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A005721
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Central quadrinomial coefficients.
(Formerly M3681)
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2
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1, 4, 44, 580, 8092, 116304, 1703636, 25288120, 379061020, 5724954544, 86981744944, 1327977811076, 20356299454276, 313095240079600, 4829571309488760, 74683398325804080, 1157402982351003420, 17971185794898859248
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Sum of squares of entries in the n-th row of triangle of quadrinomial coefficients (Pascal triangle of order 4). [From Adi Dani, Jul 03 2011]
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REFERENCES
| L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..100
Adi Dani Restricted compositions of natural numbers-section: Generalized Pascal triangle
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FORMULA
| a(n)= A005190(2*n) = A008287(2*n, 3*n).
G.f.: Let Z(x) be a solution of (-1+16*x)*(32*x-27)^2*Z^6+9*(-9+64*x)*(32*x-27)*Z^4+81*(80*x-27)*Z^2+729 = 0, with Z(0)=1. Compute a Puiseux series for Z(x) at x=0, then Z(x) in C[[x^(1/3)]]. Remove all non-integer powers of x. The result is the generating function for A005721. - Mark van Hoeij, Oct 29 2011.
G.f.: F(G^(-1)(x)) where F(t) = (t^2-1)*(6*t+t^2+1)^(1/2)/(3*t^3+13*t^2+t-1)
and G(t) = t/((t+1)^2*(6*t+t^2+1)) - Mark van Hoeij, Oct 30 2011.
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MAPLE
| F := (t^2-1)*(6*t+t^2+1)^(1/2)/(3*t^3+13*t^2+t-1); G := t/((t+1)^2*(6*t+t^2+1));
Ginv := RootOf(numer(G-x), t); series(eval(F, t=Ginv), x=0, 20);
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MATHEMATICA
| Table[Sum[(-1)^k*Binomial[2*n, k]*Binomial[5*n-4*k-1, 3*n-4*k], {k, 0, 3*n/4}], {n, 0, 25}] (* From Adi Dani, Jul 03 2011 *)
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PROG
| (PARI) a(n)={local(v=Vec((1+x+x^2+x^3)^n)); sum(k=1, #v, v[k]^2); }
(PARI) a(n)=sum(k=0, 3*n/4, (-1)^k*binomial(2*n, k)*binomial(5*n-4*k-1, 3*n-4*k));
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CROSSREFS
| Sequence in context: A087813 A082779 A053315 * A103870 A056063 A177749
Adjacent sequences: A005718 A005719 A005720 * A005722 A005723 A005724
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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