A072493 -id:A072493 - cp's OEIS frontend

A005427 Josephus problem: numbers m such that, when m people are arranged on a circle and numbered 1 through m, the final survivor when we remove every 4th person is one of the first three people.

5, 7, 9, 12, 16, 22, 29, 39, 52, 69, 92, 123, 164, 218, 291, 388, 517, 690, 920, 1226, 1635, 2180, 2907, 3876, 5168, 6890, 9187, 12249, 16332, 21776, 29035, 38713, 51618, 68824, 91765, 122353, 163138, 217517, 290023, 386697, 515596, 687461, 916615, 1222153, 1629538, 2172717, 2896956, 3862608, 5150144, 6866859, 9155812, 12207749, 16276999, 21702665, 28936887, 38582516, 51443354

Offset: 1

N. J. A. Sloane, Simon Plouffe

nonn

Is this the same as A072493 with its first 8 terms removed? See also the similar conjecture concerning A005428 and A073941.
From Petros Hadjicostas, Jul 20 2020: (Start)
We describe the counting-off game of Burde (1987) using language from Schuh (1968). Suppose m people are labeled with the numbers 1 through m (say clockwise). (Burde uses the numbers 0 through m-1 probably because he relates this problem to the representation of m in the fractional base k/(k-1) = 4/3. He actually modifies the (4/3)-representation of m to include negative coefficients. See the coefficients f(n;k) below.)
Suppose we start the counting at the person labeled 1, and we remove every 4th person. This sequence gives those numbers m for which the last survivor is one of the first three people.
When m = 5, 9, 12, 16, 218, 517, ... the last survivor is the first person.
When m = 7, 29, 69, 92, 291, 388, ... the last survivor is the second person.
When m = 22, 39, 52, 123, 164, 690, ... the last survivor is the third person.
If we know m = a(n) and the number, say i(n), of the last survivor (when there are a(n) people on the circle), we may find a(n+1) and the number i(n+1) of the new last survivor (when there are a(n+1) people on the circle) in the following way:
(a) If 0 = a(n) mod 3, then a(n+1) = (4/3)*a(n), and i(n+1) = i(n).
(b) If 1 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.
(c) If 1 = a(n) mod 3 and i(n) = 2, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1.
(d) If 1 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 2.
(e) If 2 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 2.
(f) If 2 = a(n) mod 3 and i(n) = 2, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.
(g) If 2 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1. (End)
From Petros Hadjicostas, Jul 22 2020: (Start)
In general, for k >= 2, it seems that when m people are placed on a circle, labeled 1 through m, and every k-th person is removed (starting the counting at person 1), we may determine those m for which the last survivor is in {1, 2, ..., k-1} in the following way.
Define the sequence (T(n;k): n >= 1) by T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)) for n >= 2 starting with T(1; k) = 1. Then the list of those m's for which the last survivor is in {1, 2, ..., k-1} consists of all the numbers T(n;k) >= k (thus, we exclude the cases m = 1, ..., k-1 that may be repeated more than once in the sequence (T(n;k): n >= 1)).
I do not have a general proof of this conjecture though I strongly believe that Schuh's (1968) way of solving the case k = 3 (see pp. 373-375 and 377-379, where he provides two methods of solution) may provide clues for proving the conjecture.
We have T(n; k=2) = A011782(n+1), T(n; k=3) = A073941(n), T(n; k=4) = A072493(n), T(n; k=5) = A120160(n), T(n; k=6) = A120170(n), T(n; k=7) = A120178(n), T(n; k=8) = A120186(n), T(n; k=9) = A120194(n), and T(n; k=10) = A120202(n).
We also have T(n+1;k) = floor((k/(k-1))*T(n;k)) or ceiling((k/(k-1)*T(n;k)).
To identify the last survivor that results when we place T(n; k) people on the circle (with T(n;k) >= k) in the above Josephus problem, we use a modification of Burde's algorithm due to Thériault (2000).
We use the following recursions but we start at T(k;k) (rather than at the smallest n for which T(n;k) >= k). Define the sequence (S(n;k): n >= 1) by S(n;k) = T(n+k-1; k) for n >= 1. (It is easy to prove that S(1;k) = T(k;k) = 1.)
Define also the sequences (j(n;k): n >= 1) and (f(n;k): n >= 1) by j(1;k) = 1, f(1;k) = 0, f(n+1;k) = ((j(n;k) - S(n;k) - 1) mod (k-1)) + 1 - j(n;k) and j(n+1;k) = j(n;k) + f(n+1;k) for n >= 2.
Then for all n s.t. S(n;k) >= k, j(n;k) is the number of the last survivor of the Josephus problem where every k-th person is removed (provided we start the counting at number 1). It will always be the case that j(n;k) is in {1,2,...,k-1}.
We actually have S(n+1; k) = (k*S(n;k) + f(n+1;k))/(k-1) for n >= 1.
Notice that the Burde-Thériault algorithm is a generalization of Schuh's method. (End)

