A255860 Least m > 0 such that gcd(m^n+10, (m+1)^n+10) > 1, or 0 if there is no such m.
1, 0, 20, 3, 2, 3, 320, 874, 6, 33, 1, 124, 465, 23433448460229, 81920, 3, 2, 82, 65, 2101, 1, 3, 3, 2398892314, 7270, 3, 11, 21, 2, 97546469, 1, 765170730, 6, 15, 3, 3, 23, 370460325141871548, 29206018, 3, 1
Offset: 0
Examples
For n=0, gcd(m^0+10, (m+1)^0+10) = gcd(11, 11) = 11 for any m > 0, therefore a(0)=1 is the smallest possible positive value. For n=1, gcd(m^n+10, (m+1)^n+10) = gcd(m+10, m+11) = 1, therefore a(1)=0. For n=2, we have gcd(20^2+10, 21^2+10) = gcd(410, 451) = 41, and the pair (m,m+1)=(20,21) is the smallest which yields a GCD > 1, therefore a(2)=20.
Programs
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Mathematica
A255860[n_] := Module[{m = 1}, While[GCD[m^n + 10, (m + 1)^n + 10] <= 1, m++]; m]; Join[{1, 0}, Table[A255860[n], {n, 2, 12}]] (* Robert Price, Oct 16 2018 *)
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PARI
a(n,c=10,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}
Extensions
a(13)-a(36) from Hiroaki Yamanouchi, Mar 13 2015
a(37)-a(40) from Max Alekseyev, Aug 06 2015
Comments