A007495 -id:A007495 - cp's OEIS frontend

A054995 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.

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

John W. Layman, May 30 2000

nonn

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.

			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

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.
Index entries for sequences related to the Josephus Problem

Cf. A032434, A005427, A005428, A006257, A007495, A000960, A056530.

Cf. A181281 (with s=5). - Paul Weisenhorn, Oct 10 2010

(* 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 *)

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

A032434 Triangle read by rows: last survivors of Josephus elimination process.

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 4, 1, 1, 2, 5, 3, 4, 1, 2, 6, 5, 1, 5, 1, 4, 7, 7, 4, 2, 6, 3, 5, 8, 1, 7, 6, 3, 1, 4, 4, 9, 3, 1, 1, 8, 7, 2, 3, 8, 10, 5, 4, 5, 3, 3, 9, 1, 7, 8, 11, 7, 7, 9, 8, 9, 5, 9, 5, 7, 7, 12, 9, 10, 1, 1, 3, 12, 5, 2, 5, 6, 11, 13, 11, 13, 5, 6, 9, 6, 13, 11, 2, 4, 10, 8, 14, 13, 2, 9

Offset: 1

N. J. A. Sloane

T(n,k) 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 k - 1 places clockwise from i. Repeat, counting k - 1 places from the next undeleted integer, until only one integer remains. [After John W. Layman]
From Gerhard Kirchner, Jan 08 2017: (Start)
The fast recurrence is useful for large n, if single values of T(n,k) are to be determined (not the whole sequence). The number of steps is about log(n)/log(n/(k/(k-1))) instead of n, i.e., many basic steps are bypassed by a shortcut.
Example: For computing T(10^80, 7), about 1200 steps are needed, done in a second, whereas even the age of the universe would not be sufficient for the basic recurrence. Deduction of the fast recurrence and the reason for its efficiency: See the link "Fast recurrence". (End)

			Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins
  1;
  2, 1;
  3, 3, 2;
  4, 1, 1, 2;
  5, 3, 4, 1, 2;
  6, 5, 1, 5, 1, 4;
  7, 7, 4, 2, 6, 3, 5;
  ...
Fast recurrence for n = 7 and k = 3:
m =    1 2 3 4 5 6,
z(m) = 1 2 3 4 6 9,
r(m) = 1 2 2 1 1,
z(6) > n => M = 5.
Result: T(7,3) = r(5) + 3*(n - z(5)) = 4.

W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th ed. New York: Dover, 1987, see pp. 32-36.
M. Kraitchik, "Josephus' Problem." Sec. 3.13 in Mathematical Recreations. New York: W. W. Norton, pp. 93-94, 1942.

T. D. Noe, Rows n = 1..50, flattened
Philippe Dumas, Algebraic aspects of B-regular series, inria-00074743, 1993.
Philippe Dumas, Algebraic aspects of B-regular series, in: A. Lingas, R. Karlsson, S. Carlsson S. (eds), Automata, Languages and Programming, ICALP 1993 (Lecture Notes in Computer Science, vol 700, Springer, Berlin, Heidelberg), pp. 457-468, 1993.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, preprint, 1997.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, preprint, 1997.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, Journal de Théorie des Nombres de Bordeaux 9 (1997), 303-318.
Gerhard Kirchner, Fast recurrence.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, preprint, 1991. [Cached copy, with permission]
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33 (1991), 235-240.
Eric Weisstein's World of Mathematics, Josephus Problem.
Wikipedia, Josephus problem.
Index entries for sequences related to the Josephus problem

Cf. A032435, A032436, A321781.

