OFFSET
1,1
COMMENTS
Any number having an odd number of divisors is a square, so each term in this sequence is a term of A001110 (numbers that are both triangular and square). Since A001110(k) = (A000129(k)*A001333(k))^2, A001110(k) will have exactly 9 divisors iff A000129(k) and A001333(k) are both prime (i.e., k is in both A096650 and A099088); the first 5 values of k at which this occurs are 2, 3, 5, 29, and 59.
Conjecture: a(5) is the final term of this sequence.
EXAMPLE
Writing the k-th triangular number A000217(k) as T(k):
a(1) = T(8) = 8*9/2 = 36 = 2^2 * 3^2;
a(2) = T(49) = 49*50/2 = 1225 = 5^2 * 7^2;
a(3) = T(1681) = 1681*1682/2 = 1413721 = 29^2 * 41^2.
Factorization of larger known terms:
a(4) = 44560482149^2 * 63018038201^2;
a(5) = 13558774610046711780701^2 * 19175002942688032928599^2.
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 12 2020
STATUS
approved