A378744 The least k such that prime(1+n)^k > 2 * prime(n)^k.

2, 2, 3, 2, 5, 3, 7, 4, 3, 11, 4, 7, 15, 8, 6, 7, 21, 8, 12, 25, 9, 15, 10, 9, 18, 36, 19, 38, 20, 6, 23, 16, 48, 10, 52, 18, 19, 29, 20, 21, 63, 13, 67, 34, 69, 12, 13, 39, 80, 41, 28, 84, 18, 30, 31, 31, 94, 32, 49, 98, 20, 15, 54, 109, 55, 17, 39, 24, 121, 61, 42, 32, 43, 44, 67, 45, 35, 70, 36, 29, 146, 30, 150

Offset: 1

Author

Antti Karttunen, Dec 08 2024

Keywords

• nonn

Comments

A000040(n)^a(n) = A378745(n) is always term of A337372, i.e., is primitively prime-shift abundant.

Examples

			For n=1, prime(1)=2 and prime(2)=3, and 3^1 is not larger than 2*2^1, but 3^2 > 2*2^2, therefore a(1) = 2.
For n=3, prime(3)=5 and prime(4)=7, with 7 < 2*5, 7^2 = 49 < 2*25, and 7^3 = 343 > 2*125, therefore a(3) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000
• Index entries for sequences related to prime indices in the factorization of n.
• Index entries for primes, gaps between.

Crossrefs

Cf. A000040, A001359, A003961, A337372, A341609, A378745 [= A000040(n)^a(n)].

Programs

PARI

A378744(n) = { my(p=prime(n), q=prime(1+n)); for(k=-1+floor(log(2)/log(q/p)), oo, if(q^k > 2*(p^k), return(k))); };

Formula

a(n) = ceiling(log(2) / log(A000040(n+1)/A000040(n))).

For all n >= 1, A341609(A000040(n)^a(n)) = 1.