|
| |
|
|
A242927
|
|
Numbers n such that k^n + (k+1)^n + ... + (k+n-1)^n is prime for some k.
|
|
2
|
| |
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
a(5) > 500. For n-values < 500 not listed above, k has been checked for k <= 5000.
For the first four terms, the least k that makes k^n + (k+1)^n + ... + (k+n-1)^n 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
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
k^1 = k is prime for k = 2 or any other prime (cf. A000040), so 1 is a member 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 member 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 n = 4 and n = 5 is a multiple of 2 and 5, respectively, and for all n = 7, ..., 41, the expression also shares a factor with n (and thus is a multiple of n whenever n is prime).
Index n = 110 is the smallest n > 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 n = 210 is the next one which is a similar case.
Index n = 231 is much like n = 110, but with a factor 7 instead of 5.
Index n = 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.
|
|
|
PROG
|
(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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,hard,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
a(5) from Don Reble, Mar 23 2018
|
|
|
STATUS
|
approved
|
| |
|
|
