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.

A186375 a(n) equals the sum of the squares of the expansion coefficients for (x + y + 2*z)^n.

Original entry on oeis.org

1, 6, 54, 588, 7110, 91476, 1224636, 16849944, 236523078, 3371140740, 48630906324, 708412918824, 10403176168476, 153813188724552, 2287366047735480, 34185974267420208, 513159651195396678, 7732530110414488932
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 any one of the following polynomials: (1 + 2*x^k + x^p)^n, (2 + x^k + x^p)^n, or (1 + x^k + 2*x^p)^n, for all p > n*k and fixed k > 0.
Rescaling the g.f. G(x) to T(u)=G(3*u/16) moves the singular point x=1/16 to u=1/3. Period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318245. - Bradley Klee, Aug 22 2018

Examples

			G.f.: A(x) = 1 + 6*x + 54*x^2/2!^2 + 588*x^3/3!^2 + 7110*x^4/4!^2 + ...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2]^2 *[Sum_{n>=0} (4x)^n/n!^2] where
[Sum_{n>=0} x^n/n!^2]^2 = 1 + 2*x + 6*x^2/2!^2 + 20*x^3/3!^2 + 70*x^4/4!^2 + ... + (2n)!/n!^2 *x^n/n!^2 + ...
a(4) =   256
       + 1024 + 1024
       + 576  + 2304 + 576
       + 64   + 576  + 576 + 64
       + 1    + 16   + 36  + 16 + 1  = 7110.

Links

Crossrefs

Cf. A046816, A186376, A186377, A186378. Periods: A318245, A318417.

Programs

  • Maple
    A186375 := n -> 4^n*hypergeom([1/2,-n,-n], [1,1], 1):
seq(simplify(A186375(n)), n=0..17); # Peter Luschny, May 24 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^2*Binomial[2k,k]*4^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
(* From Bradley Klee, Aug 22 2018: Start *)PyramidLevel[n_]:=If[n==0, {{1}}, Table[Coefficient[(2*x+y+z)^n,x^j*y^k*z^(n-j-k)]^2, {j,0,n}, {k,0,n-j}]]; a1[n_]:= Total[Flatten[PyramidLevel[n]]];
a1 /@ Range[0, 10]
RecurrenceTable[{4*(4*n-5)*(4*n-3)*a[n-2]-2*(10*n^2-10*n+3)*a[n-1]+n^2*a[n]==0, a[0]==1, a[1]==6},a,{n,0,1000}] (*  End *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[ (x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 27 2018 *)
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*(n-k),n-k)*4^k)}
  • PARI
    {a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^2*sum(m=0,n,(2^2*x)^m/m!^2),n)}
  • PARI
    {a(n)=local(V=Vec((1+2*x+x^(n+2))^n));V*V~}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*4^(n-k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x)^2 * B(2^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*a(n) = 2*(10*n^2-10*n+3)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+1/2)/(Pi*n). - Vaclav Kotesovec, Oct 20 2012
a(n) = 4^n*hypergeom([1/2,-n,-n], [1,1], 1). - Peter Luschny, May 24 2017
G.f.: G(x)=Sum_{n>=0}a(n)x^n, 6*(10*x-1)*G + (192*x^2-40*x+1)*G' + x*(16*x-1)*(4*x-1)*G''=0. - Bradley Klee, Aug 22 2018

Extensions

Name edited by Bradley Klee, Aug 22 2018

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
Showing 1-3 of 3 results.

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