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

Offset: 1

Author

Henry Bottomley, Jul 24 2001

Keywords

• nonn
• easy

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)

Examples

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

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.

Crossrefs

Cf. A000005, A000217.

Cf. A001108, A005383, A005384, A048161, A060778, A081978 (greedy inverse), A106483, A101755 (indices of records), A101756 (records).

Programs

Maple

seq(numtheory:-tau(n*(n+1)/2), n=1..100); # Robert Israel, Oct 26 2015

Mathematica

DivisorSigma[0,#]&/@Accumulate[Range[90]] (* Harvey P. Dale, Apr 15 2019 *)

PARI

for (n=1, 10000, write("b063440.txt", n, " ", numdiv(n*(n + 1)/2)) ) \\ Harry J. Smith, Aug 21 2009

PARI

a(n)=factorback(apply(numdiv,if(n%2,[n,(n+1)/2],[n/2,n+1]))) \\ Charles R Greathouse IV, Dec 27 2014

PARI

vector(100, n, numdiv(n*(n+1)/2)) \\ Altug Alkan, Oct 26 2015

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)