A215036 2 followed by "1,0" repeated.

2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

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.

For number of solutions see A022894, A113040; also A083309.

Links

Table of n, a(n) for n=1..94.

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

Cf. A007504, A214912, A215029, A215030; A022894, A083309, A113040.

Sequence in context: A080733 A080732 A301295 * A294448 A088568 A317161

Adjacent sequences: A215033 A215034 A215035 * A215037 A215038 A215039

Keyword

nonn,easy

Author

N. J. A. Sloane, Aug 06 2012

Status

approved