A186391 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.

1, 4, 32, 340, 4096, 52704, 705956, 9717488, 136443904, 1945097296, 28063032832, 408836809088, 6004266204964, 88779091937488, 1320294416004736, 19733192546306640, 296219343194357760, 4463668854432401280

Offset: 0

Author

Paul D. Hanna, Feb 19 2011

Keywords

• nonn

Comments

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.

Examples

			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 +...
		

Crossrefs

Cf. A082758, A186392, A186378.

Programs

Mathematica

Table[Sum[Binomial[n,k]^2 * Sum[Binomial[2*k-j,j] * Binomial[2*k, j], {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

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)}

Formula

(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.

Recurrence: n^2*(2*n-1)^2*(240*n^5 - 2720*n^4 + 12016*n^3 - 25824*n^2 + 26966*n - 10939)*a(n) = 4*(6000*n^9 - 81920*n^8 + 470480*n^7 - 1486720*n^6 + 2841766*n^5 - 3403995*n^4 + 2557086*n^3 - 1163574*n^2 + 291045*n - 30429)*a(n-1) - 24*(6720*n^9 - 103040*n^8 + 678448*n^7 - 2511512*n^6 + 5740528*n^5 - 8358208*n^4 + 7691502*n^3 - 4262963*n^2 + 1269092*n - 151182)*a(n-2) + 4*(96000*n^9 - 1633280*n^8 + 11993280*n^7 - 49661760*n^6 + 127015344*n^5 - 206357088*n^4 + 210522120*n^3 - 127943860*n^2 + 41074746*n - 5150007)*a(n-3) - 64*(n-3)^2*(4*n - 13)*(4*n - 11)*(240*n^5 - 1520*n^4 + 3536*n^3 - 3696*n^2 + 1686*n - 261)*a(n-4). - Vaclav Kotesovec, Feb 12 2015

a(n) ~ 2^(4*n + 1/2) / (sqrt(3) * Pi * n). - Vaclav Kotesovec, Feb 12 2015