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A215036 2 followed by "1,0" repeated. 2
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 (list; graph; refs; listen; history; text; internal format)
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).

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

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Last modified November 13 13:15 EST 2018. Contains 317149 sequences. (Running on oeis4.)