OFFSET
1,3
COMMENTS
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 isA075101(n)). -Reinhard Zumkeller,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 ), Apr 17 2004
Numerator of n/2^(n-1). -Alexander Adamchuk,Feb 11 2005
FromMarco Matosic,Jun 29 2005: (Start)
"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 byA000265and working to the right byA000265."(End)
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,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,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
In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) =A020988.-Eric Desbiaux,May 27 2009
This sequence appears in the analysis ofA160469andA156769,which resemble the numerator and denominator of the Taylor series for tan(x). -Johannes W. Meijer,May 24 2009
Indices n such that a(n) divides 2^n - 1 are listed inA068563.-Max Alekseyev,Aug 25 2013
FromAlexander R. Povolotsky,Dec 17 2014: (Start)
With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actualA000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:
1
1 3
1 5 3 7
1 9 5 11 3 13 7 15
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31
and so on....
The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)
Partitions natural numbers to the same equivalence classes asA064989.That is, for all i, j: a(i) = a(j) <=>A064989(i) =A064989(j). There are dozens of other such sequences (likeA003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. -Antti Karttunen,Apr 15 2017
FromPaul Curtz,Feb 19 2019: (Start)
This sequence is the truncated triangle:
1, 1;
3, 1, 5;
3, 7, 1, 9;
5, 11, 3, 13, 7;
15, 1, 17, 9, 19, 5;
21, 11, 23, 3, 25, 13, 27;
7, 29, 15, 31, 1, 33, 17, 35;
...
The first column isA069834.The second column isA213671.The main diagonal isA236999.The first upper diagonal isA125650without 0.
c(n) = ((n*(n+1)/2))/A069834= 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1,... for n > 0. n*(n+1)/2 is the rank ofA069834.(End)
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). -Peter Bala,Feb 27 2019
a(n) is also the map n ->A026741(n) applied at leastA007814(n) times. -Federico Provvedi,Dec 14 2021
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Daniel Forgues,Table of n, a(n) for n = 1..100000(first 10000 terms from T. D. Noe)
V. Daiev and J. L. Brown,Problem H-81,Fib. Quart., 6 (1968), 52.
Ralf Stephan,Some divide-and-conquer sequences...
Ralf Stephan,Table of generating functions
Eric Weisstein's World of Mathematics,Odd Part
Eric Weisstein's World of Mathematics,Trigonometry Angles
Eric Weisstein's World of Mathematics,Sphere Line Picking
FORMULA
a(n) = if n is odd then n, otherwise a(n/2). -Reinhard Zumkeller,Sep 01 2002
Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. -David W. Wilson,Aug 01 2001
a(n) = Sum_{d divides n and d is odd} phi(d). -Vladeta Jovovic,Dec 04 2002
G.f.: -x/(1 - x) + Sum_{k>=0} (2*x^(2^k)/(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))). -Ralf Stephan,Sep 05 2003
(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
a(n) = Sum_{k=0..n}A127793(n,k)*floor((k+2)/2) (conjecture). -Paul Barry,Jan 29 2007
Dirichlet g.f.: zeta(s-1)*(2^s - 2)/(2^s - 1). -Ralf Stephan,Jun 18 2007
a(n) = 2*A003602(n) - 1. -Franklin T. Adams-Watters,Jul 02 2009
a(n) = n/gcd(2^n,n). (This also shows that the true offset is 0 and a(0) = 0.) -Peter Luschny,Nov 14 2009
a(-n) = -a(n) for all n in Z. -Michael Somos,Sep 19 2011
FromReinhard Zumkeller,May 01 2012: (Start)
a((2*n-1)*2^p) = 2*n - 1, p >= 0 and n >= 1. -Johannes W. Meijer,Feb 05 2013
G.f.: G(0)/(1 - 2*x^2 + x^4) - 1/(1 - x), where G(k) = 1 + 1/(1 - x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))/(x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2))) + (1 - 2*x^(2^(k+2)) + x^(2^(k+3)))/G(k+1))); (continued fraction). -Sergei N. Gladkovskii,Aug 06 2013
Completely multiplicative with a(2) = 1 and a(p) = p for prime p > 2, i.e., the sequence b(n) = a(n) *A008683(n) for n > 0 is the Dirichlet inverse of a(n). -Werner Schulte,Jul 08 2018
FromPeter Bala,Feb 27 2019: (Start)
O.g.f.: F(x) - F(x^2) - F(x^4) - F(x^8) -..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) +..., where L(x) = log(1/(1 - x)).
Sum_{n >= 1} x^n/a(n) = 1/2*log(G(x)), where G(x) = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 +... is the o.g.f. ofA000123.(End)
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient functionA000010.-Peter Bala,Mar 22 2019
a(n) = n - (1/2)*Sum_{d|2n} (-1)^d*phi(d). -Ridouane Oudra,May 01 2019
a(n) = numerator of n/2^(floor(n/2)). -Federico Provvedi,Dec 14 2021
a(n) = Sum_{d divides n} (-1)^(d+1)*phi(2*n/d). -Peter Bala,Jan 14 2024
EXAMPLE
G.f. = x + x^2 + 3*x^3 + x^4 + 5*x^5 + 3*x^6 + 7*x^7 + x^8 + 9*x^9 + 5*x^10 + 11*x^11 +...
MAPLE
MATHEMATICA
a[n_Integer /; n > 0]:= n/2^IntegerExponent[n, 2]; Array[a, 77] (* Josh Locker *)
a[ n_]:= If[ n == 0, 0, n / 2^IntegerExponent[ n, 2]]; (*Michael Somos,Dec 17 2014 *)
PROG
(PARI)
{a(n) = n >> valuation(n, 2)}; /*Michael Somos,Aug 09 2006, edited byM. F. Hasler,Dec 18 2014 */
(Haskell)
a000265 = until odd (`div` 2)
--Reinhard Zumkeller,Jan 08 2013, Apr 08 2011, Oct 14 2010
(Scheme) (define (A000265n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2)))));;Antti Karttunen,Apr 15 2017
(Python)
from __future__ import division
defA000265(n):
while not n % 2:
n //= 2
return n #Chai Wah Wu,Mar 25 2018
(Java)
intA000265(n){
while(n%2==0) n>>=1;
return n;
}
/*Aidan Simmons,Feb 24 2019 */
(Julia)
using IntegerSequences
[OddPart(n) for n in 1:77] |> println #Peter Luschny,Sep 25 2021
(Magma)
A000265:= func< n | n/2^Valuation(n, 2) >;
[A000265(n): n in [1..120]]; //G. C. Greubel,Jul 31 2024
(SageMath)
defA000265(n): return n//2^valuation(n, 2)
[A000265(n) for n in (1..121)] #G. C. Greubel,Jul 31 2024
CROSSREFS
Cf.A195056(Dgf at s=3).
KEYWORD
mult,nonn,easy,nice
AUTHOR
EXTENSIONS
Additional comments fromHenry Bottomley,Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified byDavid A. Corneth,Apr 15 2017
STATUS
approved