A038572 -id:A038572 - cp's OEIS frontend

A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Offset: 0

N. J. A. Sloane

Write the numbers 1 through n in a circle, start at 1 and cross off every other number until only one number is left.
A version of the children's game "One potato, two potato, ...".
a(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010
By inspection, the solution to the Josephus Problem is a sequence of odd numbers (from 1) starting at each power of 2.  This yields a direct closed form expression (see formula below). - Gregory Pat Scandalis, Oct 15 2013
Also zero together with a triangle read by rows in which row n lists the first 2^(n-1) odd numbers (see A005408), n >= 1. Row lengths give A011782. Right border gives A000225. Row sums give A000302, n >= 1. See example. - Omar E. Pol, Oct 16 2013
For n > 0: a(n) = n + 1 - A080079(n). - Reinhard Zumkeller, Apr 14 2014
In binary, a(n) = ROL(n), where ROL = rotate left = remove the leftmost digit and append it to the right. For example, n = 41 = 101001_2 => a(n) = (0)10011_2 = 19. This also explains FTAW's comment above. - M. F. Hasler, Nov 02 2016
In the under-down Australian card deck separation: top card on bottom of a deck of n cards, next card separated on the table, etc., until one card is left. The position a(n), for n >= 1, from top will be the left over card. See, e.g., the Behrends reference, pp. 156-164. For the down-under case see 2*A053645(n), for n >= 3, n not a power of 2. If n >= 2 is a power of 2 the botton card survives. - Wolfdieter Lang, Jul 28 2020

			From _Omar E. Pol_, Jun 09 2009: (Start)
Written as an irregular triangle the sequence begins:
  0;
  1;
  1,3;
  1,3,5,7;
  1,3,5,7,9,11,13,15;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,
   43,45,47,49,51,53,55,57,59,61,63;
...
(End)
From _Omar E. Pol_, Nov 03 2018: (Start)
An illustration of initial terms, where a(n) is the area (or number of cells) in the n-th region of the structure:
   n   a(n)       Diagram
   0    0     _
   1    1    |_|_ _
   2    1      |_| |
   3    3      |_ _|_ _ _ _
   4    1          |_| | | |
   5    3          |_ _| | |
   6    5          |_ _ _| |
   7    7          |_ _ _ _|
(End)

Erhard Behrends, Der mathematische Zauberstab, Rowolth Taschenbuch Verlag, rororo 62902, 4. Auflage, 2019, pp. 156-164. [English version: The Math Behind the Magic, AMS, 2019.]
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 10.
M. S. Petković, "Josephus problem", Famous Puzzles of Great Mathematicians, page 179, Amer. Math. Soc. (AMS), 2009.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Paul Weisenhorn, Josephus und seine Folgen, MNU, 59(2006), pp. 18-19.

Iain Fox, Table of n, a(n) for n = 0..100000 (terms 0..1000 from T. D. Noe, terms 1001..10000 from Indranil Ghosh).
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197, ex. 34.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Paul Barry, Conjectures and results on some generalized Rueppel sequences, arXiv:2107.00442 [math.CO], 2021.
Daniel Erman and Brady Haran, The Josephus Problem, Numberphile video (2016)
Chris Groër, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 87-93.
Yuri Nikolayevsky and Ioannis Tsartsaflis, Cohomology of N-graded Lie algebras of maximal class over Z_2, arXiv:1512.87676 [math.RA], (2016), pages 2, 6.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Josephus Problem
Wikipedia, Josephus problem
Index entries for sequences related to the Josephus Problem

Cf. A005428, A035327, A038572, A049940, A049964, A053644, A053645, A088147, A088442.
Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.

