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A319037 a(n) is the length of the longest run of consecutive triangular numbers with n divisors.

%I A319037 #46 Mar 18 2023 08:08:57
%S A319037 1,1,0,4,0,2,0,3,1,1,0,3,0,1,1,6,0,2,0,3,0,1,0,7,0,1,1,3,0,2,0,13,0,1,
%T A319037 0,6,0,1,0,6,0,2,0,3,1,1,0
%N A319037 a(n) is the length of the longest run of consecutive triangular numbers with n divisors.
%C A319037 Every number with an odd number of divisors is a square, and no two consecutive positive triangular numbers are squares, so for all odd n, a(n) = 1 if a triangular number with n divisors exists, 0 otherwise.
%C A319037 T(190254647) = 18098415447674628 is the smallest triangular number that begins a run of 9 consecutive triangular numbers with 48 divisors. Does a longer run exist? - _Jon E. Schoenfield_, May 29 2022
%C A319037 From _Jon E. Schoenfield_, Mar 17 2023: (Start)
%C A319037 9 <= a(48) <= 16.
%C A319037 a(48) < 17 because a run of 17 consecutive triangular numbers with 48 divisors each would require (see the Example section at A319038) a run of 9 consecutive odd numbers 2k+1, 2k+3, ..., 2k+17 with tau2 divisors each and a run of 9 consecutive integers -- either k, k+1, ..., k+8 or k+1, k+2, ..., k+9 -- with tau1 divisors each, with tau1*tau2 = 48. There exists no run of 8 or more consecutive integers with tau1 < 12 divisors each, nor is there any run of 9 or more consecutive odd numbers with tau1 <= 4 divisors each (tau1 = 4 is impossible because among any 9 or more consecutive odd numbers there is at least one that is a multiple of 9, and the only such multiple with exactly 4 divisors is 3^3 = 27).
%C A319037 Similarly, it can be shown that if a(48) = 16, there must exist a run of 8 consecutive odd numbers 2k+1, 2k+3, ..., 2k+15 with tau2 = 4 divisors each and a run of 9 consecutive integers k, k+1, ..., k+8 with tau1 = 12 divisors each. It seems to me that such solutions should exist (although it could be very difficult to find any). E.g., the 9 consecutive integers could be numbers of the forms 2*13^2*p, 3*5^2*q, 2^5*r, s^2*t*u, 2*3^2*v, 7^2*w*x, 2^2*5*y, 3*z^2*a, and 2*b^2*c, respectively, and the 8 consecutive odd numbers could be of the forms 17*d, 7*e, 3*f, 5*g, 11*h, 3*i, 13*j, and 19*m, respectively, where each letter (other than k) represents a distinct prime. (End)
%e A319037 a(1) = 1 because there is only one triangular number T(k) = k*(k+1)/2 with 1 divisor: T(1) = 1.
%e A319037 a(2) = 1 because there is only one triangular number with 2 divisors: T(2) = 3, the only prime triangular number.
%e A319037 a(3) = 0 because there is no triangular number with 3 divisors.
%e A319037 a(4) = 4 because {6, 10, 15, 21} is the longest run of consecutive triangular numbers with 4 divisors.
%e A319037 a(16) = 6 because T(692993)..T(692998) is a run of 6 consecutive triangular numbers with 16 divisors, and no longer run of such triangular numbers exists.
%e A319037 a(24) = 7 because T(1081135121474335700644)..T(1081135121474335700650) is a run of 7 consecutive triangular numbers with 24 divisors, and no longer run of such triangular numbers exists. - _Jinyuan Wang_, Aug 23 2020
%e A319037 T(1081135121474335700644) is the smallest triangular number that begins a run of 7 consecutive triangular numbers with 24 divisors. - _Jon E. Schoenfield_, May 29 2022
%Y A319037 Cf. A000005, A000217, A081978, A276542, A319035, A319036.
%K A319037 nonn,more
%O A319037 1,4
%A A319037 _Jon E. Schoenfield_, Dec 08 2018
%E A319037 a(21)-a(24) from _Jinyuan Wang_, Aug 23 2020
%E A319037 a(25)-a(47) from _Jon E. Schoenfield_, Feb 04 2021