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URL: https://oeis.org/A255866

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A255866
Least m > 0 such that gcd(m^n+16, (m+1)^n+16) > 1, or 0 if there is no such m.
2
1, 0, 2, 22, 128, 12, 2, 81, 1, 5982, 2, 11417025, 32768, 70471611388086, 2, 26, 1, 1019, 2, 12168420936538713481747, 48, 128, 2, 788, 1, 131711329, 2, 91, 13, 2920553219286322570768516629247, 2, 237, 1, 22, 2, 108, 27, 9404578, 2, 2859801, 1, 41772125, 2
OFFSET
0,3
COMMENTS
See A118119, which is the main entry for this class of sequences.
FORMULA
a(4k+2) = 2 for k>=0, because 2^(4k+2) = 4^(2k+1), 3^(4k+2) = 9^(2k+1), and 4 = 9 = -1 (mod 5), therefore gcd(2^(4k+2)+16, 3^(4k+2)+16) >= 5.
EXAMPLE
For n=0, gcd(m^0+16, (m+1)^0+16) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+16, (m+1)^n+16) = gcd(m+15, m+16) = 1, therefore a(1)=0.
For n=2, see formula with k=0.
For n=3, gcd(22^3+16, 23^3+16) = 31 and (m, m+1) = (22, 23) is the smallest pair which yields a GCD > 1 here.
MATHEMATICA
A255866[n_] := Module[{m = 1}, While[GCD[m^n + 16, (m + 1)^n + 16] <= 1, m++]; m]; Join[{1, 0}, Table[A255866[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)
PROG
(PARI) a(n, c=16, L=10^7, S=1)={n!=1 && for(a=S, L, gcd(a^n+c, (a+1)^n+c)>1 && return(a))}
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
M. F. Hasler, Mar 09 2015
EXTENSIONS
a(11)-a(42) from Max Alekseyev, Aug 06 2015
STATUS
approved