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A230388
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a(n) = binomial(11*n+1,n)/(11*n+1).
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16
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1, 1, 11, 176, 3311, 68211, 1489488, 33870540, 793542167, 19022318084, 464333035881, 11502251937176, 288417894029200, 7306488667126803, 186719056586568660, 4807757550367267056, 124609430032127192295, 3248403420844673986345
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OFFSET
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0,3
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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=11, r=1. Interesting property when r=1, a(n+1,p,1) = a(n,p,p) for n>=0.
This is also instance k = 10 of the generalized Catalan family {C(k, n)}_{n>=0} given in a comment of A130564. - Wolfdieter Lang, Feb 05 2024
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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, with p=11, r=1.
a(n) = binomial(11*n,n)/(10*n+1) = A062993(n+9, 9).
a(0) = 1; a(n) = Sum_{i1+i2+..i11=n-1} a(i1)*a(i2)*...*a(i11) for n>=1.
(End)
O.g.f.: hypergeometric([1,...,10]/11,[2,...,9,11]/10,(11^11/10^10)*x). For the e.g.f. put an extra 1 = 10/10 into the second part. - Wolfdieter Lang, Feb 05 2024
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MAPLE
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seq(binomial(11*k+1, k)/(11*k+1), k=0..30); # Robert FERREOL, Apr 01 2015
n:=30:G:=series(RootOf(g = 1+x*g^11, g), x=0, n+1):seq(coeff(G, x, k), k=0..n); # Robert FERREOL, Apr 01 2015
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MATHEMATICA
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Table[Binomial[11 n + 1, n]/(11 n + 1), {n, 0, 30}] (* Vincenzo Librandi, Jan 01 2014 *)
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PROG
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(PARI) a(n) = binomial(11*n+1, n)/(11*n+1);
(PARI) {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11))+x*O(x^n)); polcoeff(B, n)}
(Magma) [Binomial(11*n+1, n)/(11*n+1): n in [0..30]]; // Vincenzo Librandi, Jan 01 2014
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CROSSREFS
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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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