Second column is A006257, third column is A054995. Diagonal T(n, n) is A007495.

t[1, k_] = 1; t[n_, k_] := t[n, k] = If[m = Mod[t[n-1, k] + k, n]; m != 0, m, n]; Flatten[ Table[ t[n, k], {n, 1, 14}, {k, 1, n}]] (* Jean-François Alcover, Sep 25 2012 *)

T(n,k)=local(t): if(n<2,n>0,t=(T(n-1,k)+k)%n: if(t,t,n))

Recurrence: T(1, k) = 1, T(n, k) = (T(n-1, k) + k) mod n if this is nonzero and n if not.
From Gerhard Kirchner, Jan 08 2017: (Start)
The same recurrence without these conditions:
T(1, k) = 1,  T(n, k) = 1 + (T(n-1, k) + k - 1) mod n.
This "basic" recurrence is used in the following:
Fast recurrence (n >= k):
z(1) = 1, r(1) = 1,
if z(m) < k-1 then
  z(m+1) = z(m) + 1,
  r(m+1) = 1 + (r(m)+k-1) mod z(m+1) (basic recurrence),
else
  e(m) = -z(m) mod (k-1),
  if r(m) + e(m) <= 0 then e(m) -> e(m) + k - 1,
  z(m+1) = z(m) + (z(m) + e(m))/(k-1),
  r(m+1) = r(m) + e(m).
Result for m = M with z(M) <= n < z(M+1):
T(n,k) = r(M) + k(n - z(M)). (End)
From Gerhard Kirchner, Jan 12 2017: (Start)
Another fast (and shorter) recurrence is given in "Functional iteration and the Josephus problem", p. 1, (see link):
D(m,k) = ceiling(k/(k-1)*D(m-1,k)), m >= 1; D(0,k)=1.
Result for m with D(m-1,k) <= (k-1)*n < D(m,k):
T(n,k) = k*n + 1 - D(m,k). (End)

Edited by Ralf Stephan, May 18 2004

A088333 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 3 places clockwise from i. Repeat, counting 3 places from the next undeleted integer, until only one integer remains.

Original entry on oeis.org

1, 1, 2, 2, 1, 5, 2, 6, 1, 5, 9, 1, 5, 9, 13, 1, 5, 9, 13, 17, 21, 3, 7, 11, 15, 19, 23, 27, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42

Offset: 1

N. J. A. Sloane, Nov 13 2003

If one counts only one place (resp. two places) at each stage to determine the element to be deleted, we get A006257 (resp. A054995).

See A054995 for references and links.

Index entries for sequences related to the Josephus Problem

Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.

It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n-1)+4) mod n. The conjecture is false; counterexample: a(21)=21; a(20)=17; (a(20)+4)mod 21=0; corrected formula: a(n)=(a(n-1)+3) mod n +1;
The conjecture is true. After removing the 4th number, we are reduced to the n-1 case, but starting with 5 instead of 1. - David Wasserman, Aug 08 2005
a(n) = A032434(n,4) if n>=4. - R. J. Mathar, May 04 2007

More terms from David Wasserman, Aug 08 2005

A198789 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.

Original entry on oeis.org

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

William Rex Marshall, Nov 21 2011

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.

			.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

William Rex Marshall, First 141 antidiagonals of array, flattened
Index entries for sequences related to the Josephus Problem

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.

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 *)

T(1,k) = 1; for n > 1: T(n,k) = ((T(n-1,k) + k - 1) mod n) + 1.

A128982 If in a line of n persons every n-th person is eliminated until only one person is left, which position P should one assume in the original lineup to avoid being eliminated?

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 2, 6, 6, 10, 2, 12, 2, 6, 8, 16, 2, 18, 2, 16, 18, 22, 2, 22, 12, 16, 8, 28, 2, 30, 2, 28, 18, 22, 12, 36, 2, 6, 8, 40, 2, 42, 2, 30, 42, 46, 2, 42, 14, 40, 30, 52, 2, 36, 24, 52, 54, 58, 2, 60, 2, 6, 30, 48, 24, 66, 2, 30, 18, 70, 2, 72, 2, 6, 20, 60, 18, 78, 2, 72, 78

Offset: 1

Harri Aaltonen, Apr 30 2007

nonn

The difference between this, A007495 and the diagonal of A032434 is that for each of the n-1 elimination processes, counting from 1 to n starts at the lowest position in the line that is still occupied, not right after the most recently eliminated position. Wrapping around when n exceeds the number of residual occupied positions still occurs in circular fashion as in the original Josephus problem. - R. J. Mathar, May 07 2007

			Elimination at n=6: 1,2,3,4,5,6 -> 1,2,3,4,5 -> 2,3,4,5 -> 2,4,5 -> 2,4 -> 2. After the 3 is eliminated, counting does not start at 4 but again at 2.