			From _Petros Hadjicostas_, Jul 22 2020: (Start)
We explain why 5 and 7 are in the sequence but 6 is not.
If we put m = 5 people on the circle, label them 1 through 5, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 3 -> 5 -> 2. Thus, the last survivor is 1, so m = 5 is included in this sequence.
If we put m = 6 people on a circle, label them 1 through 6, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 2 -> 1 -> 3 -> 6. Thus, the last survivor is 5 (not 1, 2, or 3), so m = 6 is not included in this sequence.
If we put m = 7 people on a circle, label them 1 through 7, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 1 -> 6 -> 5 -> 7 -> 3. Thus, the last survivor is 2, so m = 7 is included in this sequence.
Strictly speaking, m = 2 and m = 3 should have been included as well (since clearly the last survivor would be 1 or 2 or 3). In addition, m = 4 should have been included as well because the list of people removed is 4 -> 1 -> 3. The case of number 1 does create a problem since there is no survivor. Note that the numbers 1, 2, 3, 4 are all included in A072493. (End)

Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968. [This book is cited in Burde (1987). Table 18, p. 374, is related to a very similar sequence (A073941). Thus, definitely, the counting-off games described in the book are related to a similar counting-off game in Burde (1987).]
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
K. Burde, Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases], J. Number Theory 26(2) (1987), 192-209.
Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).
Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).
Index entries for sequences related to the Josephus Problem

Cf. A005428, A072493.

Similar sequences: A011782 (k = 2), A073941 (k = 3), A072493 (k = 4), A120160 (k = 5), A120170 (k = 6), A120178 (k = 7), A120186 (k = 8), A120194 (k = 9), A120202 (k = 10).

f[s_] := Append[s, Ceiling[5 + Plus@@(s/3)]]; Nest[f, {5}, 100] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)

/* Gives an n X 2 matrix w s.t. w[,1] are the terms of this sequence and w[,2] are the corresponding numbers of the last survivors (1, 2 or 3). */
lista(nn) = {my(w = matrix(nn,2)); w[1,1] = 5; w[1,2] = 1; for(n=1, nn-1,
if(0 == w[n,1] % 3, w[n+1,1] = w[n,1]*4/3; w[n+1,2] = w[n,2]);
if(1 == w[n,1] % 3 && w[n,2] == 1, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 2);
if(1 == w[n,1] % 3 && w[n,2] == 2, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 1);
if(1 == w[n,1] % 3 && w[n,2] == 3, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 1);
if(2 == w[n,1] % 3 && w[n,2] == 1, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 1);
if(2 == w[n,1] % 3 && w[n,2] == 2, w[n+1,1] = ceil(w[n,1]*4/3);  w[n+1,2] = w[n,2] + 1);
if(2 == w[n,1] % 3 && w[n,2] == 3, w[n+1,1] = floor(w[n,1]*4/3); w[n+1,2] = w[n,2] - 2);
); Vec(w[,1]);} \\ Petros Hadjicostas, Jul 21 2020