Require Import ZArith.
Fixpoint a (n : positive) : Z :=
match n with
| xH => 1
| xI n' => (2*(a n') + 1)%Z
| xO n' => (2*(a n') - 1)%Z
end.
(* Stefan Haan, Aug 27 2023 *)

a006257 n = a006257_list !! n
a006257_list =
   0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
-- Reinhard Zumkeller, Oct 06 2011

[0] cat [2*(n-2^Floor(Log(2,n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016

a(0):=0: for n from 1 to 100 do a(n):=(a(n-1)+1) mod n +1: end do:
seq(a(i),i=0..100); # Paul Weisenhorn, Oct 10 2010; corrected by Robert Israel, Jan 13 2016
A006257 := proc(n)
    convert(n,base,2) ;
    ListTools[Rotate](%,-1) ;
    add( op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, May 20 2016
A006257 := n -> 2*n  - Bits:-Iff(n, n):
seq(A006257(n), n=0..78); # Peter Luschny, Sep 24 2019

Table[ FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v, Sep 21 2003 *)
Flatten@Table[Range[1, 2^n - 1, 2], {n, 0, 5}] (* Birkas Gyorgy, Feb 07 2011 *)
m = 5; Range[2^m - 1] + 1 - Flatten@Table[Reverse@Range[2^n], {n, 0, m - 1}] (* Birkas Gyorgy, Feb 07 2011 *)

a(n)=sum(k=1,n,if(bitxor(n,k)Paul D. Hanna

a(n)=if(n, 2*n-2^logint(2*n,2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016

import math
def A006257(n):
     return 0 if n==0 else 2*(n-2**int(math.log(n,2)))+1 # Indranil Ghosh, Jan 11 2017

def A006257(n): return bool(n&(m:=1<Chai Wah Wu, Jan 22 2023
(C#)
static long cs_A006257(this long n) => n == 0 ? 0 : 1 + (1 + (n - 1).cs_A006257()) % n; // Frank Hollstein, Feb 24 2021

To get a(n), write n in binary, rotate left 1 place.
a(n) = 2*A053645(n) + 1 = 2(n-msb(n))+1. - Marc LeBrun, Jul 11 2001. [Here "msb" = "most significant bit", A053644.]
G.f.: 1 + 2/(1-x) * ((3*x-1)/(2-2*x) - Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = number of positive integers k < n such that n XOR k < n. a(n) = n - A035327(n). - Paul D. Hanna, Jan 21 2006
a(n) = n for n = 2^k - 1. - Zak Seidov, Dec 14 2006
a(n) = n - A035327(n). - K. Spage, Oct 22 2009
a(2^m+k) = 1+2*k; with 0 <= m and 0 <= k < 2^m; n = 2^m+k; m = floor(log_2(n)); k = n-2^m; a(n) = ((a(n-1)+1) mod n) + 1; a(1) = 1. E.g., n=27; m=4; k=11; a(27) = 1 + 2*11 = 23. - Paul Weisenhorn, Oct 10 2010
a(n) = 2*(n - 2^floor(log_2(n))) + 1 (see comment above). - Gregory Pat Scandalis, Oct 15 2013
a(n) = 0 if n = 0 and a(n) = 2*a(floor(n/2)) - (-1)^(n mod 2) if n > 0. - Marek A. Suchenek, Mar 31 2016
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 + x). - Ilya Gutkovskiy, Aug 31 2019
For n > 0: a(n) = 2 * A062050(n) - 1. - Frank Hollstein, Oct 25 2021

More terms from Robert G. Wilson v, Sep 21 2003

A255588 Convert n to base 3, move the least significant digit to the most significant one and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 1, 4, 7, 2, 5, 8, 3, 12, 21, 4, 13, 22, 5, 14, 23, 6, 15, 24, 7, 16, 25, 8, 17, 26, 9, 36, 63, 10, 37, 64, 11, 38, 65, 12, 39, 66, 13, 40, 67, 14, 41, 68, 15, 42, 69, 16, 43, 70, 17, 44, 71, 18, 45, 72, 19, 46, 73, 20, 47, 74, 21, 48, 75, 22, 49, 76, 23

Offset: 0

Paolo P. Lava, Feb 27 2015

a(3*n) = n.

Fixed points of the transform are listed in A048328.

			10 in base 3 is 101: moving the least significant digit to the most significant one we have 110 that is 12 in base 10.

