A186392 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^3 + x^p)^n found at sufficiently large p for some fixed k>0.

1, 21, 1005, 57117, 3515661, 227676321, 15287457741, 1054718889525, 74310865827597, 5323117605120273, 386421018984886905, 28357462296640927845, 2099749565250183356973, 156648556486910137353777

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)^3 + x^p)^n for all p>3(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 + 21*x + 1005*x^2/2!^2 + 57117*x^3/3!^2 + 3515661*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 + 20*x + 924*x^2/2!^2 + 48620*x^3/3!^2 + 2704156*x^4/4!^2 +...+ (6n)!/(3n)!^2*x^n/n!^2 +...
		

Crossrefs

Cf. A186378, A186391.

Programs

Mathematica

Table[Sum[Binomial[n,k]^2 * Binomial[6*k,3*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

PARI

{a(n)=sum(k=0,n,binomial(n,k)^2*(6*k)!/(3*k)!^2)}

PARI

{a(n)=n!^2*polcoeff(sum(m=0, n, (6*m)!/(3*m)!^2*x^m/m!^2)*sum(m=0, n, x^m/m!^2+x*O(x^n)), n)}

PARI

{a(n)=local(V=Vec(((1+x)^3+x^(3*n+4))^n)); V*V~}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(6k,3k).

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} (6n)!/(3n)!^2 *x^n/n!^2]

where (6n)!/(3n)!^2 is the sum of the squares of the coefficients in (1+x)^(3n).

Recurrence: n^2*(3*n-2)^2*(3*n-1)^2*(11664000*n^8 - 185457600*n^7 + 1268251200*n^6 - 4872227850*n^5 + 11501613345*n^4 - 17086352076*n^3 + 15601848563*n^2 - 8008019030*n + 1769497668)*a(n) = (125656272000*n^14 - 2377737892800*n^13 + 20176977398400*n^12 - 101636310193650*n^11 + 339033082048335*n^10 - 790997589023868*n^9 + 1328807234532186*n^8 - 1629746362828908*n^7 + 1463401419459585*n^6 - 955188456600918*n^5 + 445022508698326*n^4 - 143136353903096*n^3 + 29975298427288*n^2 - 3656735400000*n + 197437737600)*a(n-1) - 2*(2060573904000*n^14 - 41192487921600*n^13 + 371601043200000*n^12 - 2002776983698050*n^11 + 7194382158658545*n^10 - 18189133596160956*n^9 + 33303386556391095*n^8 - 44736636269608884*n^7 + 44153839934527497*n^6 - 31729853553647838*n^5 + 16260230748029395*n^4 - 5728259846949480*n^3 + 1303148021418356*n^2 - 170502613376352*n + 9707820967872)*a(n-2) + 18*(n-2)^2*(645497424000*n^12 - 11574979185600*n^11 + 91670927428800*n^10 - 422559236833650*n^9 + 1257324245932095*n^8 - 2530974275757936*n^7 + 3511780338639909*n^6 - 3357925240298748*n^5 + 2175657267448355*n^4 - 921464025234426*n^3 + 239290954736149*n^2 - 33846317262624*n + 1986410906748)*a(n-3) - 81*(n-3)^2*(n-2)^2*(140399568000*n^10 - 1953070315200*n^9 + 11498171428800*n^8 - 37500795421650*n^7 + 74520854931765*n^6 - 93517328384172*n^5 + 74333125575977*n^4 - 36536327729802*n^3 + 10492652783974*n^2 - 1569370920528*n + 93953212632)*a(n-4) + 321489*(n-4)^2*(n-3)^2*(n-2)^2*(11664000*n^8 - 92145600*n^7 + 296640000*n^6 - 504146250*n^5 + 489706095*n^4 - 274985196*n^3 + 85944305*n^2 - 13448002*n + 818220)*a(n-5). - Vaclav Kotesovec, Feb 12 2015

a(n) ~ 3^(4*n + 5/2) / (16 * Pi * n). - Vaclav Kotesovec, Feb 12 2015