cp's OEIS Frontend

A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.

%I A364554 #27 Sep 01 2023 21:13:18
%S A364554 1,2,3,1,3,2,2,1,3,1,2,2,1,2,4,1,0,2,1,1,4,2,2,2,1,2,2,0,1,3,1,0,4,1,
%T A364554 3,2,2,1,3,2,1,2,0,1,2,0,1,2,1,1,4,1,1,3,1,1,4,1,1,3,1,0,2,1,2,1,0,2,
%U A364554 2,2,0,0,1,1,4,1,1,2,1,0,2,1,2,2,2,2,4,0,1,4
%N A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.
%C A364554 Implementing a suggestion in the comment section of sequences A154296, ..., A154304, this sequence computes the number of primes of the form T(k)/n.
%C A364554 Equivalently, number of primes p such that 8*n*p+1 is a perfect square.
%C A364554 Let's consider, for all primes p, the set of linear recurrences {b(m)} defined as follows:
%C A364554 If p = 2, then {b(m)} = A074378 (numbers of the form x*(4*x -+ 1)); otherwise, b(m) = b(m-1) + 2*b(m-2) - 2*b(m-3) - b(m-4) + b(m-5) with initial terms b(0) = 0, b(1) = (p-1)/2, b(2) = (p+1)/2, b(3) = 2*p-1 and b(4) = 2*p+1. Numbers of the form x*(p*x -+ 1)/2.
%C A364554 Then a(n) = number of sequences {b(m)} in which n is a term.
%C A364554 This implies that:
%C A364554 i) for any n, the largest prime of the form T(k)/n is at most 2*n+1;
%C A364554 ii) if n is prime, then a(n) < 4. (3 and 5 are the only primes p such that a(p) = 3; primes p such that a(p) = 0 are A109998.)
%C A364554 Deeper in the examination of these results, we notice that the set of primes p of the form T(k)/n arises from the factorization of n. This set is exactly all primes p of the form (2*r -+ 1)/d or (r -+ 1)/(2*d), where d is some divisor of n and r is the ratio n/d. (Proof is welcome.)
%C A364554 Indices k of corresponding triangular numbers T(k) such that T(k) = n*p are then:
%C A364554   2*r     if p = (2*r + 1)/d,
%C A364554   2*r - 1 if p = (2*r - 1)/d,
%C A364554     r     if p = (r + 1)/(2*d),
%C A364554     r - 1 if p = (r - 1)/(2*d).
%C A364554 And pluging the value of p in the equivalent definition, the expression 8*n*p+1 yields respectively to following perfect squares: (4*r+1)^2, (4*r-1)^2, (2*r+1)^2 and (2*r-1)^2.
%F A364554 Conjecture: a(n) = number of primes in the union of sets {(2*r -+ 1)/d; (r -+ 1)/(2*d)}, with d divisor of n and r = n/d.
%e A364554 a(15) = 4 since there are exactly 4 triangular numbers T(k) such that T(k) = 15*p, with p prime.
%e A364554 T(9)/15 = 45/15 = 3, T(14)/15 = 105/15 = 7, T(29)/15 = 435/15 = 29 and T(30)/15 = 465/15 = 31.
%e A364554 a(17) = 0 since there is no triangular number T(k) such that T(k) = 17*p, with p prime.
%Y A364554 Cf. A000217, A154296, ..., A154304, A074378, A001318, A057569, A057570, A109998.
%Y A364554 Cf. A364555 (indices of 0's).
%K A364554 nonn
%O A364554 1,2
%A A364554 _Lamine Ngom_, Jul 28 2023