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A331234
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Triangular numbers having exactly 9 divisors.
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1
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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Intersection of A000217 (triangular numbers) and A030627 (numbers with exactly 9 divisors).
Cf. A063440 (number of divisors of n-th triangular number), A242585 (number of divisors of the n-th positive number that is both triangular and square).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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