A242603 -id:A242603 - cp's OEIS frontend

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

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, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 1, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77

Offset: 1

N. J. A. Sloane

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, 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, Feb 11 2005
From Marco 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 by A000265 and working to the right by A000265." (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 of A160469 and A156769, 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 in A068563. - Max Alekseyev, Aug 25 2013
From Alexander 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 actual A000265(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 as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) 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
From Paul 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 is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 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 of A069834. (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 least A007814(n) times. - Federico Provvedi, Dec 14 2021

			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 + ...

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).

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from T. D. Noe)
Peter Bala, A note on the sequence of numerators of a rational function
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

Cf. A000004, A000010, A000123, A000217, A000225, A000265, A001227.
Cf. A003602, A003961, A006516, A006519, A007814, A008683, A014577.
Cf. A020988, A025480, A026741, A027750, A035263, A038502, A049606.
Cf. A064989, A065330, A068563, A069834, A075101, A111929, A111930.
Cf. A111918, A111919, A111920, A111921, A111922, A111923, A125650.
Cf. A127793, A132739, A132740, A156769, A160469, A182469, A209308.
Cf. A213671, A220466, A236999, A242603.
Cf. A049606 (partial products), A135013 (partial sums), A099545 (mod 4), A326937 (Dirichlet inverse).
Cf. A026741 (map), A001511 (converging steps), A038550 (prime index).
Cf. A195056 (Dgf at s=3).

a000265 = until odd (`div` 2)
-- Reinhard Zumkeller, Jan 08 2013, Apr 08 2011, Oct 14 2010

int A000265(n){
    while(n%2==0) n>>=1;
    return n;
}
/* Aidan Simmons, Feb 24 2019 */

using IntegerSequences
[OddPart(n) for n in 1:77] |> println  # Peter Luschny, Sep 25 2021

A000265:= func< n | n/2^Valuation(n,2) >;
[A000265(n): n in [1..120]]; // G. C. Greubel, Jul 31 2024

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: seq(A000265(n), n=1..77);
A000265 := n -> n/2^padic[ordp](n,2): seq(A000265(n), n=1..77); # Peter Luschny, Nov 26 2010

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 *)

{a(n) = n >> valuation(n, 2)}; /* Michael Somos, Aug 09 2006, edited by M. F. Hasler, Dec 18 2014 */

from _future_ import division
def A000265(n):
    while not n % 2:
        n //= 2
    return n # Chai Wah Wu, Mar 25 2018

def a(n):
    while not n&1: n >>= 1
    return n
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Jun 26 2025

def A000265(n): return n//2^valuation(n,2)
[A000265(n) for n in (1..121)] # G. C. Greubel, Jul 31 2024

(define (A000265 n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2))))) ;; Antti Karttunen, Apr 15 2017

a(n) = if n is odd then n, otherwise a(n/2). - Reinhard Zumkeller, 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, 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(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller, Aug 27 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
From Reinhard Zumkeller, May 01 2012: (Start)
A182469(n, k) = A027750(a(n), k), k = 1..A001227(n).
a(n) = A182469(n, A001227(n)). (End)
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
a(n) = A003961(A064989(n)). - Antti Karttunen, Apr 15 2017
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
From Peter 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. of A000123. (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 function A000010. - Peter Bala, Mar 22 2019
a(n) = A049606(n) / A049606(n-1). - Flávio V. Fernandes, Dec 08 2020
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
a(n) = A030101(A030101(n)). - Darío Clavijo, Sep 19 2024

Additional comments from Henry Bottomley, Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified by David A. Corneth, Apr 15 2017

A038502 Remove 3's from n.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 7, 8, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 19, 20, 7, 22, 23, 8, 25, 26, 1, 28, 29, 10, 31, 32, 11, 34, 35, 4, 37, 38, 13, 40, 41, 14, 43, 44, 5, 46, 47, 16, 49, 50, 17, 52, 53, 2, 55, 56, 19, 58, 59, 20, 61, 62, 7, 64, 65, 22, 67, 68, 23, 70, 71, 8, 73, 74, 25, 76

