A182908 Rank of 2^n when all prime powers (A246655) p^n, for n>=1, are jointly ranked.
1, 3, 6, 10, 18, 27, 44, 70, 117, 198, 340, 604, 1078, 1961, 3590, 6635, 12370, 23150, 43579, 82267, 155921, 296347, 564688, 1078555, 2064589, 3958999, 7605134, 14632960, 28195586, 54403835, 105102701, 203287169, 393625231, 762951922, 1480223716, 2874422303
Offset: 1
Keywords
Examples
a(3)=6 because 2^3 has rank 6 in the sequence (2,3,4,5,7,8,9,...).
Links
- Ray Chandler, Table of n, a(n) for n = 1..92 (using b-file file from A007053)
Programs
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Mathematica
T[i_,j_]:=Sum[Floor[j*Log[Prime[i]]/Log[Prime[h]]],{h,1,PrimePi[Prime[i]^j]}]; Flatten[Table[T[i,j],{i,1,1},{j,1,22}]] f[n_] := Sum[ PrimePi[ Floor[2^(n/k)]], {k, n + 1}]; Array[f, 34] (* Robert G. Wilson v, Jul 08 2011 *) -
Python
from sympy import primepi, integer_nthroot def A182908(n): x = 1<
Chai Wah Wu, Nov 05 2024
Formula
a(n) = Sum_{i=1..n} pi(floor(2^(n/i))), where pi(n) = A000720(n). - Ridouane Oudra, Oct 26 2020
a(n) = A025528(2^n). - Pontus von Brömssen, Sep 27 2024
Extensions
Minor edits by Ray Chandler, Aug 20 2021