Indranil Ghosh, Table of n, a(n) for n = 0..19683 (terms 0..1000 Paolo P. Lava)

Cf. A030102, A038572, A048328, A255589-A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,3);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 3; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 69]) (* Michael De Vlieger, Mar 04 2015 *)
Table[FromDigits[RotateRight[IntegerDigits[n,3]],3],{n,0,70}] (* Harvey P. Dale, Feb 20 2022 *)

def A255588(n):
    x=A007089(n)
    return int(x[-1]+x[:-1], 3) # Indranil Ghosh, Feb 03 2017

A255594 Convert n to base 9, move least significant digit to most significant digit and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 1, 10, 19, 28, 37, 46, 55, 64, 73, 2, 11, 20, 29, 38, 47, 56, 65, 74, 3, 12, 21, 30, 39, 48, 57, 66, 75, 4, 13, 22, 31, 40, 49, 58, 67, 76, 5, 14, 23, 32, 41, 50, 59, 68, 77, 6, 15, 24, 33, 42, 51, 60, 69, 78, 7, 16, 25, 34, 43, 52, 61

Offset: 0

Paolo P. Lava, Mar 02 2015

Fixed points of the transform are listed in A048334.

For more than 2 base-9 digits, reversal and cyclic shifts of a number start to differ, so this sequence differs from A030108. - Alonso del Arte, Mar 22 2015

			13 in base 9 is 14: moving the least significant digit as the most significant one we have 41 that is 37 in base 10.

Paolo P. Lava, Table of n, a(n) for n = 0..1000

Cf. A030108, A038572, A048334, A255588-A255593, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,9);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 9; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 69]) (* Michael De Vlieger, Mar 04 2015 *)
Table[FromDigits[RotateRight[IntegerDigits[n,9]],9],{n,0,80}] (* Harvey P. Dale, Jun 14 2024 *)

a(9*n) = n.

a(9^n) = 9^(n-1).

A088163 Numbers for which rotating one binary place to the right less rotating one binary place to the left is equal to zero.

Original entry on oeis.org

0, 1, 2, 3, 7, 10, 15, 31, 42, 63, 127, 170, 255, 511, 682, 1023, 2047, 2730, 4095, 8191, 10922, 16383, 32767, 43690, 65535, 131071, 174762, 262143, 524287, 699050, 1048575, 2097151, 2796202, 4194303, 8388607, 11184810, 16777215, 33554431, 44739242, 67108863

Offset: 0

Robert G. Wilson v, Sep 13 2003

n is a member iff n is of the form 2^n -1 (A000225) or A000975(2n).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-4).

Cf. A000225, A000975.

f[n_] := FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2] - FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2]; Select[ Range[33560000], f[ # ] == 0 &]
(* Or *) Union[ Join[ Table[2^n - 1, {n, 0, 25}], Table[ Ceiling[2(2^n - 1)/3], {n, 2, 24, 2}]]]
LinearRecurrence[{0,0,5,0,0,-4},{0,1,2,3,7,10},40] (* Harvey P. Dale, Feb 20 2022 *)

concat(0, Vec(x*(1+x)*(1+x+2*x^2) / ((1-x)*(1+x+x^2)*(1-4*x^3)) + O(x^50))) \\ Colin Barker, May 14 2016

Numbers n such that A038572(n) - A006257(n) = A088161(n) = 0.
From Colin Barker, May 14 2016: (Start)
a(n) = 5*a(n-3)-4*a(n-6) for n>5.
G.f.: x*(1+x)*(1+x+2*x^2) / ((1-x)*(1+x+x^2)*(1-4*x^3)).
(End)

A088161 n rotated one binary place to the right less n rotated one binary place to the left.