/* Second PARI program for the general case of Josephus problem. We use the Burde-Thériault algorithm, not the formula T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)). We start with T(k;k) = 1 (and omit all previous 1's). Burde starts with the smallest T(n;k) >= k whose corresponding last survivor is 1. This, however, can be very large. To get the corresponding last survivors, modify the program to get the vector j. */
lista(nn,k) = {my(j=vector(nn)); my(f=vector(nn)); my(N=vector(nn));
j[1]=1; f[1]=0; N[1] = 1;
for(n=1, nn-1, f[n+1] = ((j[n]-N[n]-1) % (k-1)) + 1 - j[n];
j[n+1] = j[n] + f[n+1]; N[n+1] = (k*N[n] + f[n+1])/(k-1););
for(n=1, nn, if(N[n] > k-1, print1(N[n],",")));} \\ Petros Hadjicostas, Jul 23 2020

a(n) = 5 + ceiling(Sum_{k=1..n-1} a(k)/3). - Petros Hadjicostas, Jul 21 2020

More terms (from the Burde paper, p. 208) from R. J. Mathar, Sep 26 2006

Name edited by Petros Hadjicostas, Jul 20 2020

A087192 a(n) = ceiling(a(n-1)*4/3), with a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 15, 20, 27, 36, 48, 64, 86, 115, 154, 206, 275, 367, 490, 654, 872, 1163, 1551, 2068, 2758, 3678, 4904, 6539, 8719, 11626, 15502, 20670, 27560, 36747, 48996, 65328, 87104, 116139, 154852, 206470, 275294, 367059, 489412, 652550

Offset: 1

Paul D. Hanna, Aug 24 2003

nonn

If you repeatedly base 64 encode a string, starting with a single character, the length of the string at step n is 4*a(n). - Christian Perfect, Jan 06 2016

Robert Israel, Table of n, a(n) for n = 1..7994

Cf. A072493, A087165, A155167.

[n eq 1 select 1 else Ceiling(Self(n-1)*4/3): n in [1..50]]; // Vincenzo Librandi, Aug 17 2017

A[1]:= 1:
for n from 2 to 100 do A[n]:= ceil(4/3*A[n-1]) od:
seq(A[i],i=1..100); # Robert Israel, Aug 17 2017

a[1] = 1; a[n_] := a[n] = Ceiling[4 a[n - 1]/3]; Table[a@ n, {n, 45}] (* Michael De Vlieger, Jan 06 2016 *)

a(n) = if (n==1, 1, ceil(a(n-1)*4/3)) \\ Michel Marcus, Aug 01 2013

from fractions import Fraction
from functools import lru_cache
@lru_cache(maxsize=None)
def A087192(n): return int(Fraction(4*A087192(n-1),3)._ceil_()) if n>1 else 1 # Chai Wah Wu, Sep 07 2023

Partial sums of A072493. Also indices of records in A087165: A087165(a(n))=n.

A120169 a(n) = 12 + floor((1 + Sum_{j=1..n-1} a(j))/4).

Original entry on oeis.org

12, 15, 19, 23, 29, 36, 45, 57, 71, 89, 111, 139, 173, 217, 271, 339, 423, 529, 661, 827, 1033, 1292, 1615, 2018, 2523, 3154, 3942, 4928, 6160, 7700, 9625, 12031, 15039, 18798, 23498, 29372, 36715, 45894, 57368, 71710

Offset: 1

Graeme McRae, Jun 10 2006

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A072493, A073941, A112088.

Cf. A120149, A120160 - A120168, A120171.

function f(n, a, b)
  t:=0;
    for k in [1..n-1] do
      t+:= a+Floor((b+t)/4);
    end for;
  return t;
end function;
g:= func< n, a, b | f(n+1, a, b)-f(n, a, b) >;
A120169:= func< n | g(n, 12, 1) >;
[A120169(n): n in [1..60]]; // G. C. Greubel, Sep 09 2023

nxt[{t_,a_}]:=Module[{c=Floor[(t+49)/4]},{t+c,c}]; NestList[nxt,{12,12},40][[All,2]] (* Harvey P. Dale, Jun 21 2017 *)

@CachedFunction
def f(n, p, q): return p + (q +sum(f(k, p, q) for k in range(1, n)))//4
def A120169(n): return f(n, 12, 1)
[A120169(n) for n in range(1, 61)] # G. C. Greubel, Sep 09 2023

A120135 a(n) = 5 + floor((1 + Sum_{j=1..n-1} a(j)) / 2).

