A242927 Numbers m such that k^m + (k+1)^m + ... + (k+m-1)^m is prime for some k.

1, 2, 6, 42, 1806

Offset: 1

Author

Derek Orr, May 26 2014

Keywords

• nonn
• hard
• more

Comments

a(5) > 500. For m values < 500 not listed above, k has been checked for k <= 5000.

For the first four terms, the least k that makes k^m + (k+1)^m + ... + (k+m-1)^m prime is {2, 1, 4, 99} respectively.

For a(5) = 1806, k = 3081 yields a strong PRP with 6663 digits. - Don Reble, Mar 23 2018

The known terms a(1..5) coincide with the finite sequence A014117. - M. F. Hasler, May 20 2019

Examples

			k^1 = k is prime for k = 2 or any other prime (cf. A000040), so 1 is a term of this sequence.
k^2 + (k+1)^2 is prime for some k (e.g., k = 2 yields 13, see A027861 for the full list), so 2 is a term of this sequence.
k^3 + (k+1)^3 + (k+2)^3 = 3*(k+1)*(k^2+2*k+3) is never prime, therefore 3 is not a term of this sequence.
Similarly, the corresponding expression for m = 4 and m = 5 is a multiple of 2 and 5, respectively, and for all m = 7, ..., 41, the expression also shares a factor with m (and thus is a multiple of m whenever m is prime).
Index m = 110 is the smallest m > 42 for which the expression is not algebraically composite (the polynomial in k has content 1 and is irreducible over Q), but it does factor as (k(k+1)(k+2)(k+3)(k+4))^10 over Z_5, so is always a multiple of 5. Index m = 210 is the next one which is a similar case.
Index m = 231 is much like m = 110, but with a factor 7 instead of 5.
Index m = 330 again yields an irreducible polynomial with content 1, but as before one can show that it is always divisible by 5. And so on.
		

Crossrefs

Cf. A000040, A027861, A014117.

Programs

PARI

k(n)=for(k=1,5000,if(ispseudoprime(sum(i=0,n-1,(k+i)^n)),return(k)))
for(n=1,500,if(k(n),print(n)))  \\ Edited by M. F. Hasler, Mar 23 2018

Extensions

a(5) from Don Reble, Mar 23 2018

Example corrected and extended by M. F. Hasler, Apr 05 2018