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A255863
Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.
2
1, 0, 26, 1, 5, 24308100, 1, 329, 71, 1, 6, 59, 1, 135, 5, 1, 23, 7711, 1, 82, 6, 1, 8, 320594291825643656342, 1, 45, 10, 1, 755, 1107, 1, 4279, 30269, 1, 5, 205961, 1, 259, 8, 1, 9, 101975, 1, 6491, 5, 1, 8
OFFSET
0,3
COMMENTS
See A118119, which is the main entry for this class of sequences.
FORMULA
a(3k) = 1 for k>=0, because 1^(3k)+13 = 14, 2^(3k)+13 = 8^k+13 = 14 (mod 7), therefore gcd(1^(3k)+13, 2^(3k)+13) >= 7.
EXAMPLE
For n=1, gcd(m^n+13, (m+1)^n+13) = gcd(m+13, m+14) = 1, therefore a(1)=0.
For n=2, gcd(26^2+13, 27^2+13) = 53, and (m, m+1) = (26, 27) is the smallest pair which yields a GCD > 1 here.
For n=0, n=3, n=6,... see formula.
MATHEMATICA
A255863[n_] := Module[{m = 1}, While[GCD[m^n + 13, (m + 1)^n + 13] <= 1, m++]; m]; Join[{1, 0}, Table[A255863[n], {n, 2, 22}]] (* Robert Price, Oct 16 2018 *)
PROG
(PARI) a(n, c=13, 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 10 2015
EXTENSIONS
a(5)-a(46) from Hiroaki Yamanouchi, Mar 12 2015
STATUS
approved