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.

A095808 Number of ways to write n in the form m + (m+1) + ... + (m+k-1) + (m+k) + (m+k-1) + ... + (m+1) + m with integers m>= 1, k>=1. Or, number of divisors d of 4n-1 with 0 < (d-1)^2 < 4n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 2, 0, 0, 1, 1, 1, 1, 0, 0, 3, 0, 1, 2, 0, 1, 1, 0, 0, 2, 2, 0, 1, 1, 0, 3, 0, 1, 2, 0, 1, 1, 0, 0, 3, 1, 0, 2, 1, 0, 3, 1, 0, 1, 0, 2, 2, 0, 1, 1, 1, 1, 1, 0, 0, 5, 1, 1, 1, 0, 1, 1, 1, 0, 3, 1, 0, 2, 0, 1, 3, 0, 0, 2, 1, 1, 3
Offset: 1

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Author

Alfred Heiligenbrunner, Jun 15 2004

Keywords

Comments

n = m + (m+1) + ... + (m+k-1) + (m+k) + (m+k-1) + ... + (m+1) + m means n = k^2 + m*(2k+1) or 4n-1 = (2k+1)*(4m+2k-1). So if 4n-1 disparts into two odd factors a*b, then k = (a-1)/2, m=(n-k^2)/(2k+1) give the solution of the origin equation. We only count solutions with k^2 < n, such that m>0. This means we are taking into account only factors a < 2n+1.
Note that a(n) = 0 if 4n-1 is prime. - Alfred Heiligenbrunner, Mar 01 2016

Examples

			a(16) = 2 because 16 = 5+6+5 and 16 = 1+2+3+4+3+2+1.
The trivial case 16=16 (k=0, m=n) is not counted. The cases m=0, e.g. 16 = 0+1+2+3+4+3+2+1+0 are not counted. The cases m<0 e.g. 16 = -4+-3+-2+-1+0+1+2+3+4+5+6+5+4+3+2+1+0+-1+-2+-3+-4 are not counted.
		

Crossrefs

Programs

  • Maple
    seq((numtheory[tau](4*n-1)-2)/2, n=1..100); # Ridouane Oudra, Jan 18 2025
  • Mathematica
    h1 = Table[count = 0; For[k = 1, k^2 < n, k++, If[Mod[n - k^2, 2k + 1] == 0, count++ ]]; count, {n, 100}] - or - h2 = Table[Length[Select[Divisors[4n - 1], ((# - 1)^2 < 4n) &]] - 1, {n, 100}]
    a[n_] := (DivisorSigma[0, 4*n-1] - 2)/2; Array[a, 100] (* Amiram Eldar, Jan 28 2025 *)
  • PARI
    a(n) = (numdiv(4*n-1) - 2)/2; \\ Amiram Eldar, Jan 28 2025

Formula

From Ridouane Oudra, Jan 18 2025: (Start)
a(n) = (tau(4*n-1) - 2)/2.
a(n) = A070824(4*n-1)/2.
a(n) = A078703(n) - 1. (End)
Sum_{k=1..n} a(k) = (log(n) + 2*gamma - 5 + 4*log(2))*n/4 + O(n^(1/3)*log(n)), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 27 2025