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Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.
(Formerly M2216)
+10
103
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
0,4
COMMENTS
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
REFERENCES
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.
LINKS
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.
Eric Weisstein's World of Mathematics, Josephus Problem
Wikipedia, Josephus problem
FORMULA
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
EXAMPLE
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)
MAPLE
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
MATHEMATICA
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 *)
PROG
(PARI) a(n)=sum(k=1, n, if(bitxor(n, k)<n, 1, 0)) \\ Paul D. Hanna
(PARI) a(n)=if(n, 2*n-2^logint(2*n, 2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016
(Haskell)
a006257 n = a006257_list !! n
a006257_list =
0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
-- Reinhard Zumkeller, Oct 06 2011
(Magma) [0] cat [2*(n-2^Floor(Log(2, n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016
(Python)
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
(Python)
def A006257(n): return bool(n&(m:=1<<n.bit_length()-1))+((n&m-1)<<1) if n else 0 # 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
(Coq)
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 *)
CROSSREFS
Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.
KEYWORD
nonn,easy,nice
EXTENSIONS
More terms from Robert G. Wilson v, Sep 21 2003
STATUS
approved
A version of Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer two places clockwise from i. Repeat, counting two places from the next undeleted integer, until only one integer remains.
+10
19
1, 2, 2, 1, 4, 1, 4, 7, 1, 4, 7, 10, 13, 2, 5, 8, 11, 14, 17, 20, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 1, 4, 7, 10, 13, 16, 19, 22, 25
OFFSET
1,2
COMMENTS
If one counts only one place (rather than two) at each stage to determine the element to be deleted, the Josephus survivors (A006257) are obtained.
LINKS
Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
R. Baumann, Das Josephus-Problem, LOG IN, Heft Nr. 165, pp. 70-71, 2010 (in German).
Ph. Dumas, Algebraic aspects of B-regular series. [Broken link]
Philippe Dumas, Algebraic aspects of B-regular series, Research Report, RR-1931, INRIA, 1993.
Ph. Dumas, Algebraic aspects of B-regular series, in: International Colloquium on Automata, Languages and Programming, ICALP 1993 (A. Lingas, R. Karlsson, S. Carlsson, eds.), pp. 457-468, Lecture Notes in Computer Science, vol. 700, Springer, Berlin, 1993.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, J. Théor. Nombres Bordeaux 9 (1997), no. 2, 303-318.
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.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
FORMULA
a(n) = 3*n + 1 - floor(K(3)*(3/2)^(ceiling(log((2*n+1)/K(3))/log(3/2)))) where K(3) = (3/2)*K = 1.622270502884767... (K is the constant described in A061419); a(n) = 3n + 1 - A061419(k+1) where A061419(k+1) is the least integer such that A061419(k+1) > 2n.
a(1) = 1 and, for n > 1, a(n) = (a(n-1) + 3) mod n, if this value is nonzero, n otherwise.
a(n) = (a(n-1) + 2) mod n + 1. - Paul Weisenhorn, Oct 10 2010
EXAMPLE
a(5) = 4 because the elimination process gives (1^,2,3,4,5) -> (1,2,4^,5) -> (2^,4,5) -> (2^,4) -> (4), where ^ denotes the counting reference position.
a(13) = 13 => a(14) = (a(13) + 2) mod 14 + 1 = 2. - Paul Weisenhorn, Oct 10 2010
MATHEMATICA
(* First do *) Needs["Combinatorica`"] (* then *) f[n_] := Last@ InversePermutation@ Josephus[n, 3]; Array[f, 70] (* Robert G. Wilson v, Jul 31 2010 *)
Table[Nest[Rest@RotateLeft[#, 2] &, Range[n], n - 1], {n, 72}] // Flatten (* Arkadiusz Wesolowski, Jan 14 2013 *)
CROSSREFS
Cf. A181281 (with s=5). - Paul Weisenhorn, Oct 10 2010
KEYWORD
nonn
AUTHOR
John W. Layman, May 30 2000
STATUS
approved
Array T(n,k) read by antidiagonals: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, k >= 1.
+10
6
1, 1, 2, 1, 1, 3, 1, 2, 3, 4, 1, 1, 2, 1, 5, 1, 2, 2, 1, 3, 6, 1, 1, 1, 2, 4, 5, 7, 1, 2, 1, 2, 1, 1, 7, 8, 1, 1, 3, 3, 2, 5, 4, 1, 9, 1, 2, 3, 2, 4, 1, 2, 7, 3, 10, 1, 1, 2, 3, 4, 4, 6, 6, 1, 5, 11, 1, 2, 2, 3, 1, 5, 3, 3, 1, 4, 7, 12, 1, 1, 1, 4, 2, 3, 5, 1, 8, 5, 7, 9, 13
OFFSET
1,3
COMMENTS
Arrange 1, 2, 3, ..., n clockwise in a circle. Starting the count at 1, delete every k-th integer clockwise until only one remains, which is T(n,k).
