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%I #29 Oct 06 2021 12:57:11
%S 1,1,1,1,1,2,7,22,57,127,259,529,1189,3004,8009,21073,53233,129813,
%T 312733,763573,1915251,4914736,12720841,32800186,83869501,213261712,
%U 542609237,1388542312,3579043987,9273567337,24075321925,62475528190,161969731985,419914766965
%N a(n) = Sum_{k=0..floor(n/5)} binomial(n,5*k)*binomial(6*k,k)/(5*k+1).
%H Alois P. Heinz, <a href="/A226910/b226910.txt">Table of n, a(n) for n = 0..750</a>
%F Representation in terms of special values of generalized hypergeometric function of type 10F9: a(n) = hypergeom([1/6, 1/3, 1/2, 2/3, 5/6, -(1/5)*n, -(1/5)*n+4/5, -(1/5)*n+3/5, -(1/5)*n+2/5, 1/5-(1/5)*n], [1/5, 2/5, 2/5, 3/5, 3/5, 4/5, 4/5, 1, 6/5], -6^6/5^5), n>=0.
%F Recurrence: -49781*(n-9)*(n-8)*(n-7)*(n-6)*(n-5)*a(n-10) + 10*(n-8)*(n-7)*(n-6)*(n-5)*(26453*n - 123726)*a(n-9) - 15*(n-7)*(n-6)*(n-5)*(40479*n^2 - 351957*n + 782140)*a(n-8) + 120*(n-6)*(n-5)*(7013*n^3 - 87699*n^2 + 378278*n - 565577)*a(n-7) - 6*(n-5)*(148255*n^4 - 2435310*n^3 + 15491085*n^2 - 45173430*n + 50791476)*a(n-6) + 12*(69513*n^5 - 1361100*n^4 + 10838875*n^3 - 43818750*n^2 + 89776250*n - 74437500)*a(n-5) - 93750*(7*n^4 - 98*n^3 + 525*n^2 - 1274*n + 1180)*(n-3)*a(n-4) + 375000*(n-2)*(n^2-6*n+10)*(n-3)^2*a(n-3) - 46875*(n-2)*(n-1)*(3*n^2-15*n+20)*(n-3)*a(n-2) + 31250*(n-2)^2*(n-1)*n*(n-3)*a(n-1) - 3125*(n-2)*(n-1)*n*(n+1)*(n-3)*a(n) = 0. - _Vaclav Kotesovec_, Jun 28 2013
%F a(n) ~ (5+6^(1+1/5))^(n+3/2)/(5^(n+1)*6^(1+3/10)*sqrt(2*Pi)*n^(3/2)). - _Vaclav Kotesovec_, Jun 28 2013
%F G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x^5 * A(x)^6. - _Ilya Gutkovskiy_, Jul 25 2021
%F From _Peter Bala_, Sep 15 2021: (Start)
%F O.g.f.: A(x) = (1/x)*series reversion ( x*(1 - x^5)/(1 + x*(1 - x^5)) ).
%F The g.f. of the m-th binomial transform of this sequence is equal to (1/x)*series reversion ( x*(1 - x^5)/(1 + (m + 1)*x*(1 - x^5)) ). The case m = -1 gives the sequence [1, 0, 0, 0, 0, 1, 0, 0,0, 0, 6, 0, 0, 0, 0, 51, 0, 0, 0, 0, 506, ...] - an aerated version of A002295. (End)
%t Table[Sum[Binomial[n,5*k]*Binomial[6*k,k]/(5*k+1),{k,0,Floor[n/5]}],{n,0,20}] (* _Vaclav Kotesovec_, Jun 28 2013 *)
%o (PARI) a(n)=sum(k=0,n\5,binomial(n,5*k)*binomial(6*k,k)/(5*k+1)) \\ _Charles R Greathouse IV_, Jun 24 2013
%Y Cf. A002295, A049130, A212385, A227035.
%K nonn,easy
%O 0,6
%A _Karol A. Penson_, Jun 22 2013
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