Hagen von Eitzen, Table of n, a(n) for n = 1..10000
Index entries for sequences related to the Josephus Problem

Cf. A007495, A032434.

int a(int n) { if (n<3) return 1; int L=1, R=n-1, M, t, s, q, r; while (R>L+1) { s = M = (L+R)/2; t= n-1; while (s && sHagen von Eitzen, Nov 08 2022 */

A128982 := proc(n) local l ; l := [seq(i,i=1..n)] ; for i from 1 to n-1 do rm := ((n-1) mod nops(l))+1 ; l := subsop(rm=NULL,l) ; od ; RETURN(op(1,l)) ; end: for n from 1 to 85 do printf("%d, ",A128982(n)) ; od ; # R. J. Mathar, May 07 2007

a[n_] := Module[{l = Range[n]}, Do[l = Delete[l, Mod[n-1, Length[l]]+1], {n-1}]; If[l == {}, Nothing, l[[1]]]];
a /@ Range[0, 100] (* Jean-François Alcover, Apr 01 2020 *)

If n is prime then P = n - 1. If n is prime + 1 then P = 2.

This is a version of the Josephus problem. Several other versions are already in the OEIS. - N. J. A. Sloane, May 01 2007

Corrected and extended by R. J. Mathar, May 07 2007

A198788 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.

Original entry on oeis.org

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

William Rex Marshall, Nov 21 2011

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.

			.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

William Rex Marshall, First 141 antidiagonals of array, flattened
Index entries for sequences related to the Josephus Problem

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.

T(k,1) = 1;

for n > 1: T(k,n) = ((T(k,n-1) + k - 1) mod n) + 1.

A198790 Irregular table T(n,k) read by rows: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, lcm(1, 2, 3, ..., n) >= k >= 1.

Original entry on oeis.org

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

Offset: 1

William Rex Marshall, Nov 21 2011

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).

In the full table in A198789, row n repeats with a periodicity of lcm(1, 2, 3, ..., n) = A003418(n). This sequence is a scan of each row in A198789 for exactly one period length.

			n\k  1  2  3  4  5  6  7  8  9 10 11 12
---------------------------------------
1 |  1
2 |  2  1
3 |  3  3  2  2  1  1
4 |  4  1  1  2  2  3  2  3  3  4  4  1

William Rex Marshall, Rows n = 1..10, flattened
Index entries for sequences related to the Josephus Problem

Cf. A003418, A007495, A032434, A198788, A198789.

T(1,1) = 1;

for n >= 2, lcm(1, 2, ... n) >= k >=1: T(n,k) = ((T(n-1,((k-1) mod lcm(1, 2, ... n-1)) + 1) + k - 1) mod n) + 1.

A348533 Generalized Josephus problem: Let T(m,k), k>=2, m=1,2,3,.., be the number of people on a circle such that the survivor is one of the first k-1 people after every k-th person has been removed.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 8, 3, 2, 1, 16, 4, 3, 2, 1, 32, 6, 4, 3, 2, 1, 64, 9, 5, 4, 3, 2, 1, 128, 14, 7, 5, 4, 3, 2, 1, 256, 21, 9, 6, 5, 4, 3, 2, 1, 512, 31, 12, 8, 6, 5, 4, 3, 2, 1, 1024, 47, 16, 10, 7, 6, 5, 4, 3, 2, 1, 2048, 70, 22, 12, 8, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Gerhard Kirchner, Oct 21 2021

The table, see example, is read by ascending antidiagonals.

Trivial cases: T(m,k)=m for m

The recurrence in the formula section does not only yield T(m,k), but also the survivor's number S(m,k) so that the Josephus problem can be solved for any number N of people, especially for large N because T(m,k) grows exponentially, see link "Derivation of the recurrence", section II.

T(m,k) compared with other sequences ("->" means that the sequences can be made equal by removing repeated terms, see link "Derivation of the recurrence", section IV).

