login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A240747
Least number k > 0 such that n^k - (n-1)^k - ... - 3^k - 2^k - 1 is prime, or 0 if no such k exists.
1
2, 0, 2, 4, 4, 0, 0, 0, 8, 0, 0, 12
OFFSET
2,1
COMMENTS
a(n) is definitely 0 for n == 3 mod 4 or 0 mod 4 (except a(4) = 2). This is the case for n = 3; 7,8; 11,12; 15,16; 19,20; ...
Further, if n is not squarefree, then a(n) = 0. Thus, if n is in A013929, then a(n) = 0. This is the case for n = 8, 9, 12, 16, 18, 20, 24, ...
Let S = n^k - (n-1)^k - ... - 3^k - 2^k - 1. Then S is divisible by gpf(n) when k is not a multiple of gpf(n)-1, where gpf(x) denotes the greatest prime factor of x. This means that if a(n) is not 0, then a(n) must be a multiple of gpf(n)-1. Note that this holds with the previous findings.
For n <= 200, if n = {21, 26, 29, 34, 38, 58, 61, 62, 65, 70, 74, 86, 89, 97, 101, 102, 106, 110, 133, 137, 142, 146, 157, 173, 178, 181, 182, 185, 190}, there is a pattern with the factorization of S when k is a multiple of gpf(n)-1. Thus, a(n) = 0 is definite for these n-values.
For other n-values <= 200, there is not a concrete pattern when k is a multiple of gpf(n)-1. If n = 14, a(n) > 30000, if n = {33, 37, 42, 46, 53, 57, 69, 73, 77, 78, 82, 85, 93, 94}, a(n) > 8000, and if n = {105, 109, 113, 114, 118, 122, 129, 130, 134, 138, 141, 145, 149, 154, 158, 161, 165, 166, 170, 174, 177, 186, 193, 197}, a(n) > 6000. Here, a(n) could still be nonzero.
For n <= 200, it is known that a(17) = 16, a(22) = 3190, a(30) = 124, a(41) = 520, a(66) = 260, and a(194) = 288.
EXAMPLE
5^1 - 4^1 - 3^1 - 2^1 - 1 = -5 is not prime;
5^2 - 4^2 - 3^2 - 2^2 - 1 = -5 is not prime;
5^3 - 4^3 - 3^3 - 2^3 - 1 = 25 is not prime;
5^4 - 4^4 - 3^4 - 2^4 - 1 = 271 is prime. Thus, a(5) = 4.
PROG
(PARI) s(n)=for(k=1, 6000, if(ispseudoprime(n^k-sum(i=1, n-1, i^k)), return(k)))
n=1; while(n<200, print(s(n)); n+=1)
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Derek Orr, Apr 11 2014
STATUS
approved