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A242585 Number of divisors of the n-th positive number that is both triangular and square.

%I A242585 #19 Mar 20 2022 14:18:25
%S A242585 1,9,9,45,9,405,15,189,81,729,27,6075,27,1215,2025,729,81,45927,27,
%T A242585 32805,2025,6561,81,229635,243,2187,2187,18225,9,7381125,243,24057,
%U A242585 2187,19683,3645,6200145,729,19683,19683,1240029,81,22143375,243,295245,492075,19683
%N A242585 Number of divisors of the n-th positive number that is both triangular and square.
%C A242585 Number of divisors of A001110(n).
%C A242585 Since each term in A001110 is a square, each term in this sequence is an odd number.
%C A242585 Adding terms to this sequence requires obtaining the prime factorization of terms from A001110. This is facilitated by the observation that A001110(n) = (A000129(n)*A001333(n))^2, but after a few hundred terms, it becomes difficult to obtain the prime factorization of some of the values of A000129(n) and A001333(n).
%C A242585 Obviously, a(1) is the only term whose value is 1. It can be shown (see A081978) that 15 appears only at a(7).
%C A242585 Conjectures:
%C A242585 (1) There exist only 5 triangular numbers with exactly 9 divisors: the 2nd, 3rd, 5th, 29th, and 59th terms of A001110.
%C A242585 (2) A001110(4)=41616 is the only triangular number with exactly 45 divisors.
%C A242585 (3) A001110(6)=48024900 is the only triangular number with exactly 405 divisors.
%C A242585 (4) A001110(8)=55420693056 is the only triangular number with exactly 189 divisors.
%H A242585 Jon E. Schoenfield, <a href="/A242585/b242585.txt">Table of n, a(n) for n = 1..300</a>
%H A242585 Jon E. Schoenfield, <a href="/A242585/a242585.txt">Table of values of n <= 300 such that n-th square triangular number has exactly k divisors</a>
%F A242585 Conjecture: a(n) == 0 mod 9 for n different from 1 and 7 [tested up to n = 300]. - _Ivan N. Ianakiev_, May 29 2014
%e A242585 a(2) = 9 because A001110(2) = 36 = 2^2 * 3^2 has (2+1)*(2+1) = 9 divisors.
%e A242585 a(4) = 45 because A001110(4) = 41616 = 2^4 * 3^2 * 17^2 has (4+1)*(2+1)*(2+1) divisors.
%e A242585 a(6) = 405 because A001110(6) = 48024900 = 2^2 * 3^4 * 5^2 * 7^2 * 11^2 has (2+1)*(4+1)*(2+1)*(2+1)*(2+1) = 405 divisors.
%o A242585 (Magma) a:=0; b:=1; NumberOfDivisors(b); for n in [2..46 by 2] do a:=34*b-a+2; NumberOfDivisors(a); b:=34*a-b+2; NumberOfDivisors(b); end for;
%Y A242585 Cf. A001110, A000217, A081978.
%K A242585 nonn
%O A242585 1,2
%A A242585 _Jon E. Schoenfield_, May 25 2014