Mihai Prunescu - cp's OEIS frontend

User: Mihai Prunescu

Mihai Prunescu has authored 1 sequences.

A344949 a(n) is the smallest square s > 0 such that s*(2n+1) is a triangular number.

1, 1, 9, 4, 4, 441, 25, 1, 9, 9, 1, 3218436, 49, 1089, 1656369, 16, 16, 225, 46225, 9, 81, 314721, 1, 12217323024, 25, 25, 2427192623025, 1, 2304, 199572129, 121, 400, 81225, 39727809, 4, 36, 36, 4, 736164, 94864, 592900, 4357032433168041, 169, 3025, 3600, 1

Offset: 0

Mihai Prunescu, Jun 03 2021

nonn

Proof that every odd natural number 2n+1 is a triangular number divided by a square. As the number 4n + 2 is never a square, Pell's equation x^2 - (4n+2)*y^2 = 1 has solutions in integers with y != 0 for every n. It is immediate that x has to be odd. We replace x = 2b+1 and we observe that y must be then even. We replace y = 2a and it follows that b(b+1)/2 = (2n+1)*a^2. So (2n+1) is a triangular number divided by a square. Of course, for given n, there are infinitely many such pairs (b,a).

Cf. A000217, A000290, A061782.

Table[k=1;While[!IntegerQ[Sqrt[8k^2(2n+1)+1]],k++];k^2,{n,0,22}] (* Giorgos Kalogeropoulos, Jun 03 2021 *)

a(n) = my(k=1); while (!ispolygonal(k^2*(2*n+1), 3), k++); k^2; \\ Michel Marcus, Jun 06 2021

from sympy.solvers.diophantine.diophantine import diop_DN
def A344949(n): return min(d[1]**2 for d in diop_DN(4*n+2, 1))//4 # Chai Wah Wu, Jun 21 2021

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User: Mihai Prunescu

A344949 a(n) is the smallest square s > 0 such that s*(2n+1) is a triangular number.

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