuc.edu/documenta/vol-15/28.pdf">Fuss-Catalan Numbers in Noncommutative Probability</a>, Docum. Mathm. 15: 939-955.

%F G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=8, r=7.

%F E.g.f.: hypergeom([7, 9, 10, 11, 12, 13, 14]/8, [8, 9, 10, 11, 12, 13, 14]/7, (8^8/7^7)*x). Cf.: _Ilya Gutkovskiy_ in A118971. - _Wolfdieter Lang_, Feb 06 2020

%F D-finite with recurrence: +7*(7*n+4)*(7*n+1)*(7*n+5)*(7*n+2)*(7*n+6)*(7*n+3)*(n+1)*a(n) -128*(8*n+3)*(4*n+3)*(8*n+1)*(2*n+1)*(8*n-1)*(4*n+1)*(8*n+5)*a(n-1)=0. - _R. J. Mathar_, Feb 21 2020

%F From _Wolfdieter Lang_, Feb 15 2024: (Start)

%F a(n) = binomial(8*n + 6, n+1)/(7*n + 6). This is instance k = 7 of c(k, n+1) given in a comment in A130564.

%F The compositional inverse of y*(1 - y)^7 is x*G(x), where G is the o.g.f.. That is, G(x)*(1 - x*G(x))^7 = 1. This is equivalent to the formula of the first line above with B = G. Take A = B^(1/7) then A*(1 - x*B) = 1 or B*(1 - x*B)^7 = 1.

%F The o.g.f is G(x) = 8F7([7..14]/8, [8..14]/7; (8^8/7^7)*x) = (7/(8*x))*(1 - 7F6([-1,1,2,3,4,5,6]/8, [1,2,3,4,5,6]/7; (8^8/7^7)*x)). See the e.g.f. above.(End)

%t Table[7 Binomial[8 n + 7, n]/(8 n + 7), {n, 0, 40}] (* _Vincenzo Librandi_, Dec 26 2013 *)

%o (PARI) a(n) = 7*binomial(8*n+7,n)/(8*n+7);

%o (PARI) {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(8/7))^7+x*O(x^n)); polcoeff(B, n)}

%o (Magma) [7*Binomial(8*n+7, n)/(8*n+7): n in [0..30]]; // _Vincenzo Librandi_, Dec 26 2013

%Y Cf. A000108, A007556, A118971, A130564, A234461, A234462, A234463, A234464, A234465, A234467, A230390.

%K nonn,easy

%O 0,2

%A _Tim Fulford_, Dec 26 2013