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 1-2 of 2 results.

A090678 a(n) = A088567(n) mod 2.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Dec 20 2003

Keywords

Comments

a(n) = -(-1)^[n/2]*A110036(n)/2 for n>=2, where A110036 gives the partial quotients of the continued fraction expansion of 1 + Sum_{n>=0} 1/x^(2^n). - Paul D. Hanna, Jul 09 2005

Crossrefs

b(8m) is (apart from the first term) A038189(m).

Programs

  • Mathematica
    nmax = 104; f = 1 + x/(1 - x) + Sum[x^(3*2^(k - 1))/Product[1 - x^(2^j), {j, 0, k}], {k, 1, Log[2, nmax]}];
    a[n_] := Mod[SeriesCoefficient[f, {x, 0, n}], 2];
    Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Jul 26 2018 *)
  • PARI
    {a(n)=(-1)^(n\2)*polcoeff(1+x-x^2*(1+x)/(1+x^2)+ sum(k=1,#binary(n),x^(3*2^(k-1))/prod(j=0,k,1+x^(2^j)+x*O(x^n))),n)} /* Paul D. Hanna */

Formula

b(0) == 1; if n is odd, b(n) == b(n-1) + 1; b(8m+2) == 1; b(8m+6) == 0; b(16m+4) == 0; b(16m+12) == 1; for m>0, b(16m) == b(8m), b(32m+8) == 0, b(32m+24) == 1. In other words, for m>0, b(8m) is the value of the bit immediately to the left of the rightmost 1 when m is written in binary.
a(n) = (-1)^floor(n/2)*A110037(n). - Paul D. Hanna, Jul 09 2005

A110037 Signed version of A090678 and congruent to A088567 mod 2.

Original entry on oeis.org

1, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 0
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2005

Keywords

Comments

a(n) = (-1)^[n/2]*A090678(n) = A088567(n) mod 2, where A088567(n) equals the number of "non-squashing" partitions of n. a(n) = -A110036(n)/2 for n>=2, where the A110036 gives the partial quotients of the continued fraction expansion of 1 + Sum_{n>=0} 1/x^(2^n).

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(A=1+x-x^2*(1+x)/(1+x^2)+ sum(k=1,#binary(n),x^(3*2^(k-1))/prod(j=0,k,1+x^(2^j)+x*O(x^n))),n)}

Formula

G.f.: A(x) = 1+x - x^2*(1+x)/(1+x^2) + Sum_{k>=1} x^(3*2^(k-1))/Product_{j=0..k} (1+x^(2^j)).
Conjecture: a(n) = A073089(n) - A073089(n+1) for n >= 2. - Alan Michael Gómez Calderón, Aug 19 2025
Showing 1-2 of 2 results.