A127854 Largest number k such that k^2 divides A007781(6n+1).

19, 61, 127, 217, 331, 469, 631, 817, 1027, 1261, 1519, 1801, 2107, 2437, 2791, 3169, 3571, 3997, 4447, 4921, 5419, 5941, 6487, 7057, 7651, 8269, 8911, 9577, 10267, 10981, 11719, 12481, 13267, 14077, 14911, 15769, 16651, 17557, 18487, 19441

Offset: 1

Author

Alexander Adamchuk, Apr 05 2007

Keywords

• nonn

Comments

A007781(n) = (n+1)^(n+1) - n^n. A007781(6n+1) is not squarefree for n > 0. a(n) is the largest square divisor of A007781(6n+1). All terms belong to A003215 Hex (or centered hexagonal) numbers: 3n(n+1)+1 (crystal ball sequence for hexagonal lattice). It appears that a(n) = A003215(2n) = 6n(2n+1)+1. A007781(6n+1)/A003215(2n)^2 = ((6n+2)^(6n+2)-(6n+1)^(6n+1))/(6n(2n+1)+1)^2 = {44193, 2904899682603, 6378521938392937343349, 128847538453506016002947264859159, 13183819636551142123977274666051092130410345, ...}. Prime terms of a(n) belong to A002407. Factorizations of the terms of a(n) are {19, 61, 127, 7*31, 331, 7*67, 631, 19*43, 13*79, 13*97, 7*7*31, 1801, 7*7*43, 2437, 2791, 3169, 3571, 7*571, 4447, 7*19*37, 5419, 13*457, 13*499, 7067, 7*1093, 8269, 7*19*67, 61*157, 10267, 79*139, ...}. All prime factors of a(n) are of the form 6k+1.

Crossrefs

Cf. A007781 = (n+1)^(n+1) - n^n. Cf. A000312, A068955, A003215, A002407.

Formula

Conjecture: a(n) = 12n^2 + 6n + 1.

Conjecture: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); g.f.: x*(19 + 4*x + x^2)/(1-x)^3. - Colin Barker, Mar 16 2012

These conjectures are false. For n=74, 12*n^2 + 6*n + 1 = 66157 but A007781(6*74+1) is divisible by 5491031^2. - Robert Israel, Nov 19 2017

Extensions

a(24) corrected by T. D. Noe, Mar 14 2008