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

0,3

Comments

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).

Links

Table of n, a(n) for n=0..45.

Mathematica

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

Prog

(C#)
static BigInteger a(int n)
        {
            // The next lines solve the Pell equation x^2 - D y^2 = 1
            int D = 4 * n + 2;
            BigInteger num = 0;
            BigInteger den = 0;
            if (n < 0)
                return 0;
            BigInteger limit = (int)Math.Sqrt(D);
            BigInteger m = 0;
            BigInteger d = 1;
            BigInteger a = limit;
            BigInteger numm1 = 1;
            num = a;
            BigInteger denm1 = 0;
            den = 1;
            while (num * num - D * den * den != 1)
            {
                m = d * a - m;
                d = (D - m * m) / d;
                a = (limit + m) / d;
                BigInteger numm2 = numm1;
                numm1 = num;
                BigInteger denm2 = denm1;
                denm1 = den;
                num = a * numm1 + numm2;
                den = a * denm1 + denm2;
            }
            // The list square is computed
            BigInteger square = (den * den) / 4;
            return square;
        }
(PARI) a(n) = my(k=1); while (!ispolygonal(k^2*(2*n+1), 3), k++); k^2; \\ Michel Marcus, Jun 06 2021
(Python)
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

Crossrefs

Cf. A000217, A000290, A061782.

Sequence in context: A340578 A374225 A199179 * A300072 A062546 A245887

Adjacent sequences: A344946 A344947 A344948 * A344950 A344951 A344952

Keyword

nonn

Author

Mihai Prunescu, Jun 03 2021

Status

approved