

A255832


Least m > 0 such that gcd(m^(2n+1)+2, (m+1)^(2n+1)+2) > 1.


19



51, 40333, 434, 16, 1234, 78607, 8310, 817172, 473, 116, 22650, 736546059, 22, 1080982, 252, 7809, 644, 1786225573
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OFFSET

1,1


COMMENTS

For n=0 one has gcd(m+2, m+3) = 1 for any m.
See A255852 for the sequence including also even exponents, for which the GCD is > 1 already for m=1 (because gcd(1^2k+2, 2^2k+2) = gcd(3, 2^2k1) = gcd(3, 4^k1) = 3), and also for m=4 (because gcd(4^2k+2, 5^2k+2) = gcd(4^2k+2, (5^k4^k)(5^k+4^k)) >= 3), etc.


LINKS

Table of n, a(n) for n=1..18.


FORMULA

a(n) = A255852(2n+1).


MATHEMATICA

A255832[n_] := Module[{m = 1}, While[GCD[m^(2 n + 1) + 2, (m + 1)^(2 n + 1) + 2] <= 1, m++]; m]; Table[A255832[n], {n, 1, 10}] (* Robert Price, Oct 15 2018 *)


PROG

(PARI) a(n, c=2, L=10^6)={n=n*2+1; for(a=1, L, gcd(a^n+c, (a+1)^n+c)>1&&return(n))}


CROSSREFS

Cf. A118119, A255853, A255853, ... for other variants, corresponding to different constant offsets (+1, +3, ...) in the arguments of gcd.
Sequence in context: A208172 A200798 A241117 * A208198 A208405 A208412
Adjacent sequences: A255829 A255830 A255831 * A255833 A255834 A255835


KEYWORD

nonn,more,hard


AUTHOR

M. F. Hasler, Mar 08 2015


EXTENSIONS

a(12)a(18) from Max Alekseyev, Aug 06 2015


STATUS

approved



