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.

A104405 A084203 (the cube root of A083953) read mod 3.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 2, 1, 2, 0, 2, 0, 0, 2, 2, 0, 2, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 2, 0, 0, 2, 0, 1, 2, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 0, 2, 2, 2, 1, 2, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane and Nadia Heninger, Jul 19 2005

Keywords

A084202 G.f. A(x) defined by: A(x)^2 consists entirely of integer coefficients between 1 and 2 (A083952); A(x) is the unique power series solution with A(0)=1.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, -1, 2, -2, 4, -6, 10, -16, 27, -44, 75, -127, 218, -375, 650, -1130, 1974, -3460, 6086, -10736, 18993, -33685, 59882, -106683, 190446, -340611, 610243, -1095102, 1968200, -3542468, 6384518, -11521308, 20815942, -37651528, 68176596, -123574852, 224204708, -407153894
Offset: 0

Views

Author

Paul D. Hanna, May 19 2003

Keywords

Comments

Limit a(n)/a(n+1) -> r = -0.530852489019085 where A(r)=0.
Let A_n(x) be the power series formed from the first n terms of this sequence. Then a(0) = 1, a(n) = floor(1 - [x^n] (A_(n-1)(x))^2/2). Replacing 2 with a larger integer k generates the related sequences A084203-A084212. - Charlie Neder, Jan 16 2019

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = Block[{s = Sum[a[i]*x^i, {i, 0, n - 1}]}, If[ IntegerQ@ Last@ CoefficientList[ Series[ Sqrt[s + x^n], {x, 0, n}], x], 1, 2]]; Table[a[n], {n, 0, 42}]; CoefficientList[ Series[ Sqrt[ Sum[ a[i]*x^i, {i, 0, 42}]], {x, 0, 42}], x] (* Robert G. Wilson v, Nov 11 2007 *)
  • PARI
    /* Using Charlie Neder's formula */
    {a(n) = my(A=[1]); for(i=0,n, A=concat(A,0); A[#A] = floor(1 - polcoeff( Ser(A)^2, #A-1)/2) ); A[n+1]}
    for(n=0,50, print1(a(n),", ")) \\ Paul D. Hanna, Jan 17 2019

A084204 G.f. A(x) defined by: A(x)^4 consists entirely of integer coefficients between 1 and 4 (A083954); A(x) is the unique power series solution with A(0)=1.

Original entry on oeis.org

1, 1, -1, 3, -7, 20, -58, 177, -554, 1769, -5739, 18866, -62684, 210146, -709882, 2413743, -8253995, 28366316, -97916761, 339326189, -1180068800, 4116957243, -14404398636, 50530280752, -177684095927, 626181400993, -2211215950469, 7823025701314, -27724997048327
Offset: 0

Views

Author

Paul D. Hanna, May 20 2003

Keywords

Comments

Limit a(n)/a(n+1) -> r = -0.269562488839799 where A(r)=0.

Crossrefs

Programs

  • Maple
    g:= 1: a[0]:= 1:
    for n from 1 to 50 do
      a[n]:= -floor((coeff(g^4,x,n)-1)/4);
      g:= g + a[n]*x^n;
    od:
    seq(a[n],n=0..50); # Robert Israel, Sep 04 2019
  • Mathematica
    kmax = 30;
    A[x_] = Sum[a[k] x^k, {k, 0, kmax}];
    coes = CoefficientList[A[x]^4 + O[x]^(kmax + 1), x];
    r = {a[0] -> 1, a[1] -> 1}; coes = coes /. r;
    Do[r = Flatten @ Append[r, Reduce[1 <= coes[[k]] <= 4, a[k-1], Integers] // ToRules];
    coes = coes /. r, {k, 3, kmax + 1}];
    Table[a[k], {k, 0, kmax}] /. r (* Jean-François Alcover, Jul 26 2018 *)

A110628 Trisection of A083953 such that the self-convolution cube is congruent modulo 9 to A083953, which consists entirely of 1's, 2's and 3's.

Original entry on oeis.org

1, 1, 3, 3, 1, 2, 2, 1, 2, 3, 2, 3, 3, 2, 2, 3, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 2, 3, 3, 2, 3, 1, 2, 1, 3, 3, 2, 3, 3, 1, 2, 3, 3, 1, 3, 3, 2, 2, 2, 1, 2, 3, 3, 3, 3, 1, 2, 2, 3, 2, 1, 2, 2, 1, 2, 3, 3, 2, 2, 1, 1, 2, 1, 3, 2, 2, 2, 1, 3, 2, 2, 3, 3, 2, 3, 1, 1, 1, 1, 3, 3
Offset: 0

Views

Author

Keywords

Comments

Congruent modulo 3 to A084203 and A104405; the self-convolution cube of A084203 equals A083953.

Crossrefs

Programs

  • PARI
    {a(n)=local(p=3,A,C,X=x+x*O(x^(p*n)));if(n==0,1, A=sum(i=0,n-1,a(i)*x^(p*i))+p*x*((1-x^(p-1))/(1-X))/(1-X^p); for(k=1,p,C=polcoeff((A+k*x^(p*n))^(1/p),p*n); if(denominator(C)==1,return(k);break)))}

Formula

a(n) = A083953(3*n) for n>=0. G.f. satisfies: A(x^3) = G(x) - 3*x*(1+x)/(1-x^3), where G(x) is the g.f. of A083953. G.f. satisfies: A(x)^3 = A(x^3) + 3*x*(1+x)/(1-x^3) + 9*x^2*H(x) where H(x) is the g.f. of A111582.
Showing 1-4 of 4 results.