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A255864
Least m > 0 such that gcd(m^n+14, (m+1)^n+14) > 1, or 0 if there is no such m.
2
1, 0, 1, 12, 1, 15, 1, 2, 1, 1929501, 1, 13228907223310811104028677, 1, 94, 1, 11, 1, 85364353, 1, 1563, 1, 49, 1, 9258095644888888790279763522646107297983, 1, 23, 1, 66, 1
OFFSET
0,4
COMMENTS
See A118119, which is the main entry for this class of sequences.
a(29) with 141 decimal digits is too large to include here (see b-file).
LINKS
FORMULA
a(2k) = 1 for k>=0, because gcd(1^(2k)+14, 2^(2k)+14) = gcd(15, 4^k-1) >= 3, since 4^k-1 = 1-1 = 0 (mod 3).
EXAMPLE
For n=1, gcd(m^n+14, (m+1)^n+14) = gcd(m+14, m+15) = 1, therefore a(1)=0.
For n=0 and n=2, see formula with k=0 and k=1.
For n=3, gcd(12^3+14, 13^3+14) = 67, and (m, m+1) = (12, 13) is the smallest pair which yields a GCD > 1 here.
MATHEMATICA
A255864[n_] := Module[{m = 1}, While[GCD[m^n + 14, (m + 1)^n + 14] <= 1, m++]; m]; Join[{1, 0}, Table[A255864[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)
PROG
(PARI) a(n, c=14, 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(40) from Max Alekseyev, Aug 06 2015
STATUS
approved