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.

A353367 Sum of A110963 and its Dirichlet inverse.

Original entry on oeis.org

2, 0, 0, 1, 0, 2, 0, 1, 1, 4, 0, 1, 0, 2, 4, 1, 0, 5, 0, 2, 2, 4, 0, 1, 4, 8, 5, 1, 0, -2, 0, 1, 4, 10, 4, 3, 0, 6, 8, 2, 0, 10, 0, 2, 8, 4, 0, 1, 1, 10, 10, 4, 0, 3, 8, 1, 6, 16, 0, 1, 0, 2, 15, 1, 16, 14, 0, 5, 4, 6, 0, 3, 0, 20, 6, 3, 4, -2, 0, 2, 9, 22, 0, 6, 20, 12, 16, 2, 0, 16, 8, 2, 2, 4, 12, 1, 0, 25, 24
Offset: 1

Views

Author

Antti Karttunen, Apr 18 2022

Keywords

Comments

Note the negative terms, in contrast to A349135, which apparently has none.

Crossrefs

Programs

  • PARI
    up_to = 65537;
    DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(dA003602(n) = (1+(n>>valuation(n,2)))/2;
    A110963(n) = if(n%2, A003602((1+n)/2), A110963(n/2));
    v353366 = DirInverseCorrect(vector(up_to,n,A110963(n)));
    A353366(n) = v353366[n];
    A353367(n) = (A110963(n)+A353366(n));

Formula

a(n) = A110963(n) + A353366(n).
For n > 1, a(n) = -Sum_{d|n, 1A110963(d) * A353366(n/d).
For all n >= 1, a(4*n) = A110963(n), and a(8*n-4) = A003602(n).

A353368 Dirichlet inverse of A103391, "even fractal sequence".

Original entry on oeis.org

1, -2, -2, 1, -2, 4, -3, -1, 2, 2, -4, -3, -3, 4, 3, 0, -2, -10, -6, 1, 8, 4, -7, 3, 1, -2, -8, -1, -5, -4, -9, -1, 14, -10, 2, 17, -6, 4, 1, -1, -4, -22, -12, 1, -3, 4, -13, -1, 6, -14, -6, 11, -8, 28, 1, 1, 19, -10, -16, 3, -9, 4, -25, -1, 10, -42, -18, 25, 18, 0, -19, -17, -6, -14, -12, 5, 13, 12, -21, 3, 24
Offset: 1

Views

Author

Antti Karttunen, Apr 18 2022

Keywords

Crossrefs

Programs

  • PARI
    up_to = 65537;
    DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(dA003602(n) = (n/2^valuation(n, 2)+1)/2; \\ From A003602
    A103391(n) = if(1==n,1,(1+A003602(n-1)));
    v353368 = DirInverseCorrect(vector(up_to,n,A103391(n)));
    A353368(n) = v353368[n];

Formula

a(1) = 1; a(n) = -Sum_{d|n, d < n} A103391(n/d) * a(d).
a(n) = A353369(n) - A103391(n).
Showing 1-2 of 2 results.