OFFSET
1,1
COMMENTS
Sorting by k/2^bigomega(k) would give the same sequence.
It appears that this sequence can be used to approximate the imaginary parts of the nontrivial zeta zeros, that is, A002410(n) is roughly equal to 2*Pi*a(n)/2^bigomega(a(n)) - n/2 + sqrt(n)/2.
Calculations show that the relative error approaches 1.0+-0.005 for the first 3800 zeros (z=2000 in Mathematica code). For further zeros, a better approximation may be useful, e.g. 2*Pi*a(n)/2^bigomega(a(n)) - n/2 + (1/Pi) * n/log(n+1) +- (...).
LINKS
Friedjof Tellkamp, Table of n, a(n) for n = 1..20000
Friedjof Tellkamp, Plots showing approximation and exact values for the first 3800 zeros
FORMULA
A374022(a(n)/2^(bigomega(a(n))-1)) ~ n.
EXAMPLE
The odd composite numbers (A071904) are: 9, 15, 21, 25, 27, ... .
Divide by 2^(bigomega()-1): 9/2, 15/2, 21/2, 25/2, 27/4, ... .
Sort: 9/2, 27/4, 15/2, 81/8, ... .
Take numerator: this sequence = 9, 27, 15, 81, ... .
MATHEMATICA
(*This algorithm calculates and sorts all noninteger rationals of the form x = k/2^(bigomega(k)-1) up to z, where the numerators of x are returned as the sequence*)
z = 100; op = Prime[Range[2, PrimePi[z]]]; (*A065091, odd primes*)
x = Select[Union[Flatten[Outer[Times, op, op/2]]], # <= z &];
For[i = 1, i < Max[1, Floor[Log[3/2, z/2]]] - 1, i++, x = Select[Union[x, Flatten[Outer[Times, x, op/2]]], # <= z &]]
a = Numerator[x] (*sequence*)
zzaprx = N@Table[2 Pi a[[i]]/2^PrimeOmega[a[[i]]] - i/2 + Sqrt[i]/2, {i, 1, Length[a]}] (*approximation for zeta zeros*)
CROSSREFS
KEYWORD
nonn
AUTHOR
Friedjof Tellkamp, Jun 27 2024
STATUS
approved