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A255862 Least m > 0 such that gcd(m^n+12, (m+1)^n+12) > 1, or 0 if there is no such m. 2
1, 0, 3, 1926, 96, 6, 2, 26, 3, 320, 538, 27, 1, 145, 3, 6, 393216, 982, 3, 2557, 3, 2, 30, 18781248, 1, 6, 3, 188, 14, 145, 3, 2808, 3, 16, 24340653915, 6, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
See A118119, which is the main entry for this class of sequences.
LINKS
FORMULA
a(6k+2) = 3 for k>=0, because 3^(6k+2) = 9^(3k+1), 4^(6k+2) = 16^(3k+1), and 9 = 16 = 2 (mod 7), 2^3 = 1 (mod 7) and 12 = -2 (mod 7), therefore 3^(6k+2)+12 = 4^(6k+2)+12 = 0 (mod 7) and gcd(3^(6k+2)+12, 4^(6k+2)+12) >= 7.
EXAMPLE
For n=0, gcd(m^0+12, (m+1)^0+12) = gcd(13, 13) = 13, therefore a(1)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+12, (m+1)^n+12) = gcd(m+12, m+13) = 1, therefore a(1)=0.
For n=2, see formula with k=0.
MATHEMATICA
A255862[n_] := Module[{m = 1}, While[GCD[m^n + 12, (m + 1)^n + 12] <= 1, m++]; m]; Join[{1, 0}, Table[A255862[n], {n, 2, 22}]] (* Robert Price, Oct 16 2018 *)
PROG
(PARI) a(n, c=12, L=10^7, S=1)={n!=1 && for(a=S, L, gcd(a^n+c, (a+1)^n+c)>1 && return(a))}
CROSSREFS
Sequence in context: A367542 A172940 A096730 * A367941 A236481 A307928
KEYWORD
nonn,hard,more
AUTHOR
M. F. Hasler, Mar 09 2015
EXTENSIONS
a(23)-a(33) from Hiroaki Yamanouchi, Mar 13 2015
a(34)-a(36) from Max Alekseyev, Aug 07 2015
STATUS
approved

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Last modified March 19 01:34 EDT 2024. Contains 370952 sequences. (Running on oeis4.)