The main diagonal (1, 1, 2, 2, 2, 4, 5, 4, ...) is A007495.
Concatenation of consecutive rows (up to the main diagonal) gives A032434.
The periods of the rows, (1, 2, 6, 12, 60, 60, 420, 840, ...), is given by A003418.
FORMULA
T(1,k) = 1; for n > 1: T(n,k) = ((T(n-1,k) + k - 1) mod n) + 1.
EXAMPLE
.n\k 1 2 3 4 5 6 7 8 9 10
----------------------------------
.1 | 1 1 1 1 1 1 1 1 1 1
.2 | 2 1 2 1 2 1 2 1 2 1
.3 | 3 3 2 2 1 1 3 3 2 2
.4 | 4 1 1 2 2 3 2 3 3 4
.5 | 5 3 4 1 2 4 4 1 2 4
.6 | 6 5 1 5 1 4 5 3 5 2
.7 | 7 7 4 2 6 3 5 4 7 5
.8 | 8 1 7 6 3 1 4 4 8 7
.9 | 9 3 1 1 8 7 2 3 8 8
10 | 10 5 4 5 3 3 9 1 7 8
MATHEMATICA
T[n_, k_] := T[n, k] = If[n == 1, 1, Mod[T[n-1, k]+k-1, n]+1];
Table[T[n-k+1, k], {n, 1, 13}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 04 2023 *)
CROSSREFS
Cf. A000027 (k = 1), A006257 (k = 2), A054995 (k = 3), A088333 (k = 4), A181281 (k = 5), A360268 (k = 6), A178853 (k = 7), A109630 (k = 8).
Cf. A003418, A007495 (main diagonal), A032434, A198788, A198790.
KEYWORD
nonn,easy,tabl
AUTHOR
STATUS
approved
Array T(k,n) read by descending antidiagonals: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, k >= 1.
+10
4
1, 2, 1, 3, 1, 1, 4, 3, 2, 1, 5, 1, 2, 1, 1, 6, 3, 1, 2, 2, 1, 7, 5, 4, 2, 1, 1, 1, 8, 7, 1, 1, 2, 1, 2, 1, 9, 1, 4, 5, 2, 3, 3, 1, 1, 10, 3, 7, 2, 1, 4, 2, 3, 2, 1, 11, 5, 1, 6, 6, 4, 4, 3, 2, 1, 1, 12, 7, 4, 1, 3, 3, 5, 1, 3, 2, 2, 1, 13, 9, 7, 5, 8, 1, 5, 3
OFFSET
1,2
COMMENTS
Arrange 1, 2, 3, ... n clockwise in a circle. Starting the count at 1, delete every k-th integer clockwise until only one remains, which is T(k,n).
The main diagonal of the array (1, 1, 2, 2, 2, 4, 5, 4, ...) is A007495.
Consecutive columns down to the main diagonal (1, 2, 1, 3, 3, 2, 4, 1, 1, 2, ...) is A032434.
Period lengths of columns (1, 2, 6, 12, 60, 60, 420, 840, ...) is A003418.
FORMULA
T(k,1) = 1;
for n > 1: T(k,n) = ((T(k,n-1) + k - 1) mod n) + 1.
EXAMPLE
.k\n 1 2 3 4 5 6 7 8 9 10
----------------------------------
.1 | 1 2 3 4 5 6 7 8 9 10 A000027
.2 | 1 1 3 1 3 5 7 1 3 5 A006257
.3 | 1 2 2 1 4 1 4 7 1 4 A054995
.4 | 1 1 2 2 1 5 2 6 1 5 A088333
.5 | 1 2 1 2 2 1 6 3 8 3 A181281
.6 | 1 1 1 3 4 4 3 1 7 3
.7 | 1 2 3 2 4 5 5 4 2 9 A178853
.8 | 1 1 3 3 1 3 4 4 3 1 A109630
.9 | 1 2 2 3 2 5 7 8 8 7
10 | 1 1 2 4 4 2 5 7 8 8
CROSSREFS
Cf. A000027 (k = 1), A006257 (k = 2), A054995 (k = 3), A088333 (k = 4), A181281 (k = 5), A178853 (k = 7), A109630 (k = 8).
Cf. A003418, A007495 (main diagonal), A032434, A198789, A198790.
KEYWORD
nonn,easy,tabl
AUTHOR
STATUS
approved
Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the Josephus elimination process for n people and a count of 2, 1 <= k <= n.
