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 A201250 The number of primes <= n^2 can be arranged as a telescoping sum Pi[n^2]-Pi[(n-1)^2] + Pi[(n-1)^2]- Pi[(n-2)^2]+...We divide the sum into alternating sums A and B, with A = (Pi[n^2]-Pi[(n-1)^2]) + (Pi[(n-2)^2]- Pi[(n-3}^2])+..., and B = (Pi[(n-1)^2]-Pi[(n-2)^2])+(Pi[(n-3)^2]-Pi[(n-4)^2])...We form the sequence {A_n - B_n} = C_n, and finally the sequence S_k = {k such that C_k = 0}. 0
 1, 3, 8, 16, 36, 38, 70, 108, 116, 148, 251, 280 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: the sequence is infinite. The graph of A(i)-B(i) provides a visual suggestion but no hint of a proof. In general there is no reason to suppose A_n - B_n = 0. LINKS FORMULA A = Sum[{i=1 to n-1}(-1)^(i+1)*Pi[(n-i+1)^2]; B = Sum[{i=1 to n-1}(-1)^(i+1)*Pi[(n-i)^2]; Sequence is S_n = {index(A_n - B_n) such that A_n - B_n = 0}. EXAMPLE For n = 3, A = [Pi(3^2)-Pi(2^2)]; B = [Pi(2^2)-Pi(1^2)]; A-B = (4-2) - (2 -0) = 2 - 2 = 0. MATHEMATICA A=Sum[(-1)^(i+1)PrimePi[(j-i+1)^2], {i, 1, j-1}]; B=Sum[(-1)^(i+1)PrimePi[(j-i)^2], {i, 1, j-1}]; D = Table[A-B, {j, 2, 100}]; CROSSREFS Sequence in context: A077552 A171497 A024623 * A196373 A027291 A048952 Adjacent sequences:  A201247 A201248 A201249 * A201251 A201252 A201253 KEYWORD nonn,easy AUTHOR Daniel Tisdale, Nov 28 2011 STATUS approved

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Last modified November 22 10:59 EST 2019. Contains 329389 sequences. (Running on oeis4.)