Original entry on oeis.org

0, 0, 0, 0, 1, 3, -2, 0, 3, 9, 0, 6, -3, 3, -6, 0, 7, 21, 4, 18, 1, 15, -2, 12, -5, 9, -8, 6, -11, 3, -14, 0, 15, 45, 12, 42, 9, 39, 6, 36, 3, 33, 0, 30, -3, 27, -6, 24, -9, 21, -12, 18, -15, 15, -18, 12, -21, 9, -24, 6, -27, 3, -30, 0, 31, 93, 28, 90, 25, 87, 22, 84, 19, 81, 16, 78, 13, 75, 10, 72, 7, 69, 4

Offset: 0

Robert G. Wilson v, Sep 13 2003

f(n) is negative about 2/7 of the time to 10^7. f(n) is zero, see A088163.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A006257, A038572, A088163.

f:= proc(n) local a,b;
  if n::even then a:= n/2 else a:= 2^ilog2(n) + (n-1)/2 fi;
  if n = 0 then b:= 0 else b:= 2*n + 1 - 2^(ilog2(n)+1) fi;
  a-b
end proc:
map(f, [$0..200]); # Robert Israel, Mar 03 2025

f[n_] := FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2] - FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2]; Table[ f[n], {n, 0, 82}]

a(n) = A038572(n) - A006257(n).

A088146 n-th prime rotated one binary place to the right.

Original entry on oeis.org

1, 3, 6, 7, 13, 14, 24, 25, 27, 30, 31, 50, 52, 53, 55, 58, 61, 62, 97, 99, 100, 103, 105, 108, 112, 114, 115, 117, 118, 120, 127, 193, 196, 197, 202, 203, 206, 209, 211, 214, 217, 218, 223, 224, 226, 227, 233, 239, 241, 242, 244, 247, 248, 253, 384, 387, 390

Offset: 1

Robert G. Wilson v, Sep 13 2003

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A038572, A088147.

f:= proc(p) 2^ilog2(p)+(p-1)/2 end proc:
f(2):= 1:
map(f, [seq(ithprime(i),i=1..100)]); # Robert Israel, Sep 19 2019

f[n_] := FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2]; Table[ f[ Prime[i]], {i, 1, 60}]

A255589 Convert n to base 4, move the least significant digit to the most significant one and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 3, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 15, 4, 20, 36, 52, 5, 21, 37, 53, 6, 22, 38, 54, 7, 23, 39, 55, 8, 24, 40, 56, 9, 25, 41, 57, 10, 26, 42, 58, 11, 27, 43, 59, 12, 28, 44, 60, 13, 29, 45, 61, 14, 30, 46, 62, 15, 31, 47, 63, 16, 80, 144, 208, 17, 81

Offset: 0

Paolo P. Lava, Mar 02 2015

Fixed points of the transform are listed in A048329.

			11 in base 4 is 23: moving the least significant digit to the most significant one we have 32 that is 14 in base 10.

Indranil Ghosh, Table of n, a(n) for n = 0..16384 (terms 0..1000 from Paolo P. Lava)

Cf. A030103, A038572, A048329, A255588, A255590-A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,4);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 4; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 69]) (* Michael De Vlieger, Mar 04 2015 *)
Array[FromDigits[RotateRight[IntegerDigits[#,4]],4]&,70,0] (* Harvey P. Dale, Mar 01 2016 *)

def A255589(n):
    x=str(A007090(n))
    return int(x[-1]+x[:-1],4) # Indranil Ghosh, Feb 03 2017

a(4*k) = k.

a(4^k) = 4^(k-1).

A255590 Convert n to base 5, move the least significant digit to the most significant one and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 6, 11, 16, 21, 2, 7, 12, 17, 22, 3, 8, 13, 18, 23, 4, 9, 14, 19, 24, 5, 30, 55, 80, 105, 6, 31, 56, 81, 106, 7, 32, 57, 82, 107, 8, 33, 58, 83, 108, 9, 34, 59, 84, 109, 10, 35, 60, 85, 110, 11, 36, 61, 86, 111, 12, 37, 62, 87, 112, 13, 38, 63, 88

Offset: 0

Paolo P. Lava, Mar 02 2015

a(5*n) = n.
a(5^n) = 5^(n-1).
Fixed points of the transform are listed in A048330.

			14 in base 5 is 24: moving the least significant digit to the most significant one we have 42 that is 22 in base 10.

