OFFSET
1,1
COMMENTS
Take the first n primes and combine them with coefficients +1 and -1; then a(n) is the smallest number (in absolute value) that can be obtained.
For example, a(1) = 2, a(2) = 1 from 3-2 = 1; a(3) = 0 from -2-3+5 = 0; a(11) = 0 from 2-3-5-7+11-13+17+19-23-29+31 = 0.
Comment from Franklin T. Adams-Watters, Aug 05 2012: Sketch of proof that the above sum of primes results in this sequence. If S_n is the set of possible values of the signed sums for the first n primes, then S_{n+1} = S_n U (S_n + prime(n+1)) U (S_n - prime(n+1)). Beyond about n=4, this will be everything even or everything odd in an interval around zero, and then a fringe on either side; the size of the interval will be 2 * A007504(n) - k for some small k. Recursively, since prime(n) << A007504(n), this will continue to hold. Hence the sequence continues to alternate 0's and 1's. A quite modest estimate on the distribution of primes suffices to complete the proof.
LINKS
StackExchange, A prime number pattern, Jul 29 2012.
Index entries for linear recurrences with constant coefficients, signature (0,1).
MATHEMATICA
PadRight[{2}, 120, {0, 1}] (* Harvey P. Dale, Nov 29 2018 *)
PROG
(PARI) if(n%2, 2*(n<2), 1) \\ Charles R Greathouse IV, Aug 06 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Aug 06 2012
STATUS
approved