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A306615 Least positive k such that 2n - p is prime where p is some prime divisor of n^k - 1, or 0 if no such k exists. 0
0, 0, 0, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 6, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 4, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 3, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 10, 3, 3, 2, 1, 1, 2, 2, 1, 1, 1, 2, 10, 1, 1, 2, 1, 3, 2, 1, 6, 2, 2, 1, 1, 1, 1, 1, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,5
COMMENTS
Conjecture: a(n) >= 1 for n >= 4.
Records: a(4) = 1, a(5) = 2, a(19) = 6, a(62) = 10, a(166) = 18, ...
For n >= 4, a(n) < b(n) where b(n) is the smallest m > 1 such that q(2n - q) is some semiprime divisor of n^m - 1, or 0 if no such m exists: 0, 0, 0, 2, 6, 2, 10, 4, 6, 6, 6, 2, 11, 22, 22, 7, 4, 2, 30, 35, 18, 30, 20, 42, 9, 40, 8, 13, 26, 2, 42, 12, 20, 10, 52, 21, 3, 36, 42, 11, 26, 2, 24, 82, 21, 12, 44, 88, 39, 8, 32, 25, 88, 24, 30, 25, 20, 96, 88, 2, 54, 220, 48, 6, ... (from Goldbach's problem).
LINKS
EXAMPLE
a(4) = 1 because 4^1 - 1 = 3 where 3 is some prime divisor of 3 and 2*4 - 3 = 5 is prime;
a(5) = 2 because 5^2 - 1 = 24 where 3 is some prime divisor of 24 and 2*5 - 3 = 7 is prime.
MATHEMATICA
Table[If[n < 4, 0, Block[{k = 1}, While[NoneTrue[FactorInteger[n^k - 1][[All, 1]], PrimeQ[2 n - #] &], k++]; k]], {n, 104}] (* Michael De Vlieger, Mar 11 2019 *)
PROG
(PARI) isok(n, k) = {my(pf=factor(n^k-1, 2*n)[, 1]); for (j=1, #pf, if (isprime(2*n-pf[j]), return (1)); ); }
a(n) = {if (n < 4, return(0)); my(k=1); while (!isok(n, k), k++); k; } \\ Michel Marcus, Mar 02 2019
CROSSREFS
Cf. A306261.
Sequence in context: A218622 A337621 A091243 * A037826 A079882 A362415
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified April 24 00:30 EDT 2024. Contains 371917 sequences. (Running on oeis4.)