A136107 Number of representations of n as the difference of two positive triangular numbers.

0, 1, 1, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 4, 1, 2, 4, 2, 1, 4, 2, 4, 2, 2, 2, 4, 2, 2, 4, 2, 2, 5, 2, 2, 2, 3, 3, 4, 2, 2, 4, 3, 2, 4, 2, 2, 4, 2, 2, 6, 1, 4, 3, 2, 2, 4, 4, 2, 3, 2, 2, 6, 2, 4, 3, 2, 2, 5, 2, 2, 4, 4, 2, 4, 2, 2, 6, 3, 2, 4, 2, 4, 2, 2, 3, 6, 3, 2, 4, 2, 2, 7

Offset

1,5

Comments

a(n) is also the number of partitions of n into consecutive parts greater than 1. - Omar E. Pol, Feb 07 2022

a(n) is the number of solutions of the equations 2(x-1)y-(x-3)x=2(n+1) for 0<x<=y, x-values in A351284; y-values in A351285. Also the number of times n+1 appears in A351153. - Stefano Spezia, Feb 12 2022

Equivalence with Stefano Spezia solutions: The equation 2(x-1)y-(x-3)x=2(n+1) can be rewritten (y+1-x/2)(x+1)=n; proof by solving both for y. So solutions factorize n, and since x+1 must be an integer and y+1-x/2 must be an integer, x must be even. So (x+1)|n means we are looking for odd divisors of n, which is the A001227 term of the Alekseyev formula. The correction by A010054 in the Alekseyev formula means: if n is a triangular number, the solution x=y+1 where x>y is not counted by Spezia. - R. J. Mathar, Feb 12 2022

Links

Robert G. Wilson v, Table of n, a(n) for n = 1..54000.

Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson, The Comma Sequence: A Simple Sequence With Bizarre Properties, arXiv:2401.14346, Youtube

Robert Dougherty-Bliss and Natalya Ter-Saakov, The Comma Sequence is Finite in Other Bases, arXiv:2408.03434 [math.NT], 2024.

Formula

G.f.: Sum_{n>=1} x^((n^2+3*n)/2)/(1-x^n). - Vladeta Jovovic, May 13 2008

a(n) = A001227(n) - A010054(n). - Max Alekseyev, May 13 2009

Example

a(2) = 1 because 3 - 1 = 2,
a(5) = 2 because 6 - 1 = 15 - 10 = 5,
a(9) = 3 because 10 - 1 = 15 - 6 = 45 - 36 = 9, etc.
For n = 21 the four partitions of 21 into consecutive parts are [21], [11, 10], [8, 7, 6] and [6, 5, 4, 3, 2, 1]. The last partition contains 1 as a part, hence there are only three partitions of 21 into consecutive parts whose parts are greater than 1, so a(21) = 3. - Omar E. Pol, Feb 07 2022

Mathematica

f[n_] := Block[{c = 0, k = 1}, While[k < n, If[ IntegerQ[ Sqrt[8 n + 4 k (k + 1) + 1]], c++ ]; k++ ]; c]; Table[f@n, {n, 105}]

Prog

(PARI) a(n) = numdiv(n>>valuation(n, 2)) - ispolygonal(n, 3); \\ Michel Marcus, Jan 08 2024

Crossrefs

Cf. A000217, A001227, A010054, A136108, A351153, A351284, A351285.

Sequence in context: A318831 A303710 A263280 * A178691 A329313 A329312

Adjacent sequences: A136104 A136105 A136106 * A136108 A136109 A136110

Keyword

nonn

Author

John W. Layman and Robert G. Wilson v, Dec 12 2007

Status

approved