

A136107


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


12



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
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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(x1)y(x3)x=2(n+1) for 0<x<=y, xvalues in A351284; yvalues 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(x1)y(x3)x=2(n+1) can be rewritten (y+1x/2)(x+1)=n; proof by solving both for y. So solutions factorize n, and since x+1 must be integer and y+1x/2 must be 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



FORMULA



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}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