Offset: 1

N. J. A. Sloane

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 21 2019

The largest divisor of n not divisible by 3. - Amiram Eldar, Sep 15 2020

			From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - (2*3)*G(x^3) - (2*9)*G(x^9) - (2*27)*G(x^27) - ..., where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (2/3)*H(x^3) - (2/9)*H(x^9) - (2/27)*H(x^27) - ..., where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (2/3^2)*L(x^3) - (2/9^2)*L(x^9) - (2/27^2)*L(x^27) - ..., where L(x) = Log(1/(1 - x)).
Also, Sum_{n >= 1} 1/a(n)*x^n = L(x) + (2/3)*L(x^3) + (2/3)*L(x^9) + (2/3)*L(x^27) + ... .
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Peter Bala, A note on the sequence of numerators of a rational function .

Cf. A007949, A038500, A065330.

Result of iterative removal of other factors: A000265 (2), A065883 (4), A132739 (5), A244414 (6), A242603 (7), A004151 (10).

a038502 n = if m > 0 then n else a038502 n'  where (n', m) = divMod n 3
-- Reinhard Zumkeller, Jan 03 2011

[n/3^Valuation(n,3): n in [1..80]]; // Bruno Berselli, May 21 2013

f[n_] := Times @@ (First@#^Last@# & /@ Select[ FactorInteger@n, First@# != 3 &]); Array[f, 76] (* Robert G. Wilson v, Jul 31 2006 *)
Table[n/3^IntegerExponent[n, 3], {n, 100}] (* Amiram Eldar, Sep 15 2020 *)

a(n)=if(n<1, 0, n/3^valuation(n,3)) /* Michael Somos, Nov 10 2005 */

Multiplicative with a(p^e) = 1 if p = 3, otherwise p^e. - Mitch Harris, Apr 19 2005
a(0) = 0, a(3*n) = a(n), a(3*n+1) = 3*n+1, a(3*n+2) = 3*n+2.
Dirichlet g.f. zeta(s-1)*(3^s-3)/(3^s-1). - R. J. Mathar, Feb 11 2011
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,3^n).
O.g.f.: F(x) - 2*F(x^3) - 2*F(x^9) - 2*F(x^27) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers. More generally, for m >= 1,
Sum_{n >= 0} a(n)^m*x^n = F(m,x) - (3^m - 1)( F(m,x^3) + F(m,x^9) + F(m,x^27) + ... ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence produces generating functions for the sequences n^m*a(n), m in Z. Some examples are given below. (End)
Sum_{k=1..n} a(k) ~ (3/8) * n^2. - Amiram Eldar, Oct 29 2022
a(n) = n / A038500(n). - R. J. Mathar, Mar 13 2024

A132739 Largest divisor of n not divisible by 5.

Original entry on oeis.org

1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 11, 12, 13, 14, 3, 16, 17, 18, 19, 4, 21, 22, 23, 24, 1, 26, 27, 28, 29, 6, 31, 32, 33, 34, 7, 36, 37, 38, 39, 8, 41, 42, 43, 44, 9, 46, 47, 48, 49, 2, 51, 52, 53, 54, 11, 56, 57, 58, 59, 12, 61, 62, 63, 64, 13, 66, 67, 68, 69, 14, 71, 72, 73, 74, 3, 76, 77

Offset: 1

Reinhard Zumkeller, Aug 27 2007

A000265(a(n)) = a(A000265(n)) = A132740(n).
a(n) = A060791(n) when n is not divisible by 5. When n is divisible by 5 a(n) divides A060791(n). Tom Edgar, Feb 08 2014
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 21 2019

			From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - (4*5)*G(x^5) - (4*25)*G(x^25) - (4*125)*G(x^125) - ..., where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (4/5)*H(x^5) - (4/25)*H(x^25) - (4/125)*H(x^125) - ..., where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (4/5^2)*L(x^5) - (4/25^2)*L(x^25) - (4/125^2)*L(x^125) - ..., where L(x) = Log(1/(1 - x)).
Also, Sum_{n >= 1} 1/a(n)*x^n = L(x) + (4/5)*L(x^5) + (4/5)*L(x^25) + (4/5)*L(x^125) + ....
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.

