A275341 Positions of ones in A275737.

1, 2, 3, 5, 6, 8, 11, 13, 15, 18, 20, 23, 25, 28, 30, 33, 37, 39, 42, 46, 49, 52, 55, 58, 61, 64, 68, 71, 74, 78, 82, 85, 89, 92, 95, 99, 103, 107, 110, 114, 118, 121, 126, 129, 133, 137, 140, 144, 148, 153, 156, 160, 165, 168, 172, 176, 180, 184, 189, 193, 197

Offset: 1

Author

Mats Granvik, Jul 28 2016

Keywords

• nonn

Comments

a(n) appears to be asymptotic to log((n+1)!). The question at MathOverflow discusses a related but more complicated sequence.

Examples

			The sequence A275737: round(im(zetazero(n + 1))/(2*Pi)) - round(im(zetazero(n))/(2*Pi)) where n=1,2,3,4,5, starts:
1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1,...
The positions of ones in that sequence are a(n):
1, 2, 3, 5, 6, 8, 11, 13, 15, 18, 20, 23, 25, 28, 30,...
Compare this to round(log((n+1)!)) A046654:
1, 2, 3, 5, 7, 9, 11, 13, 15, 18, 20, 23, 25, 28, 31,...
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..10000 (terms 1..1000 from G. C. Greubel)
• Mats Granvik, What explains the asymptotic and the pattern in this sequence related to Riemann zeta zeros?.

Crossrefs

Cf. A046654, A275579, A275737.

Programs

Mathematica

Flatten[Position[Differences[Round[Im[ZetaZero[Range[135]]]/(2*Pi)]], 1]]

Formula

a(n) is the positions of ones in round(im(zetazero(n + 1))/(2*Pi)) - round(im(zetazero(n))/(2*Pi)), where n starts at 1.