cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A186378 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^2 + x^p)^n found at sufficiently large p for some fixed k>0.

Original entry on oeis.org

1, 7, 95, 1609, 30271, 606057, 12636689, 271026455, 5934011839, 131978406553, 2971793928145, 67586972435495, 1549805136840625, 35783848365934663, 831089570101489423, 19400246240227360809, 454864027237803296703
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2011

Keywords

Comments

Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: (1 + 2*x + x^2 + x^p)^n for all p>2(n+1).
...
More generally, let B(x) equal the g.f. for the least sum of the squares of the coefficients in (F(x^k) + 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} 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 + 7*x + 95*x^2/2!^2 + 1609*x^3/3!^2 + 30271*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = C(x) * BesselI(0, 2*sqrt(x)) where
C(x)= 1 + 6*x + 70*x^2/2!^2 + 924*x^3/3!^2 + 12870*x^4/4!^2 +...+ (4n)!/(2n)!^2*x^n/n!^2

Crossrefs

Cf. A186375, A186376, A186377, A186391, A186392.

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^2 * Binomial[4*k,2*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)
  • PARI
    {a(n)=local(V=Vec((1+2*x+x^2+x^(2*n+3))^n));V*V~}
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*(4*k)!/(2*k)!^2)}
  • PARI
    {a(n)=n!^2*polcoeff(sum(m=0,n,(4*m)!/(2*m)!^2*x^m/m!^2)*sum(m=0,n,x^m/m!^2+x*O(x^n)),n)}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(4k,2k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} (4n)!/(2n)!^2 *x^n/n!^2] *[Sum_{n>=0} x^n/n!^2].
Recurrence: n^2*(2*n-1)^2*(1152*n^4 - 8160*n^3 + 21040*n^2 - 23376*n + 9467)*a(n) = 3*(55296*n^8 - 503808*n^7 + 1891456*n^6 - 3812256*n^5 + 4504864*n^4 - 3193428*n^3 + 1326995*n^2 - 296732*n + 27900)*a(n-1) - 3*(451584*n^8 - 4607232*n^7 + 19744768*n^6 - 46227488*n^5 + 64243016*n^4 - 53731348*n^3 + 26049967*n^2 - 6596672*n + 675000)*a(n-2) + (n-2)^2*(2230272*n^6 - 16991232*n^5 + 49582720*n^4 - 69169056*n^3 + 46825856*n^2 - 13847412*n + 1451547)*a(n-3) - 900*(n-3)^2*(n-2)^2*(1152*n^4 - 3552*n^3 + 3472*n^2 - 1168*n + 123)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 5^(2*n+3/2) / (2^(7/2) * Pi * n). - Vaclav Kotesovec, Feb 12 2015

A186376 a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n.

Original entry on oeis.org

1, 14, 294, 7292, 198310, 5717124, 171485916, 5290989816, 166688596998, 5335746337460, 172951272017524, 5662840724506056, 186960502253087836, 6215612039963043368, 207865952390729881080, 6987002286567227550192
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2011

Keywords

Comments

a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + 3*x^p)^n found at sufficiently large p for some fixed k>0.
Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:
. (1 + 3*x^k + 2*x^p)^n, or
. (2 + x^k + 3*x^p)^n, or
. (3 + 2*x^k + x^p)^n, etc.,
for all p>(n+1)k and fixed k>0.
a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n. - Michael Somos, Aug 25 2018

Examples

			G.f.: A(x) = 1 + 14*x + 294*x^2/2!^2 + 7292*x^3/3!^2 +...
The g.f. may be expressed as:
[Sum_{n>=0}x^n/n!^2]*[Sum_{n>=0}(4x)^n/n!^2]*[Sum_{n>=0}(9x)^n/n!^2].

Crossrefs

Cf. A186375, A186377, A186378, A186391.

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^2 * Sum[Binomial[k,j]^2 * 4^(k-j) * 9^j, {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[(x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 2, z -> 3}]]; (* Michael Somos, Aug 25 2018 *)
  • PARI
    {a(n)=local(V=Vec((1+2*x+3*x^(n+2))^n));V*V~}
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*sum(j=0,k,binomial(k,j)^2*4^(k-j)*9^j))}
  • PARI
    {a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)*sum(m=0,n,(2^2*x)^m/m!^2)*sum(m=0,n,(3^2*x)^m/m!^2),n)}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2 *Sum_{j=0..k} C(k,j)^2*4^(k-j)*9^j.
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x) * B(2^2*x) * B(3^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*(4*n - 7)*a(n) = 14*(16*n^3 - 44*n^2 + 34*n - 9)*a(n-1) - 196*(2*n - 3)^2*(4*n - 3)*a(n-2) + 144*(4*n - 9)*(4*n - 7)*(4*n - 3)*a(n-3). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 2^(2*n-1) * 3^(2*n+1) / (Pi*n). - Vaclav Kotesovec, Feb 12 2015

Extensions

Name changed to match the definition given by Michael Somos. - Paul D. Hanna, Sep 05 2018

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.

Original entry on oeis.org

1, 21, 1005, 57117, 3515661, 227676321, 15287457741, 1054718889525, 74310865827597, 5323117605120273, 386421018984886905, 28357462296640927845, 2099749565250183356973, 156648556486910137353777
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2011

Keywords

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
Showing 1-3 of 3 results.

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