Cf. A000265, A038502, A060791, A060904, A112765, A242603.

a132739 n | r > 0     = n
          | otherwise = a132739 n' where (n',r) = divMod n 5
-- Reinhard Zumkeller, Apr 08 2011

f[n_]:=Denominator[5^n/n];Array[f,100] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011*)
Table[n/5^IntegerExponent[n, 5], {n, 100}] (* Amiram Eldar, Sep 15 2020 *)

a(n)=n/5^valuation(n, 5) /* Simon Strandgaard, Nov 01 2021 */

def A132739(n):
    a, b = divmod(n, 5)
    while b == 0:
        a, b = divmod(a,5)
    return 5*a+b # Chai Wah Wu, Dec 05 2021

p (1..50).map { |n| n /= 5 while (n % 5) == 0; n } # Simon Strandgaard, Nov 01 2021

a(n) = n/A060904(n). Dirichlet g.f.: zeta(s-1)*(5^s-5)/(5^s-1). - R. J. Mathar, Jul 12 2012
a(n) = n/5^A112765(n). See A060904. - Wolfdieter Lang, Jun 18 2014
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,5^n).
O.g.f.: F(x) - 4*F(x^5) - 4*F(x^25) - 4*F(x^125) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers. More generally, for m >= 1,
Sum_{n >= 0} a(n)^m*x^n = F(m,x) - (5^m - 1)(F(m,x^5) + F(m,x^25) + F(m,x^125) + ...), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence produces generating functions for the sequences n^m*a(n), m in Z. Some examples are given below. (End)
Sum_{k=1..n} a(k) ~ (5/12) * n^2. - Amiram Eldar, Nov 28 2022

A065883 Remove factors of 4 from n (i.e., write n in base 4, drop final zeros, then rewrite in decimal).

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 9, 10, 11, 3, 13, 14, 15, 1, 17, 18, 19, 5, 21, 22, 23, 6, 25, 26, 27, 7, 29, 30, 31, 2, 33, 34, 35, 9, 37, 38, 39, 10, 41, 42, 43, 11, 45, 46, 47, 3, 49, 50, 51, 13, 53, 54, 55, 14, 57, 58, 59, 15, 61, 62, 63, 1, 65, 66, 67, 17, 69, 70, 71, 18, 73, 74, 75

Offset: 1

Henry Bottomley, Nov 26 2001

			a(7)=7, a(14)=14, a(28)=a(4*7)=7, a(56)=a(4*14)=14, a(112)=a(4^2*7)=7.

Indranil Ghosh, Table of n, a(n) for n = 1..20000 (First 1000 terms from Harry J. Smith)

Cf. A214392, A235127, A350091 (drop final 2's).

Remove other factors: A000265, A038502, A132739, A244414, A242603, A004151.

A065883:= n -> n/4^floor(padic:-ordp(n,2)/2):
map(A065883, [$1..1000]); # Robert Israel, Dec 08 2015

If[Divisible[#,4],#/4^IntegerExponent[#,4],#]&/@Range[80] (* Harvey P. Dale, Aug 31 2013 *)

a(n)=n/4^valuation(n,4); \\ Joerg Arndt, Dec 09 2015

def A065883(n): return n>>((~n&n-1).bit_length()&-2) # Chai Wah Wu, Jul 09 2022

If n mod 4 = 0 then a(n) = a(n/4), otherwise a(n) = n.
Multiplicative with a(p^e) = 2^(e (mod 2)) if p = 2 and a(p^e) = p^e if p is an odd prime.
a(n) = n/4^A235127(n).
a(n) = A214392(n) if n mod 16 != 0. - Peter Kagey, Sep 02 2015
From Robert Israel, Dec 08 2015: (Start)
G.f.: x/(1-x)^2 - 3 Sum_{j>=1} x^(4^j)/(1-x^(4^j))^2.
G.f. satisfies G(x) = G(x^4) + x/(1-x)^2 - 4 x^4/(1-x^4)^2. (End)
Sum_{k=1..n} a(k) ~ (2/5) * n^2. - Amiram Eldar, Nov 20 2022
Dirichlet g.f.: zeta(s-1)*(4^s-4)/(4^s-1). - Amiram Eldar, Jan 04 2023