Indranil Ghosh, Table of n, a(n) for n = 0..15625 (terms 0..1000 from Paolo P. Lava)

Cf. A030104, A038572, A048330, A255588, A255589, A255591-A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,5);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 5; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 68]) (* Michael De Vlieger, Mar 04 2015 *)
Table[FromDigits[RotateRight[IntegerDigits[n,5]],5],{n,0,100}] (* Harvey P. Dale, Jun 11 2025 *)

def A255590(n):
    x=str(A007091(n))
    return int(x[-1]+x[:-1], 5) # Indranil Ghosh, Feb 03 2017

A255591 Convert n to base 6, move least significant digit to most significant digit and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 1, 7, 13, 19, 25, 31, 2, 8, 14, 20, 26, 32, 3, 9, 15, 21, 27, 33, 4, 10, 16, 22, 28, 34, 5, 11, 17, 23, 29, 35, 6, 42, 78, 114, 150, 186, 7, 43, 79, 115, 151, 187, 8, 44, 80, 116, 152, 188, 9, 45, 81, 117, 153, 189, 10, 46, 82, 118, 154, 190, 11

Offset: 0

Paolo P. Lava, Mar 02 2015

a(6*n) = n.

Fixed points of the transform are listed in A048331.

			16 in base 6 is 24: moving the least significant digit to the most significant one we have 42 that is 26 in base 10.

Indranil Ghosh, Table of n, a(n) for n = 0..46656 (terms 0..1000 from Paolo P. Lava)

Cf. A030105, A038572, A048331, A255588-A255590, A255592-A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,6);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 6; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 66]) (* Michael De Vlieger, Mar 04 2015 *)

def A255591(n):
    x=A007092(n)
    return int(x[-1]+x[:-1],6) # Indranil Ghosh, Feb 03 2017

A255592 Convert n to base 7, move least significant digit to most significant digit and convert back to base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 1, 8, 15, 22, 29, 36, 43, 2, 9, 16, 23, 30, 37, 44, 3, 10, 17, 24, 31, 38, 45, 4, 11, 18, 25, 32, 39, 46, 5, 12, 19, 26, 33, 40, 47, 6, 13, 20, 27, 34, 41, 48, 7, 56, 105, 154, 203, 252, 301, 8, 57, 106, 155, 204, 253, 302, 9, 58, 107, 156

Offset: 0

Paolo P. Lava, Mar 02 2015

a(7*n) = n.
a(7^n) = 7^(n-1).
Fixed points of the transform are listed in A048332.

			11 in base 7 is 14: moving the least significant digit to the most significant one we have 41 that is 29 in base 10.

Paolo P. Lava, Table of n, a(n) for n = 0..1000

Cf. A030106, A038572, A048332, A255588-A255591, A255593, A255594, A048787, A255689-A255693.

with(numtheory): P:=proc(q,h) local a,b,k,n; print(0);
for n from 1 to q do
a:=convert(n,base,h); b:=[]; for k from 2 to nops(a) do b:=[op(b),a[k]]; od; a:=[op(b),a[1]];
a:=convert(a,base,h,10); b:=0; for k from nops(a) by -1 to 1 do b:=10*b+a[k]; od;
print(b); od; end: P(10^4,7);

roll[n_, b_] := Block[{w = IntegerDigits[n, b]}, Prepend[Most@ w, Last@ w]]; b = 7; FromDigits[#, b] & /@ (roll[#, b] & /@ Range[0, 66]) (* Michael De Vlieger, Mar 04 2015 *)
Table[FromDigits[RotateRight[IntegerDigits[n,7]],7],{n,0,100}] (* Harvey P. Dale, Jan 27 2023 *)

cp's OEIS Frontend

A006257 Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

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A255588 Convert n to base 3, move the least significant digit to the most significant one and convert back to base 10.

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A255594 Convert n to base 9, move least significant digit to most significant digit and convert back to base 10.

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A088163 Numbers for which rotating one binary place to the right less rotating one binary place to the left is equal to zero.

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A088161 n rotated one binary place to the right less n rotated one binary place to the left.

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A088146 n-th prime rotated one binary place to the right.

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A255589 Convert n to base 4, move the least significant digit to the most significant one and convert back to base 10.

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A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.