1,3

When n>0 is written as k*2^j with k odd then k=A000265(n) and j=A007814(n), so: when n is written as k*2^j-1 with k odd then k=A000265(n+1) and j=A007814(n+1), when n>1 is written as k*2^j+1 with k odd then k=A000265(n-1) and j=A007814(n-1)

Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Sep 01 2002

Slope of line connecting (o,a(o)) where o=(2^k)(n-1)+1 is 2^k and (by design) starts at (1,1) - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004

Numerator of n/2^(n-1). - Alexander Adamchuk (alex(AT)kolmogorov.com), Feb 11 2005

Comment from Marco Matosic (marcomatosic(AT)hotmail.com), Jun 29 2005:

"The sequence can be arranged in a table:

...................................1

................................1..3..1

............................1...5..3..7...1

....................1...9...5..11..3..13..7...15..1

......1..17..9..19..5..21..11..23..3..25..13..27..7..29..15..31..1

Every new row is the previous row interspaced with the continuation of the odd numbers.

Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265."

(a(k),a(2k),a(3k),...)=a(k)*(a(1),a(2),a(3),...) In general, a[n*m]=a[n]*a[m] - Josh Locker (jlocker(AT)mail.rochester.edu), Oct 04 2005

This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell (lkmitch(AT)gmail.com), Dec 07 2005

2k+1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Dec 30 2005

It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005

a(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 27 2007

Problem H-81, Fib. Quart., 6 (1968), 52.

T. D. Noe, Table of n, a(n) for n=1..10000

R. Stephan, Some divide-and-conquer sequences ...

R. Stephan, Table of generating functions

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Trigonometry Angles

Eric Weisstein's World of Mathematics, Sphere Line Picking

Eric Weisstein's World of Mathematics, Odd Part

a(n) = if n is odd then n else a(n/2). - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Sep 01 2002

a(n) = n/A006519(n) = 2*A025480(n-1)+1

Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. - David W. Wilson (davidwwilson(AT)comcast.net), Aug 01, 2001.

a(n) = Sum_{d divides n and d is odd} phi(d). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Dec 04 2002

G.f.: -1/(1-x) + sum(k>=0, 2x^2^k/(1-2x^2^(k+1)+x^2^(k+2))). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Sep 05 2003

Dirichlet g.f.: zeta(s-1)*(2^s-2)/(2^s-1). - R. Stephan, Jun 18 2007

a(n)=sum{k=0..n, A127793(n,k)*floor((k+2)/2)} (conjecture). - Paul Barry (pbarry(AT)wit.ie), Jan 29 2007

A000265:=proc(n) local t1, d; t1:=1; for d from 1 by 2 to n do if n mod d = 0 then t1:=d; fi; od; t1; end;

Table[Times@@(#[[1]]^#[[2]]&/@Select[FactorInteger[i], #[[1]]!=2&]), {i, 90}] (from Harvey Dale)

a[n_Integer /; n > 0] := n/2^IntegerExponent[n, 2] (Josh Locker)

(PARI) a(n)=if(n<1, 0, n/2^valuation(n, 2)) /* Michael Somos Aug 09 2006 */

Cf. A111929, A111930, A111918, A111919, A111920, A111921, A111922, A111923.

Cf. A038502, A065330, A135013.

Sequence in context: A036233 A098985 A072963 this_sequence A106617 A040026 A106609

Adjacent sequences: A000262 A000263 A000264 this_sequence A000266 A000267 A000268

mult,nonn,easy,nice

njas

Additional comments from Henry Bottomley (se16(AT)btinternet.com), Mar 02 2000. More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000.