A085238 Sort the numbers 2^i and 3^j. Then a(n) is the exponent of the n-th term.

0, 1, 1, 2, 3, 2, 4, 3, 5, 6, 4, 7, 5, 8, 9, 6, 10, 11, 7, 12, 8, 13, 14, 9, 15, 10, 16, 17, 11, 18, 19, 12, 20, 13, 21, 22, 14, 23, 15, 24, 25, 16, 26, 17, 27, 28, 18, 29, 30, 19, 31, 20, 32, 33, 21, 34, 22, 35, 36, 23, 37, 38, 24, 39, 25, 40, 41, 26, 42, 27, 43, 44, 28

Offset: 1

Author

Reinhard Zumkeller, Jun 22 2003

Keywords

• nonn

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..500 from T. D. Noe)

Crossrefs

Cf. A006899, A085239, A085240.

Programs

Haskell

a085238 n = e (mod x 2 + 2) x where
   x = a006899 n
   e b p = if p == 1 then 0 else 1 + e b (p `div` b)
-- Reinhard Zumkeller, Oct 09 2013

Mathematica

seq[lim_] := Module[{r2 = Range[0, Floor[Log2[lim]]], r3 = Range[0, Floor[Log[3, lim]]]}, Rest@ SortBy[Join[{#, 2^#} & /@ r2, {#, 3^#} & /@ r3], Last][[;; , 1]]]; seq[10^14] (* Amiram Eldar, Mar 25 2025 *)

PARI

do(lim)=my(v=List(vector(logint(lim\=1,2),i,1<my(t=valuation(n,2)); if(t, t, valuation(n,3)), Set(v)) \\ Charles R Greathouse IV, Sep 02 2015

Python

from sympy import integer_log
def A085238(n): return k+1 if 6**(k:=integer_log(m:=3**(n-1),6)[0])<<1Chai Wah Wu, Feb 04 2025

Formula

A006899(n) = A085239(n)^a(n).

a(A085240(n)) = a(n).