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A191476
Values of j in the numbers 2^i*3^j, i >= 1, j >= 1, arranged in increasing order (A033845).
7
1, 1, 2, 1, 2, 1, 3, 2, 1, 3, 2, 4, 1, 3, 2, 4, 1, 3, 5, 2, 4, 1, 3, 5, 2, 4, 6, 1, 3, 5, 2, 4, 6, 1, 3, 5, 7, 2, 4, 6, 1, 3, 5, 7, 2, 4, 6, 1, 8, 3, 5, 7, 2, 4, 6, 1, 8, 3, 5, 7, 2, 9, 4, 6, 1, 8, 3, 5, 7, 2, 9, 4, 6, 1, 8, 3, 10, 5, 7, 2, 9, 4, 6, 1, 8, 3
OFFSET
1,3
COMMENTS
This is the signature sequence of log(2)/log(3) (compare A022328). - N. J. A. Sloane, May 26 2024
EXAMPLE
a(10) = 3 because A033845(10) = 108 = 2^2*3^3.
a(100) = 7 because A033845(100) = 59872 = 2^8*3^7.
a(1000) = 1 because A033845(1000) = 216172782113783808 = 2^56*3^1.
MATHEMATICA
mx = 1000000; t = Select[Sort[Flatten[Table[2^i 3^j, {i, Log[2, mx]}, {j, Log[3, mx]}]]], # <= mx &]; Table[FactorInteger[i][[2, 2]], {i, t}] (* T. D. Noe, Aug 31 2012 *)
PROG
(Python)
from sympy import integer_log
def A191476(n):
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x): return n+x-sum((x//3**i).bit_length() for i in range(integer_log(x, 3)[0]+1))
return 1+integer_log((m:=bisection(f, n, n))>>(~m&m-1).bit_length(), 3)[0] # Chai Wah Wu, Sep 15 2024
CROSSREFS
Cf. A033845 (numbers 2^i*3^j), A191475 (values of i).
A022329 (= a(n)-1) is an essentially identical sequence.
See also A022328.
Sequence in context: A292224 A023130 A084532 * A134583 A087467 A231568
KEYWORD
nonn
AUTHOR
Zak Seidov, Aug 30 2012
EXTENSIONS
Edited by N. J. A. Sloane, May 26 2024
STATUS
approved