A358440 a(n) is the largest prime that divides any two successive terms of the sequence b(m) = m^2 + n with m >= 1.

5, 3, 13, 17, 7, 5, 29, 11, 37, 41, 5, 7, 53, 19, 61, 13, 23, 73, 11, 3, 17, 89, 31, 97, 101, 7, 109, 113, 13, 11, 5, 43, 19, 137, 47, 29, 149, 17, 157, 23, 11, 13, 173, 59, 181, 37, 7, 193, 197, 67, 41, 19, 71, 31, 17, 5, 229, 233, 79, 241

Offset

1,1

Comments

Inspired by Project Euler, Problem 659 (see link).

For every sequence b(m) = m^2 + n, m >= 1 and n fixed >= 1, there is a largest prime dividing any two successive terms of this sequence.

Links

Table of n, a(n) for n=1..60.

Project Euler, Problem 659: Largest prime.

Formula

a(n) is the largest prime factor of 4n+1. - Charles R Greathouse IV, Nov 16 2022

3 <= a(n) <= A280818(n) <= 4n+1. Both bounds are sharp; a(n) = 3 if and only if n is of the form (9^n-1)/4, while a(n) = 4n+1 if and only if 4n+1 is prime. - Charles R Greathouse IV, Nov 16 2022

Example

With A117950 (m^2+3, m>=1), the two successive terms A117950(6) = 39 and A117950(7) = 52 are both multiples of 13; the second such pair is A117950(19) = 364 and A117950(20) = 403, ... As 13 is the largest prime that divides two any successive integers of A117950, a(3) = 13.

Mathematica

a[n_] := FactorInteger[4*n + 1][[-1, 1]]; Array[a, 60] (* Amiram Eldar, Nov 16 2022 *)

Prog

(PARI) a(n)=my(f=factor(4*n+1)[, 1]); f[#f] \\ Charles R Greathouse IV, Nov 16 2022
(Python)
from sympy import primefactors
def A358440(n): return max(primefactors((n<<2)+1)) # Chai Wah Wu, Nov 16 2022

Crossrefs

Cf. A051541, A117950.

Sequence in context: A206435 A080797 A376975 * A280818 A085910 A093544

Adjacent sequences: A358437 A358438 A358439 * A358441 A358442 A358443

Keyword

nonn

Author

Bernard Schott, Nov 16 2022

Status

approved