A308935 a(n) is the smallest m > n such that n^2*(n^2 + 1) divides m^2*(m^2 + 1).

2, 8, 12, 64, 18, 216, 35, 112, 360, 818, 660, 348, 208, 2744, 693, 4096, 493, 450, 3420, 4832, 1071, 2112, 1242, 13824, 7800, 17576, 1998, 4368, 10133, 1560, 1178, 1280, 3597, 3060, 8582, 46656, 5032, 1292, 29640, 12768, 1189, 14868, 3182, 13112, 36468, 6670

Offset: 1

Author

Rémy Sigrist, Jul 01 2019

Keywords

• nonn

Comments

For any n > 0, a(n) exists as n^2*(n^2+1) divides (n^3)^2*((n^3)^2+1).

Tsz Ho Chan proved that a(n) >> n*log(n)^(1/8)/log(log(n))^12.

Examples

			For n = 2:
- A071253(3) mod A071253(2) = 10,
- A071253(4) mod A071253(2) = 12,
- A071253(5) mod A071253(2) = 10,
- A071253(6) mod A071253(2) = 12,
- A071253(7) mod A071253(2) = 10,
- A071253(8) mod A071253(2) = 0,
- hence a(2) = 8.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..2376
• Tsz Ho Chan, Gaps between divisible terms in a^2*(a^2+1), arXiv:1906.11128 [math.NT], 2019.
• Tsz Ho Chan, The Diophantine equation $b (b+1) (b+2) = t a (a + 1) (a + 2)$ and gap principle, arXiv preprint (2024). arXiv:2408.01306 [math.NT]

Crossrefs

Cf. A065876, A071253.

Programs

Magma

a:=[]; for n in [1..50] do m:=n+1; while not IsIntegral( (m^2*(m^2 + 1))/(n^2*(n^2 + 1) ))  do m:=m+1; end while; Append(~a,m); end for; a; // Marius A. Burtea, Dec 20 2019

Mathematica

a[n_] := With[{n2 = n^2(n^2+1)}, For[m = n+1, True, m++, If[Divisible[ m^2(m^2+1), n2], Print[n, " ", m]; Return[m]]]];
a /@ Range[100] (* Jean-François Alcover, Dec 20 2019 *)

PARI

a(n, f = x->x^2*(x^2+1)) = my (fn=f(n)); for (m=n+1, oo, if (f(m)%fn==0, return (m)))

Python

def A308935(n):
    n2, m, m2 = n**2*(n**2+1), n+1, ((n+1)**2*((n+1)**2+1)) % (n**2*(n**2+1))
    while m2:
        m2, m = (m2 + 2*(2*m+1)*(m**2+m+1)) % n2, (m+1) % n2
    return m # Chai Wah Wu, Jul 01 2019

Formula

a(n) <= n^3.