A234800 First occurrence of n in A234323: Number of nontrivial zeros of the Riemann zeta function in the interval 1/2 + i[n,n+1).

1, 14, 111, 5826, 85865, 4580009, 290820868

Offset: 0

Author

Simon Plouffe and Robert G. Wilson v, Dec 30 2013

Keywords

• nonn
• hard
• more

Comments

k

0: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, ..., A120401

1: 14, 21, 25, 30, 32, 37, 40, 43, 48, 49, 52, 56, 59, 60, 65, ..., A234802

2: 111, 150, 169, 224, 231, 329, 357, 373, 415, 478, 493, 540, ..., A234803

3: 5826, 5978, 6494, 7563, 8106, 8942, 9601, 9856, 9976, 10000, ..., A234804

4: 85865, 193997, 245986, 276125, 283519, 297624, 298486, 311014, ..., A234805

5: 4580009, 7149902, 8120618, 10002309, 10597386, 11333337, 11432756, ..., A234806

6: 290820868, 317905108, 334924359, 386701579, 410462993, 430633085, ..., A234807

The occurrence of <0> is probably finite since the average height of a(n) is A234323 is log(n)/(2*Pi). This is a conjecture.

Links

• Simon Plouffe, Table for A234323(n) for n = 1..1000000
• LMFDB David Platt's Table for A234323(n) of the first 103 billion zeros.
• Simon Plouffe, Table for A234323(n) up to 2 billion - note the file is 25 gigabytes uncompressed
• Wikipedia, Riemann zeta function zeros

Crossrefs

Cf. A234323, A002410, A122526.

Programs

Mathematica

t = Table[0, {100}]; k = 1; cnt = 1; a = 0; t[[1]] = 14; While[k < 1000001, b = Floor[ Im[ N[ ZetaZero[ k]] ]]; If[b == a, cnt++; If[t[[cnt]] == 0, t[[cnt]] = b; Print[{cnt, b}]], cnt = 1]; a = b; k++]