A255861 Least m > 0 such that gcd(m^n+11, (m+1)^n+11) > 1, or 0 if there is no such m.
1, 0, 1, 2, 1, 23, 1, 19010820161, 1, 7, 1, 360, 1, 41953103, 1, 4, 1, 638386957517954762853, 1, 38884, 1, 2, 1, 2852, 1, 23, 1, 102, 1, 8384, 1, 36556, 1, 33, 1, 37, 1, 336, 1, 2, 1, 1123, 1, 19734, 1, 9, 1, 135356, 1, 399351, 1, 33, 1
Offset: 0
Keywords
Examples
For n=1, gcd(m^n+11, (m+1)^n+11) = gcd(m+11, m+12) = 1, therefore a(1)=0. For n=2, we have gcd(2^2+11, 3^2+11) = gcd(15, 20) = 5, and the pair (m,m+1)=(2,3) is the smallest which yields a GCD > 1, therefore a(2)=2.
Programs
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Mathematica
A255861[n_] := Module[{m = 1}, While[GCD[m^n + 11, (m + 1)^n + 11] <= 1, m++]; m]; Join[{1, 0}, Table[A255861[n], {n, 2, 6}]] (* Robert Price, Oct 16 2018 *)
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PARI
a(n,c=11,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}
Formula
a(2k)=1 for k>=0, because gcd(1^(2k)+11, 2^(2k)+11) = gcd(12, 4^k-1) = 3.
Extensions
a(7)-a(48) from Hiroaki Yamanouchi, Mar 12 2015
a(49)-a(52) from Max Alekseyev, Aug 06 2015
Comments