T(m,2) = A000079(m)=2^(m-1)

T(m,3) -> A073941

T(m,4) -> A072493

T(m+4,4)= A005427(m)

T(m,5) -> A120160

T(m,6) -> A120170

T(m,7) -> A120178

T(m,8) -> A120186

T(m,9) -> A120194

T(m,10)-> A120202

			k=4: 7 people, survivors number 2 <4.
k=4: 6 people, survivors number 5>=4, counterexample.
Table T(m,k) begins:
  m\k____2____3____4____5
   1:    1    1    1    1
   2:    2    2    2    2
   3:    4    3    3    3
   4:    8    4    4    4
   5:   16    6    5    5
   6:   32    9    7    6
   7:   64   14    9    8
   8:  128   21   12   10
   9:  256   31   16   12
  10:  512   47   22   15

Gerhard Kirchner, Derivation of the recurrence
Gerhard Kirchner, Table for 2<=k<=10 and 1<=m<=30
Index entries for sequences related to the Josephus Problem

Cf. A005428, A032434, A054995, A007495.

block(k:10, mmax:30, t:1, s:1, T:[1],
/*Terms T(m,k), m=1 thru mmax */
for m from 1 thru mmax-1 do(
    p:  mod(t, k-1),
    if s>p then e:-p else e:k-1-p,
    t: (k*t+e)/(k-1), s: 1+mod(s+e-1, t),
    T:append(T,[t])),
return (T));

Recurrence for T(m,k) and S(m,k), the survivor's number.
Start: T(1,k)=S(1,k)=1.
T(m+1,k)=(k*T(m,k)+e)/(k-1),
S(m+1,k)=1 + (S(m,k)+e-1) mod T(m+1,k),
with e=-p if S(m,k)>p and e=k-1-p otherwise, p = T(m,k) mod (k-1).

A128529 Survivor of the Josephus problem, counting direction reversed after each step.

Original entry on oeis.org

1, 1, 1, 1, 3, 4, 1, 3, 5, 1, 9, 8, 3, 3, 11, 1, 15, 7, 7, 18, 19, 16, 3, 7, 15, 24, 25, 18, 9, 28, 19, 24, 7, 13, 21, 5, 31, 20, 11, 15, 21, 32, 3, 11, 31, 7, 39, 23, 25, 15, 35, 1, 47, 32, 15, 54, 55, 48, 9, 19, 39, 60, 59, 58, 63, 7, 49, 50, 11, 40, 27, 70, 63, 48, 23, 27, 47, 74, 67

Offset: 1

R. J. Mathar, May 07 2007

nonn

As in A007495, counting for elimination starts clockwise for the first elimination, then continues counterclockwise from the eliminated place for the second, then toggles again to clockwise for the third elimination and changes direction in that manner after each elimination. Sequence shows original place of the survivor.

			n=5 start with 1,2,3,4,5, count upwards to eliminate 5: 1,2,3,4. Count backwards from 4 over 3 over 2 over 1 to eliminate 4: 1,2,3. Then count forwards from 1 (wrapping around and upwards of 4) over 2 etc. to eliminate 2: 1,3. Count backwards starting at 1 (left of eliminated 2) to eliminate 1 and to leave a(5)=3.

Index entries for sequences related to the Josephus Problem

Cf. A007495, A128982.

a := proc(n) local l,dir,pos,i,c ; dir := 1 ; pos := 0 ; l := [seq(i,i=1..n)] ; for i from 1 to n-1 do pos := pos+n*dir ; pos := 1+((pos-1) mod nops(l)) ; l := subsop(pos=NULL,l) ; dir := -dir ; if dir > 0 then pos := pos-1 ; fi ; od ; RETURN(op(1,l)) ; end: for n from 1 to 85 do printf("%d, ",a(n)) ; od;

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A032434 Triangle read by rows: last survivors of Josephus elimination process.

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A198789 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.

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A128982 If in a line of n persons every n-th person is eliminated until only one person is left, which position P should one assume in the original lineup to avoid being eliminated?

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A198788 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.

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A198790 Irregular table T(n,k) read by rows: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, lcm(1, 2, 3, ..., n) >= k >= 1.

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A348533 Generalized Josephus problem: Let T(m,k), k>=2, m=1,2,3,.., be the number of people on a circle such that the survivor is one of the first k-1 people after every k-th person has been removed.

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