OFFSET
0,1
COMMENTS
a(n) = 0 iff n is prime.
a(n) = 2 only for n=0; the only possible sums for k=2 are n+(n+2) = 2n+2, divisible by 2, and n+(n+1)+(n+2) = 3n+3, divisible by 3.
There are infinitely many 1's in the sequence; if p > 5 is a prime == 1 (mod 4), a((p-1)/2) = 1.
Conjecture: every nonnegative integer except 2 occurs infinitely often in the sequence.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..100000
EXAMPLE
If n is prime, sum({n}) is prime, so we can take k = 0, whence n+1..n+0 is empty, so a(n) = 0.
6 is not prime, but 6+7 = 13 is prime, so a(6) = 1.
4 is not prime, and 4+5 is not prime, but 4+7 = 11 and 4+6+7 = 17 are prime; either of these suffices to make a(4) = 3.
MAPLE
b:= (n, i, t)-> isprime(n) or t>0 and
(b(n, i+1, t-1) or b(n+i, i+1, t-1)):
a:= proc(n) local k;
for k from 0 while not b(n, n+1, k) do od; k
end:
seq(a(n), n=0..100); # Alois P. Heinz, Feb 07 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Franklin T. Adams-Watters, Feb 03 2014
STATUS
approved