A078414 a(n) = (a(n-1)+a(n-2))/7^k, where 7^k is the highest power of 7 dividing a(n-1)+a(n-2).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 3, 16, 19, 5, 24, 29, 53, 82, 135, 31, 166, 197, 363, 80, 443, 523, 138, 661, 799, 1460, 2259, 3719, 122, 3841, 3963, 7804, 1681, 1355, 3036, 4391, 1061, 5452, 6513, 11965, 18478, 4349, 3261, 7610, 1553, 187, 1740, 1927, 3667, 5594, 27

Offset: 1

Yasutoshi Kohmoto, Dec 28 2002

From Vladimir Shevelev, Apr 01 2013; edited by Danny Rorabaugh, Feb 19 2016: (Start)
If we consider Fibonacci-like numbers {F_p(n)} without positive multiples of p, where p is a fixed prime, then {F_2(n)} has period of length 1, {F_3(n)} has period of length 3, {F_5(n)} has period of length 6. This sequence is the first which does not have a trivial period and, probably, even is non-periodic.
An open question: Is this sequence bounded?
Consider Fibonacci-like sequences without multiples of several primes, defined analogously: e.g., for {F_(p,q)(n)}, a(0)=0, a(1)=1, for n>=2, a(n)=a(n-1)+a(n-2) divided by the maximal possible powers of p and q.
Problem: For what sets of primes is the corresponding Fibonacci-like sequence without multiples of these primes periodic?
Examples: sequence {F_(7,11,13)(n)} has period of length 12: 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 19, 29, 48, 1, 1, 2, 3, 5,...; {F_(11,13,19)(n)} has period of length 9; {F_(13,19,23)(n)} has period of length 12; {F_(17,19,23,29)(n)} has period of length 15; {F_(19,23,31,53,59,89)(n)} has period of length 24; {F_(23,29,73,233)(n)} has period of length 18.
Don Reble noted that lengths of all such periods could only be multiples of 3 because every Fibonacci-like sequence considered here modulo 2 has the form 0,1,1,0,1,1,... .
(End)

Alois P. Heinz, Table of n, a(n) for n = 1..4000
B. Avila, T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.

Cf. A000045, A078412, A214094, A214156, A216231, A216275, A216835.

a:= proc(n) option remember; local t, j;
      if n<3 then 1
    else t:= a(n-1)+a(n-2);
         while irem(t, 7, 'j')=0 do t:=j od; t
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 25 2012

nxt[{a_,b_}]:=Module[{n=IntegerExponent[a+b,7]},{b,(a+b)/7^n}]; Transpose[ NestList[nxt,{1,1},60]][[1]] (* Harvey P. Dale, Jul 23 2012 *)

a(n) = A242603(a(n-1)+a(n-2)). - R. J. Mathar, Mar 13 2024

Corrected by Harvey P. Dale, Jul 23 2012

A268354 Highest power of 7 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7

Offset: 1

Tom Edgar, Feb 02 2016

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 7.

			Since 14 = 7 * 2, a(14) = 7. Likewise, since 7 does not divide 13, a(13) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..20790
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A000420, A001620, A006519, A038500, A234957, A214411, A234959, A060904, A242603, A268357.

