A247835 Indices of prime(n)^3 in A247834, or a(n)=0 if prime(n)^3 is not in A247834.

1, 0, 3, 4, 7, 9, 13, 15, 21, 27, 29, 37, 43, 47, 52, 61, 71, 74, 83, 89, 94, 105, 111, 123, 138, 145, 149, 158, 161, 168, 196, 208, 220, 226, 246, 248, 261, 276, 287, 299, 316, 319, 340, 345, 358, 364, 392, 422, 432, 436, 447, 464, 470, 496, 512, 530, 544, 549

Offset: 1

Author

Vladimir Shevelev, Sep 24 2014

Keywords

• nonn

Comments

Conjecture: all a(n)>0, except for n=2.

Examples

			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.
		

Crossrefs

Cf. A030078, A156759, A247393, A247394, A247395, A247396, A247509, A247510, A247511, A247606, A247834, A247867.

Programs

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

Formula

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).

Extensions

More terms from Jinyuan Wang, Feb 17 2021