

A201250


The number of primes <= n^2 can be arranged as a telescoping sum Pi[n^2]Pi[(n1)^2] + Pi[(n1)^2] Pi[(n2)^2]+...We divide the sum into alternating sums A and B, with A = (Pi[n^2]Pi[(n1)^2]) + (Pi[(n2)^2] Pi[(n3}^2])+..., and B = (Pi[(n1)^2]Pi[(n2)^2])+(Pi[(n3)^2]Pi[(n4)^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
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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

Table of n, a(n) for n=1..12.


FORMULA

A = Sum[{i=1 to n1}(1)^(i+1)*Pi[(ni+1)^2];
B = Sum[{i=1 to n1}(1)^(i+1)*Pi[(ni)^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)];
AB = (42)  (2 0) = 2  2 = 0.


MATHEMATICA

A=Sum[(1)^(i+1)PrimePi[(ji+1)^2], {i, 1, j1}]; B=Sum[(1)^(i+1)PrimePi[(ji)^2], {i, 1, j1}]; D = Table[AB, {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