[7^Valuation(n,7): n in [1..150]]; // Vincenzo Librandi, Feb 03 2016

7^Table[IntegerExponent[n, 7], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *)

a(n) = 7^valuation(n, 7) \\ Michel Marcus, Feb 05 2016

Sage
```
[7^valuation(i, 7) for i in [1..100]]
```

a(n) = 7^valuation(n,7).
a(n) = 7^A214411(n).
Completely multiplicative with a(7) = 7, a(p) = 1 for prime p and p <> 7. - Andrew Howroyd, Jul 20 2018
From Peter Bala, Feb 21 2019: (Start)
a(n) = gcd(n,7^n).
a(n) = n/A242603(n).
O.g.f.: x/(1 - x) + 6*Sum_{n >= 1} 7^(n-1)*x^(7^n)/ (1 - x^(7^n)). (End)
Sum_{k=1..n} a(k) ~ (6/(7*log(7)))*n*log(n) + (4/7 + 6*(gamma-1)/(7*log(7)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(7^s-1)/(7^s-7). - Amiram Eldar, Jan 03 2023

More terms from Antti Karttunen, Dec 22 2017

A244414 Remove highest power of 6 from n.

Original entry on oeis.org

1, 2, 3, 4, 5, 1, 7, 8, 9, 10, 11, 2, 13, 14, 15, 16, 17, 3, 19, 20, 21, 22, 23, 4, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 1, 37, 38, 39, 40, 41, 7, 43, 44, 45, 46, 47, 8, 49, 50, 51, 52, 53, 9, 55, 56, 57, 58, 59, 10, 61, 62, 63, 64, 65, 11

Offset: 1

Wolfdieter Lang, Jun 27 2014

This is instance g = 6 of the g-family of sequences, call it r(g,n), where for g >= 2 the highest power of g is removed from n. See the crossrefs.

The present sequence is not multiplicative: a(6) = 1 not a(2)*a(3) = 6. In the prime factor decomposition one has to consider a(2^e2*3^e^3) as one entity, also for e2 >= 0, e3 >= 0 with a(1) = 1, and apply the rule given in the formula section. With this rule the sequence will be multiplicative in an unusual sense. - Wolfdieter Lang, Feb 12 2018

			a(1) = 1 = 1/6^A122841(1) = 1/6^0.
a(9) = a(2^0*3^2), min(0,2) = 0, a(9) = 2^(0-0)*3^(2-0) = 1*9 = 9.
a(12) = a(2^2*3^1), m = min(2,1) = 1, a(12) = 2^(2-1)*3^(1-1) = 2^1*1 = 2.
a(30) = a(2*3*5) = a(2^1*3^1)*a(5) = 1*a(5) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A122841, A000265, A038502, A065883, A132739, A242603.

A007310, A007913, A008833 are used to express relationship between terms of this sequence.

a[n_] := n/6^IntegerExponent[n, 6]; Array[a, 66] (* Robert G. Wilson v, Feb 12 2018 *)

a(n) = n/6^valuation(n,6); \\ Joerg Arndt, Jun 28 2014

a(n) = n/6^A122841(n), n >= 1.
For n >= 2, a(n) is sort of multiplicative if a(2^e2*3^e3) = 2^(e2 - m)*3^(e3 - m) with m = m(e2, e3) = min(e2, e3), for e2, e3 >= 0, a(1) = 1, and a(p^e) = p^e for primes p >= 5.
From Peter Munn, Jun 04 2020: (Start)
Proximity to being multiplicative may be expressed as follows:
a(n * A007310(k)) = a(n) * a(A007310(k));
a(n^2) = a(n)^2;
a(n) = a(A007913(n)) * a(A008833(n)).
(End)
Sum_{k=1..n} a(k) ~ (3/7) * n^2. - Amiram Eldar, Nov 20 2022

Incorrect multiplicity claim corrected by Wolfdieter Lang, Feb 12 2018

A277545 a(n) = n/7^m mod 7, where 7^m is the greatest power of 7 that divides n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4, 5, 6, 2, 1, 2, 3, 4, 5, 6, 3, 1, 2, 3, 4, 5, 6, 4, 1, 2, 3, 4, 5, 6, 5, 1, 2, 3, 4, 5, 6, 6, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4, 5, 6, 2, 1, 2, 3, 4, 5, 6, 3, 1, 2, 3, 4, 5, 6, 4, 1, 2, 3, 4, 5, 6, 5, 1, 2

Offset: 1

Clark Kimberling, Oct 19 2016

a(n) is the rightmost nonzero digit in the base 7 expansion of n.

			a(9) = (9/7 mod 7) = 2.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A010876, A242603.

