

A306431


Least number x > 1 such that n*x divides 1 + Sum_{k=1..x1} k^(x1).


1



2, 3, 13, 7, 19, 31, 41, 31, 13, 19, 43, 31, 23, 83, 139, 31, 61, 67, 113, 79, 251, 43, 19, 31, 199, 23, 13, 167, 53, 139, 83, 127, 157, 67, 293, 431, 443, 151, 103, 79, 61, 251, 113, 47, 337, 19, 179, 31, 41, 199, 67, 23, 19, 499, 181, 367, 607, 139, 257, 359
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OFFSET

1,1


COMMENTS

If n = 1, all the solutions of x  1 + Sum_{k=1..x1} k^(x1) should be prime numbers, according to Giuga's conjecture.
If n*x  1 + Sum_{k=1..x1} k^(x1), then certainly x does, so Giuga's conjecture would say x must be prime. Similarly if x^n divides it, so does x, so again Giuga would say x is prime.  Robert Israel, Apr 26 2019
E.g., the first solution for x^2  1 + Sum_{k=1..x1} k^(x1) is x = 1277, that is prime.


LINKS

Robert Israel, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Giuga's Conjecture


FORMULA

Least solution of n*x  1 + Sum_{k=1..x1} k^(x1), for n = 1, 2, 3, ...


EXAMPLE

a(4) = 7 because (1 + 1^6 + 2^6 + 3^6 + 4^6 + 5^6 + 6^6) / (4*7) = 67172 / 28 = 2399 and it is the least prime to have this property.


MAPLE

P:=proc(j) local k, n; for n from 2 to 10^6 do
if frac((add(k^(n1), k=1..n1)+1)/(j*n))=0
then RETURN(n); break; fi; od; end: seq(P(i), i=1..60);


PROG

(PARI) a(n) = my(x=2); while (((1 + sum(k=1, x1, k^(x1))) % (n*x)), x++); x; \\ Michel Marcus, Apr 27 2019


CROSSREFS

Cf. A191677. All the solutions for n = m: A000040 (m=1), A002145 (m=2), A007522 (m=4), A127576 (m=8), A141887 (m=10), A127578 (m=16), A142198 (m=20), A127579 (m=32), A095995 (m=50).
Sequence in context: A128369 A087568 A087564 * A192362 A057776 A110362
Adjacent sequences: A306428 A306429 A306430 * A306432 A306433 A306434


KEYWORD

nonn,easy


AUTHOR

Paolo P. Lava, Apr 05 2019


STATUS

approved



