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

Author

N. J. A. Sloane, Aug 06 2012

Keywords

• nonn
• easy

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

• StackExchange, A prime number pattern, Jul 29 2012.
• Index entries for linear recurrences with constant coefficients, signature (0,1).

Crossrefs

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

Essentially the same as A135528, A059841, A000035.

Programs

Mathematica

PadRight[{2},120,{0,1}] (* Harvey P. Dale, Nov 29 2018 *)

PARI

if(n%2,2*(n<2),1) \\ Charles R Greathouse IV, Aug 06 2012