+10
3
1, 2, 1, 2, 1, 3, 2, 4, 3, 1, 2, 4, 1, 5, 3, 2, 4, 6, 3, 1, 5, 2, 4, 6, 1, 5, 3, 7, 2, 4, 6, 8, 3, 7, 5, 1, 2, 4, 6, 8, 1, 5, 9, 7, 3, 2, 4, 6, 8, 10, 3, 7, 1, 9, 5, 2, 4, 6, 8, 10, 1, 5, 9, 3, 11, 7, 2, 4, 6, 8, 10, 12, 3, 7, 11, 5, 1, 9, 2, 4, 6, 8, 10, 12, 1, 5, 9, 13, 7, 3, 11, 2, 4, 6, 8, 10, 12, 14
OFFSET
1,2
COMMENTS
In the Josephus elimination process for n and k, the numbers 1 through n are written in a circle. A pointer starts at position 1. Each turn, the pointer skips (k-1) non-eliminated number(s) going around the circle and eliminates the k-th number, until no numbers remain. This sequence represents the triangle J(n, i), where n is the number of people in the circle, i is the turn number, and k is fixed at 2 (every other number is eliminated).
FORMULA
From Pontus von Brömssen, Sep 18 2022: (Start)
The terms are uniquely determined by the following recursive formulas:
T(n,k) = 2*k if k <= n/2;
T(2*n,k) = 2*T(n,k-n)-1 if k > n;
T(2*n+1,k) = 2*T(n,k-n-1)+1 if k > n+1;
T(2*n+1,n+1) = 1.
(End)
EXAMPLE
Triangle begins:
1;
2, 1;
2, 1, 3;
2, 4, 3, 1;
2, 4, 1, 5, 3;
2, 4, 6, 3, 1, 5;
2, 4, 6, 1, 5, 3, 7;
2, 4, 6, 8, 3, 7, 5, 1;
2, 4, 6, 8, 1, 5, 9, 7, 3;
2, 4, 6, 8, 10, 3, 7, 1, 9, 5;
2, 4, 6, 8, 10, 1, 5, 9, 3, 11, 7;
2, 4, 6, 8, 10, 12, 3, 7, 11, 5, 1, 9;
2, 4, 6, 8, 10, 12, 1, 5, 9, 13, 7, 3, 11;
...
For n = 5, to get the entries in 5th row from left to right, start with (^1, 2, 3, 4, 5) and the pointer at position 1, indicated by the caret. 1 is skipped and 2 is eliminated to get (1, ^3, 4, 5). (The pointer moves ahead to the next "live" number.) On the next turn, 3 is skipped and 4 is eliminated to get (1, 3, ^5). Then 1, 5, and 3 are eliminated in that order (going through (^3, 5) and (^3)). This gives row 5 of the triangle and entries a(11) through a(15) in this sequence.
MATHEMATICA
Table[Rest@ Nest[Append[#1, {Delete[#2, #3 + 1], #2[[#3 + 1]], #3}] & @@ {#, #[[-1, 1]], Mod[#[[-1, -1]] + 1, Length@ #[[-1, 1]]]} &, {{Range@ n, 0, 0}}, n][[All, 2]], {n, 14}] // Flatten (* Michael De Vlieger, Nov 13 2018 *)
PROG
(Python)
def A321298(n, k):
if 2*k<=n: return 2*k
n2, r=divmod(n, 2)
if r==0: return 2*A321298(n2, k-n2)-1
if k==n2+1: return 1
return 2*A321298(n2, k-n2-1)+1 # Pontus von Brömssen, Sep 18 2022
CROSSREFS
The right border of this triangle is A006257.
KEYWORD
easy,nonn,tabl
AUTHOR
Zeph L. Turner, Nov 02 2018
EXTENSIONS
Name clarified by Pontus von Brömssen, Sep 18 2022
STATUS
approved
A version of the Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer 5 places clockwise from i. Repeat, counting 5 places from the next undeleted integer, until only one integer remains.
+10
1
1, 1, 1, 3, 4, 4, 3, 1, 7, 3, 9, 3, 9, 1, 7, 13, 2, 8, 14, 20, 5, 11, 17, 23, 4, 10, 16, 22, 28, 4, 10, 16, 22, 28, 34, 4, 10, 16, 22, 28, 34, 40, 3, 9, 15, 21, 27, 33, 39, 45, 51, 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 1, 7, 13, 19, 25
OFFSET
1,4
FORMULA
a(n) = (a(n-1) + 5) mod n + 1 if n > 1, a(1) = 1.
EXAMPLE
a(7) = 3 because the elimination process gives (^1,2,3,4,5,6,7) -> (1,2,3,4,5,^7) -> (1,2,3,4,^7) -> (^1,2,3,4) -> (1,^3,4) -> (^3,4) -> (3), where ^ denotes the counting reference position.
a(13) = 9 => a(14) = (a(13) + 5) mod 14 + 1 = 1.
MATHEMATICA
a[1] = 1; a[n_] := a[n] = Mod[a[n - 1] + 5, n] + 1; Array[a, 100] (* Amiram Eldar, Feb 03 2023 *)
PROG
(Python)
def A360268_up_to_n(n):
val = 1
return [val := (val + 5) % i + 1 for i in range(1, n+1)]
CROSSREFS
6th column of A198789.
KEYWORD
nonn
AUTHOR
Benjamin Lilley, Jan 31 2023
STATUS
approved

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