|
%I #24 Feb 18 2021 00:49:37
%S 1,0,3,4,7,9,13,15,21,27,29,37,43,47,52,61,71,74,83,89,94,105,111,123,
%T 138,145,149,158,161,168,196,208,220,226,246,248,261,276,287,299,316,
%U 319,340,345,358,364,392,422,432,436,447,464,470,496,512,530,544,549
%N Indices of prime(n)^3 in A247834, or a(n)=0 if prime(n)^3 is not in A247834.
%C Conjecture: all a(n)>0, except for n=2.
%F If prime(n)^3 is in A247834, then a(n) = pi(prime(n)^(3/2)) - n + 1, where pi(x) is the prime counting function (A000720).
%e Using the formula, let us find the position in A247834, in which should be 17^3, if 17^3 belongs to A247834. Since 17 = prime(7), then we have a(7) = pi(17^(3/2)) - 6 = pi(70) - 6 = 13. Indeed, A247834(13) = 4913 = 17^3.
%o (PARI) a(n) = my(p=prime(n)); if(nextprime(ceil(p*sqrt(p))) > p*sqrt(prime(n+1)), 0, primepi(prime(n)^(3/2)) - n + 1); \\ _Jinyuan Wang_, Feb 17 2021
%Y Cf. A030078, A156759, A247393, A247394, A247395, A247396, A247509, A247510, A247511, A247606, A247834, A247867.
%K nonn
%O 1,3
%A _Vladimir Shevelev_, Sep 24 2014
%E More terms from _Jinyuan Wang_, Feb 17 2021
|
