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A234466
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a(n) = 7*binomial(8*n+7,n)/(8*n+7).
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13
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1, 7, 77, 1015, 14763, 228459, 3689595, 61474519, 1048927880, 18236463245, 321899509386, 5753527081211, 103922382296180, 1893943017506925, 34783258504651434, 643111366544129175, 11960812088346090200, 223614812152492437432, 4200107505573406222425
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OFFSET
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0,2
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COMMENTS
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Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=8, r=7.
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LINKS
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FORMULA
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G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=8, r=7.
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
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
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.
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.
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)
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MATHEMATICA
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Table[7 Binomial[8 n + 7, n]/(8 n + 7), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)
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PROG
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(PARI) a(n) = 7*binomial(8*n+7, n)/(8*n+7);
(PARI) {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(8/7))^7+x*O(x^n)); polcoeff(B, n)}
(Magma) [7*Binomial(8*n+7, n)/(8*n+7): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
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CROSSREFS
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Cf. A000108, A007556, A118971, A130564, A234461, A234462, A234463, A234464, A234465, A234467, A230390.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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