Original entry on oeis.org

5, 8, 12, 18, 27, 40, 60, 90, 135, 203, 304, 456, 684, 1026, 1539, 2309, 3463, 5195, 7792, 11688, 17532, 26298, 39447, 59171, 88756, 133134, 199701, 299552, 449328, 673992, 1010988, 1516482, 2274723, 3412084, 5118126, 7677189, 11515784

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134, A120136 - A120209.

a[n_]:= a[n]= 5 +Floor[(1+Sum[a[k], {k,n-1}])/2];
Table[a[n], {n,60}] (* G. C. Greubel, May 07 2023 *)

@CachedFunction
def A120135(n): return 5 + (1 + sum(A120135(k) for k in range(1,n)))//2
[A120135(n) for n in range(1,61)] # G. C. Greubel, May 07 2023

a(n) ~ c * (3/2)^n, where c = 3.514931952760438754899508881646642282344325354834703833076259269449577... - Vaclav Kotesovec, May 07 2023

A120136 a(n) = 7 + floor(Sum_{j=1..n-1} a(j) / 2).

Original entry on oeis.org

7, 10, 15, 23, 34, 51, 77, 115, 173, 259, 389, 583, 875, 1312, 1968, 2952, 4428, 6642, 9963, 14945, 22417, 33626, 50439, 75658, 113487, 170231, 255346, 383019, 574529, 861793, 1292690, 1939035, 2908552, 4362828, 6544242, 9816363, 14724545

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134, A120135, A120137 - A120209.

nxt[{t_,a_}] := Module[{c=7+Floor[t/2]},{t+c,c}];
NestList[nxt,{7,7},40][[All,2]] (* Harvey P. Dale, Jan 13 2017 *)

@CachedFunction
def A120136(n): return 7 +sum(A120136(k) for k in range(1,n))//2
[A120136(n) for n in range(1,60)] # G. C. Greubel, May 08 2023

A120137 a(n) = 8 + floor( (1 + Sum_{j=1..n-1} a(j)) / 2).

Original entry on oeis.org

8, 12, 18, 27, 41, 61, 92, 138, 207, 310, 465, 698, 1047, 1570, 2355, 3533, 5299, 7949, 11923, 17885, 26827, 40241, 60361, 90542, 135813, 203719, 305579, 458368, 687552, 1031328, 1546992, 2320488, 3480732, 5221098, 7831647, 11747471, 17621206

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134 - A120136, A120138 - A120209.

a[n_]:= a[n]= 8 +Floor[(1 +Sum[a[k], {k,n-1}])/2];
Table[a[n], {n,60}] (* G. C. Greubel, May 08 2023 *)
nxt[{t_,a_}]:=Module[{c=8+Floor[(1+t)/2]},{t+c,c}]; NestList[nxt,{8,8},40][[;;,2]] (* Harvey P. Dale, Sep 10 2023 *)

@CachedFunction
def A120137(n): return 8 +(1 +sum(A120137(k) for k in range(1,n)))//2
[A120137(n) for n in range(1,60)] # G. C. Greubel, May 08 2023

A120138 a(n) = 10 + floor(Sum_{j=1..n-1} a(j) / 2).

Original entry on oeis.org

10, 15, 22, 33, 50, 75, 112, 168, 252, 378, 567, 851, 1276, 1914, 2871, 4307, 6460, 9690, 14535, 21803, 32704, 49056, 73584, 110376, 165564, 248346, 372519, 558779, 838168, 1257252, 1885878, 2828817, 4243226, 6364839, 9547258, 14320887

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134 - A120137, A120139 - A120209.

a[n_]:= a[n]= 10 +Quotient[Sum[a[k], {k,n-1}],2];
Table[a[n], {n,60}] (* G. C. Greubel, May 08 2023 *)

@CachedFunction
def A120138(n): return 10 +sum(A120138(k) for k in range(1,n))//2
[A120138(n) for n in range(1,60)] # G. C. Greubel, May 08 2023

A120139 a(n) = 11 + floor( (1 + Sum_{j=1..n-1} a(j)) / 2).