Table[Mod[n/7^IntegerExponent[n, 7], 7], {n, 1, 160}]

a(n) = n/7^valuation(n, 7) % 7; \\ Michel Marcus, Oct 20 2016

a(n) = A242603(n) mod 7. - Michel Marcus, Oct 20 2016

A370757 a(n) is the least k > 0 such that 1/n and 1/k have equivalent repeating decimal digits.

Original entry on oeis.org

1, 1, 3, 1, 1, 6, 7, 1, 9, 1, 11, 3, 13, 7, 6, 1, 17, 18, 19, 1, 21, 22, 23, 6, 1, 26, 27, 7, 29, 3, 31, 1, 33, 17, 7, 36, 37, 19, 39, 1, 41, 42, 43, 44, 45, 23, 47, 3, 49, 1, 51, 13, 53, 54, 55, 7, 57, 29, 59, 6, 61, 31, 63, 1, 26, 66, 67, 17, 69, 7, 71, 72

Offset: 1

Rémy Sigrist, Feb 29 2024

In other words, a(n) is the least k > 0 such that the fractional parts of (10^i)/n and (10^j)/k are equal for some integers i, j.

a(n) is not always a divisor of n. For example, a(65) = 26 is not a divisor of 65. - David A. Corneth, Mar 01 2024

			The first terms, alongside the decimal expansion of 1/n with its repeating decimal digits in parentheses, are:
  n   a(n)  1/n
  --  ----  -----------
   1     1  1.(0)
   2     1  0.5(0)
   3     3  0.(3)
   4     1  0.25(0)
   5     1  0.2(0)
   6     6  0.1(6)
   7     7  0.(142857)
   8     1  0.125(0)
   9     9  0.(1)
  10     1  0.1(0)
  11    11  0.(09)
  12     3  0.08(3)
  13    13  0.(076923)
  14     7  0.07(142857)
  15     6  0.0(6)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program
Index entries for sequences related to decimal expansion of 1/n

Cf. A000265 (base-2 analog), A038502 (base-3 analog), A132739 (base-5 analog), A242603 (base-7 analog).

Cf. A003592, A007732, A132740.

PARI
```
\\ See Links section.
```

from itertools import count
from sympy import multiplicity, n_order
def A370757(n):
    m2, m5 = (~n & n-1).bit_length(), multiplicity(5,n)
    r = max(m2,m5)
    w, m = 10**r, 10**(t:=n_order(10,n2) if (n2:=(n>>m2)//5**m5)>1 else 1)-1
    c = w//n
    s = str(m*w//n-c*m).zfill(t)
    l = len(s)
    for k in count(1):
        m2, m5 = (~k & k-1).bit_length(), multiplicity(5,k)
        r = max(m2,m5)
        w, m = 10**r, 10**(t:=n_order(10,k2) if (k2:=(k>>m2)//5**m5)>1 else 1)-1
        c = w//k
        if any(s[i:]+s[:i] == str(m*w//k-c*m).zfill(t) for i in range(l)):
            return k # Chai Wah Wu, Mar 03 2024

a(n) = 1 iff n belongs to A003592.
a(10*n) = a(n).
A007732(a(n)) = A007732(n).

cp's OEIS Frontend

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

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A038502 Remove 3's from n.

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A132739 Largest divisor of n not divisible by 5.

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A065883 Remove factors of 4 from n (i.e., write n in base 4, drop final zeros, then rewrite in decimal).

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A078414 a(n) = (a(n-1)+a(n-2))/7^k, where 7^k is the highest power of 7 dividing a(n-1)+a(n-2).

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A268354 Highest power of 7 dividing n.

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