A379697 - OEIS

A379697

For n >= 3, a(n) is the least k >= 0 such that (k + 1)*(2*n + k) / 2 is a triangular number (A000217).

0, 2, 5, 0, 1, 20, 4, 0, 2, 6, 8, 2, 0, 10, 119, 3, 4, 20, 0, 1, 5, 14, 2, 5, 1, 0, 32, 6, 7, 464, 20, 2, 8, 0, 24, 8, 2, 4, 65, 9, 10, 47, 0, 3, 11, 30, 17, 2, 3, 1, 59, 12, 0, 2, 21, 4, 14, 38, 40, 14, 4, 42, 101, 0, 16, 74, 2, 5, 17, 46, 48, 17, 5, 1, 11, 0, 19, 125, 10, 6, 20, 54, 1, 20, 6, 44, 272, 21, 0

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OFFSET

3,2

COMMENTS

Also for n >= 3, a(n) is the least k >= 0 such that the Sum_{i = 0..k} (n + i) is a triangular number (A000217). For n = 0, 1 the Sum is a triangular number for all n. For n = 2, there is no solution.

LINKS

Robert Israel, Table of n, a(n) for n = 3..10000

FORMULA

a(n) = 0 for n from A000217.

a(n) = 1 for n from A074377 AND n is not a triangular number.

EXAMPLE

n = 4: the least k >= 0 such that (k + 1)*(8 + k)/2 is a triangular number is k = 2, thus a(4) = 2.

n = 6: the least k >= 0 such that (k + 1)*(12 + k)/2 is a triangular number is k = 0, thus a(6) = 0.

MAPLE

f:= proc(n) local t, alpha, beta;

t:= n^2-n;

alpha:= convert(select(type, numtheory:-divisors(t), odd), list);