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Rank of 2^n when all prime powers (A246655) p^n, for n>=1, are jointly ranked.
7

%I #31 Nov 05 2024 20:39:55

%S 1,3,6,10,18,27,44,70,117,198,340,604,1078,1961,3590,6635,12370,23150,

%T 43579,82267,155921,296347,564688,1078555,2064589,3958999,7605134,

%U 14632960,28195586,54403835,105102701,203287169,393625231,762951922,1480223716,2874422303

%N Rank of 2^n when all prime powers (A246655) p^n, for n>=1, are jointly ranked.

%H Ray Chandler, <a href="/A182908/b182908.txt">Table of n, a(n) for n = 1..92</a> (using b-file file from A007053)

%F a(n) = A182908(n) = A024622(n) - 1 for n>=1.

%F a(n) = Sum_{i=1..n} pi(floor(2^(n/i))), where pi(n) = A000720(n). - _Ridouane Oudra_, Oct 26 2020

%F a(n) = A025528(2^n). - _Pontus von Brömssen_, Sep 27 2024

%e a(3)=6 because 2^3 has rank 6 in the sequence (2,3,4,5,7,8,9,...).

%t 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}]]

%t f[n_] := Sum[ PrimePi[ Floor[2^(n/k)]], {k, n + 1}]; Array[f, 34] (* _Robert G. Wilson v_, Jul 08 2011 *)

%o (Python)

%o from sympy import primepi, integer_nthroot

%o def A182908(n):

%o x = 1<<n

%o return int(sum(primepi(integer_nthroot(x, k)[0]) for k in range(1, n+1))) # _Chai Wah Wu_, Nov 05 2024

%Y Cf. A000720, A024622, A025528, A246655.

%Y Row 1 of A182869. Complement of A182909.

%K nonn

%O 1,2

%A _Clark Kimberling_, Dec 13 2010

%E Minor edits by _Ray Chandler_, Aug 20 2021