Original entry on oeis.org

11, 17, 25, 38, 57, 85, 128, 192, 288, 432, 648, 972, 1458, 2187, 3280, 4920, 7380, 11070, 16605, 24908, 37362, 56043, 84064, 126096, 189144, 283716, 425574, 638361, 957542, 1436313, 2154469, 3231704, 4847556, 7271334, 10907001, 16360501

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134 - A120138, A120140 - A120209.

a[n_]:= a[n]= 11 +Quotient[1 + Sum[a[k], {k,n-1}], 2];
Table[a[n], {n,60}] (* G. C. Greubel, May 08 2023 *)

@CachedFunction
def A120139(n): return 11 +(1 +sum(A120139(k) for k in range(1,n)))//2
[A120139(n) for n in range(1,60)]  # G. C. Greubel, May 08 2023

A120140 a(n) = 13 + floor(Sum_{j=1..n-1} a(j)/2).

Original entry on oeis.org

13, 19, 29, 43, 65, 97, 146, 219, 328, 492, 738, 1107, 1661, 2491, 3737, 5605, 8408, 12612, 18918, 28377, 42565, 63848, 95772, 143658, 215487, 323230, 484845, 727268, 1090902, 1636353, 2454529, 3681794, 5522691, 8284036, 12426054, 18639081

Offset: 1

Graeme McRae, Jun 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134 - A120139, A120141 - A120209.

a[n_]:= a[n]= 13 +Quotient[Sum[a[k], {k,n-1}], 2];
Table[a[n], {n,60}] (* G. C. Greubel, May 11 2023 *)

@CachedFunction
def A120140(n): return 13 + sum(A120140(k) for k in range(1,n))//2
[A120140(n) for n in range(1,60)] # G. C. Greubel, May 11 2023

A120141 a(n) = 14 + floor( (1 + Sum_{j=0..n-1} a(j)) / 2).

Original entry on oeis.org

14, 21, 32, 48, 72, 108, 162, 243, 364, 546, 819, 1229, 1843, 2765, 4147, 6221, 9331, 13997, 20995, 31493, 47239, 70859, 106288, 159432, 239148, 358722, 538083, 807125, 1210687, 1816031, 2724046, 4086069, 6129104, 9193656, 13790484

Offset: 1

Graeme McRae, Jun 10 2006

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A073941, A072493, A112088, A120134 - A120140, A120142 - A120209.

nxt[{t_,a_}]:=Module[{a2=14+Floor[(1+t)/2]},{t+a2,a2}]; NestList[nxt,{0,14},60][[All,2]]//Rest (* Harvey P. Dale, Nov 28 2018 *)

@CachedFunction
def A120141(n): return 14 +(1 +sum(A120141(k) for k in range(1,n)))//2
[A120141(n) for n in range(1,60)] # G. C. Greubel, May 11 2023

cp's OEIS Frontend

A005427 Josephus problem: numbers m such that, when m people are arranged on a circle and numbered 1 through m, the final survivor when we remove every 4th person is one of the first three people.

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A087192 a(n) = ceiling(a(n-1)*4/3), with a(1) = 1.

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A120169 a(n) = 12 + floor((1 + Sum_{j=1..n-1} a(j))/4).

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A120135 a(n) = 5 + floor((1 + Sum_{j=1..n-1} a(j)) / 2).

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A120136 a(n) = 7 + floor(Sum_{j=1..n-1} a(j) / 2).

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A120137 a(n) = 8 + floor( (1 + Sum_{j=1..n-1} a(j)) / 2).

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A120138 a(n) = 10 + floor(Sum_{j=1..n-1} a(j) / 2).

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A120139 a(n) = 11 + floor( (1 + Sum_{j=1..n-1} a(j)) / 2).

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