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A215036 2 followed by "1,0" repeated. 2

%I #28 Nov 29 2018 15:33:12

%S 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,

%T 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,

%U 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

%N 2 followed by "1,0" repeated.

%C 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.

%C 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.

%C 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.

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

%H StackExchange, <a href="http://math.stackexchange.com/questions/176394/a-prime-number-pattern">A prime number pattern</a>, Jul 29 2012.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,1).

%t PadRight[{2},120,{0,1}] (* _Harvey P. Dale_, Nov 29 2018 *)

%o (PARI) if(n%2,2*(n<2),1) \\ _Charles R Greathouse IV_, Aug 06 2012

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

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_, Aug 06 2012

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)