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A186391
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a(n) equals the least sum of the squares of the coefficients in (1 + x^k + x^(2k) + x^p)^n found at sufficiently large p for some fixed k>0.
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3
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1, 4, 32, 340, 4096, 52704, 705956, 9717488, 136443904, 1945097296, 28063032832, 408836809088, 6004266204964, 88779091937488, 1320294416004736, 19733192546306640, 296219343194357760, 4463668854432401280
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OFFSET
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0,2
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COMMENTS
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Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: (1+x+x^2 + x^p)^n for all p>2(n+1).
...
More generally, let B(x) = Sum_{n>=0} b(n)*x^n/n!^2 such that b(n) is the least sum of the squares of the coefficients in (F(x^k) + t*x^p)^n where F(x) is a finite polynomial in x with degree d and p>(n+1)dk for some fixed k>0,
then B(x) = [Sum_{n>=0} (t^2*x)^n/n!^2]*[Sum_{n>=0} c(n)/n!^2] where c(n) equals the sum of the squares of the coefficients in the polynomial F(x)^n.
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LINKS
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Table of n, a(n) for n=0..17.
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FORMULA
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(1) a(n) = Sum_{k=0..n} C(n,k)^2*A082758(k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} A082758(n)*x^n/n!^2]
where A082758(n) is the sum of the squares of the trinomial coefficients in (1+x+x^2)^n.
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EXAMPLE
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G.f.: A(x) = 1 + 4*x + 32*x^2/2!^2 + 340*x^3/3!^2 + 4096*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2] * C(x) where
C(x)= 1 + 3*x + 19*x^2/2!^2 + 141*x^3/3!^2 + 1107*x^4/4!^2 + 8953*x^5/5!^2 + 73789*x^6/6!^2 +...+ A082758(n)*x^n/n!^2 +...
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PROG
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(PARI) {a(n)=local(V=Vec((1+x+x^2+x^(2*n+3))^n)); V*V~}
(PARI) {a(n)=sum(k=0, n, binomial(n, k)^2*sum(j=0, k, binomial(2*k-j, j)*binomial(2*k, j)))}
(PARI) {a(n)=n!^2*polcoeff(sum(m=0, n, x^m/m!^2+x*O(x^n)) *sum(m=0, n, sum(k=0, m, binomial(2*m-k, k)*binomial(2*m, k))*x^m/m!^2), n)}
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CROSSREFS
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Cf. A082758, A186392, A186378.
Sequence in context: A099912 A002005 A123309 * A137432 A177750 A224993
Adjacent sequences: A186388 A186389 A186390 * A186392 A186393 A186394
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna, Feb 19 2011
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STATUS
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approved
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