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 1-4 of 4 results.

A087222 G.f. satisfies A(x) = 1 + x*A(x)*f(x)^3, where f(x) = Sum_{k>=0} x^((4^k-1)/3).

Original entry on oeis.org

1, 1, 4, 10, 26, 69, 184, 488, 1294, 3436, 9116, 24190, 64190, 170334, 451994, 1199400, 3182706, 8445556, 22410946, 59469200, 157806184, 418751069, 1111188772, 2948626472, 7824411358, 20762688580, 55095420880, 146200015984
Offset: 0

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Author

Paul D. Hanna, Aug 27 2003

Keywords

Examples

			Given f(x) = 1 + x + x^5 + x^21 + x^85 + x^341 + ...
so that f(x)^3 = 1 + 3x + 3x^2 + x^3 + 3x^5 + 6x^6 + 3x^7 + 3x^10 + ...
then A(x) = 1 + x*A(x)*(1 + 3x + 3x^2 + x^3 + 3x^5 + 6x^6 + ...)
= 1 + x + 4x^2 + 10x^3 + 26x^4 + 69x^5 + 184x^6 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 30; CoefficientList[Series[1/(1 - Sum[x^((4^k - 1)/3), {k, 0, nmax}]^3*x), {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 11 2020 *)
  • PARI
    a(n)=local(A,m); if(n<1,n==0,m=1; A=1+O(x); while(m<=3*n+3,m*=4; A=1/(1/subst(A,x,x^4)-x)); polcoeff(A,3*n))

Formula

a(n) = A087221(3n).

A087224 G.f. satisfies A(x) = f(x)^2 + x*A(x)*f(x)^3, where f(x) = Sum_{k>=0} x^((4^k-1)/3).

Original entry on oeis.org

1, 3, 7, 19, 50, 133, 352, 935, 2482, 6584, 17473, 46365, 123034, 326478, 866338, 2298895, 6100296, 16187616, 42955106, 113984740, 302467434, 802621041, 2129817812, 5651638433, 14997065388, 39795888008, 105601506802
Offset: 0

Views

Author

Paul D. Hanna, Aug 27 2003

Keywords

Examples

			Given f(x) = 1 + x + x^5 + x^21 + x^85 + x^341 + ...
so that f(x)^2 = 1 + 2x + x^2 + 2x^5 + 2x^6 + x^10 + 2x^21 + ...
and f(x)^3 = 1 + 3x + 3x^2 + x^3 + 3x^5 + 6x^6 + 3x^7 + 3x^10 + ...
then A(x) = (1 + 2x + x^2 + 2x^5 + ...) + x*A(x)*(1 + 3x + 3x^2 + x^3 + 3x^5 + ...)
= 1 + 3x + 7x^2 + 19x^3 + 50x^4 + 133x^5 + 352x^6 + ...
		

Crossrefs

Programs

  • PARI
    a(n)=local(A,m); if(n<1,n==0,m=1; A=1+O(x); while(m<=3*n+3,m*=4; A=1/(1/subst(A,x,x^4)-x)); polcoeff(A,3*n+2))

Formula

a(n) = A087221(3n+2).

A087223 G.f. satisfies A(x) = f(x) + x*A(x)*f(x)^3, where f(x) = Sum_{k>=0} x^((4^k-1)/3).

Original entry on oeis.org

1, 2, 5, 14, 36, 96, 254, 676, 1792, 4756, 12621, 33490, 88868, 235818, 625764, 1660510, 4406296, 11692452, 31026836, 82332140, 218474784, 579739960, 1538385398, 4082226194, 10832507040, 28744906148, 76276860598, 202406625820
Offset: 0

Views

Author

Paul D. Hanna, Aug 27 2003

Keywords

Examples

			Given f(x) = 1 + x + x^5 + x^21 + x^85 + x^341 + ...
so that f(x)^3 = 1 + 3x + 3x^2 + x^3 + 3x^5 + 6x^6 + 3x^7 + 3x^10 + ...
then A(x) = (1 + x + x^5 + ...) + x*A(x)*(1 + 3x + 3x^2 + x^3 + 3x^5 + 6x^6 + ...)
= 1 + 2x + 5x^2 + 14x^3 + 36x^4 + 96x^5 + 254x^6 + ...
		

Crossrefs

Programs

  • PARI
    a(n)=local(A,m); if(n<1,n==0,m=1; A=1+O(x); while(m<=3*n+3,m*=4; A=1/(1/subst(A,x,x^4)-x)); polcoeff(A,3*n+1))

Formula

a(n) = A087221(3n+1).

A346565 Number of compositions (ordered partitions) of 4^n into powers of 4.

Original entry on oeis.org

1, 2, 96, 579739960, 773527571233557154337704151068262296
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 23 2021

Keywords

Comments

The next term is too large to include.

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[1/(1 - Sum[x^(4^k), {k, 0, n}]), {x, 0, 4^n}], {n, 0, 4}]

Formula

a(n) = [x^(4^n)] 1 / (1 - Sum_{k>=0} x^(4^k)).
a(n) = A087221(A000302(n)).
Showing 1-4 of 4 results.