A063440 - OEIS

A063440

Number of divisors of n-th triangular number.

1, 2, 4, 4, 4, 4, 6, 9, 6, 4, 8, 8, 4, 8, 16, 8, 6, 6, 8, 16, 8, 4, 12, 18, 6, 8, 16, 8, 8, 8, 10, 20, 8, 8, 24, 12, 4, 8, 24, 12, 8, 8, 8, 24, 12, 4, 16, 24, 9, 12, 16, 8, 8, 16, 24, 24, 8, 4, 16, 16, 4, 12, 36, 24, 16, 8, 8, 16, 16, 8, 18, 18, 4, 12, 24, 16, 16, 8, 16, 40, 10, 4, 16

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OFFSET

1,2

COMMENTS

a(n) = 4 iff either n is in A005383 or n/2 is in A005384.

a(n) is odd iff n is in A001108.

a(n) = 6 if either n = 18 or n = q^2 where q is in A048161 or n = 2 q^2 - 1 where q is in A106483. - Robert Israel, Oct 26 2015

From Bernard Schott, Aug 29 2020: (Start)

a(n-1) is the number of solutions in positive integers (x, y, z) to the simultaneous equations (x + y - z = n, x^2 + y^2 - z^2 = n) for n > 1. See the British Mathematical Olympiad link. In this case, one always has z > x and z > y.

For n = 12 as in the Olympiad problem, the a(11) = 8 solutions are (13,78,79), (14,45,47), (15,34,37), (18,23,29), (23,18,29), (34,15,37), (45,14,47), (78,13,79). (End)

REFERENCES

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 2 of the British Mathematical Olympiad 2007, page 28.

LINKS

Ray Chandler, Table of n, a(n) for n = 1..10000

British Mathematical Olympiad 2007/2008, Round 1, Problem 2.

Index to sequences related to Olympiads.

FORMULA

a(n) = A000005(A000217(n)).

From Robert Israel, Oct 26 2015: (Start)

a(2k) = A000005(k)*A000005(2k+1).

a(2k+1) = A000005(2k+1)*A000005(k+1).

gcd(a(2k), a(2k+1)) = A000005(2k+1) * A060778(k). (End)

EXAMPLE

a(6) = 4 since 1+2+3+4+5+6 = 21 has four divisors {1,3,7,21}.