A005427 -id:A005427 - cp's OEIS frontend

A072493 a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.

1, 1, 1, 1, 2, 2, 3, 4, 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

Offset: 1

Benoit Cloitre, Nov 22 2002

nonn

Is this sequence, with its first 8 terms removed, the same as A005427? See also the similar conjecture with A005428/A073941. - Ralf Stephan, Apr 04 2003
Yes; the first 8 terms sum to 15, so upon dividing by 3 they are equivalent to the +5 in the formula for A005427. - Charlie Neder, Jan 10 2019
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture 1: a(n) equals the number of multiples of 3 whose representation in base 4/3 (see A024631) has n-1 digits. For example, a(8) = 4 because there are four multiples of 3 with n-1 = 7 digits in their representation in base 4/3: 33 = 3210201, 36 = 3210230, 39 = 3210233, and 42 = 3213122.
Conjecture 2: a(n) equals 1/4 times the number of nonnegative integers with the property that their 4/3-expansion has n digits (assuming that the 4/3-expansion of 0 has 1 digit). For example, a(7)*4 = 12 because the following 12 numbers have 4/3 expansions with n = 7 digits: 32 = 3210200, 33 = 3210201, 34 = 3210202, ..., 42 = 3213122, 43 = 3213123. (End)

Michel Marcus, 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. [The author deals with the representation of n in fractional bases k/(k-1) and its relation to counting-off games (variations of Josephus problem). Here k = 4. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33(2) (1991), 235-240.
Index entries for sequences related to the Josephus Problem

Cf. A005427, A005428, A024631, A073941, A083286.

f[s_] := Append[s, Ceiling[Plus @@ s/3]]; Nest[f, {1}, 52] (* Robert G. Wilson v, Jul 07 2006 *)

lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = ceil(sum(k=1, n-1, va[k])/3);); va;} \\ Michel Marcus, Apr 16 2015

a(n) = ceiling(c*(4/3)^n - 1/2) where c = 0.389324199524937508840138455...
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture: The constant c above equals (3/16)*K(4), where K(q) = C(q/(q-1)) (q > 1) is described in Odlyzko and Wilf (1991).
For a > 1, the constant C(a) = limit_{n -> infinity} f_n(a)/a^n, where f_{n+1}(a) = ceiling(a*f_n(a)) for n >= 0 and f_0(a) = 1.
Thus, K(4) = limit_{n -> infinity} f_n(4/3)/(4/3)^n = 2.076395730799666... We have K(2) = 1 and K(3) = A083286 = 1.622270502884767315... (End)

A005428 a(n) = ceiling((1 + sum of preceding terms) / 2) starting with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 9, 14, 21, 31, 47, 70, 105, 158, 237, 355, 533, 799, 1199, 1798, 2697, 4046, 6069, 9103, 13655, 20482, 30723, 46085, 69127, 103691, 155536, 233304, 349956, 524934, 787401, 1181102, 1771653, 2657479, 3986219, 5979328, 8968992, 13453488, 20180232, 30270348, 45405522, 68108283, 102162425, 153243637, 229865456, 344798184

Offset: 0

N. J. A. Sloane and Simon Plouffe

Original definition: a(0) = 1, state(0) = 2; for n >= 1, if a(n-1) is even then a(n) = 3*a(n-1)/2 and state(n) = state(n-1); if a(n-1) is odd and state(n-1) = 1 then a(n) = ceiling( 3*a(n-1)/2) and state(n) = 3 - state(n-1) and if a(n-1) is odd and state(n-1) = 2 then a(n) = floor( 3*a(n-1)/2) and state(n) = 3 - state(n-1). [See formula by M. Alekseyev for a simpler equivalent. - Ed.]
Arises from a version of the Josephus problem: sequence gives set of n where, if you start with n people and every 3rd person drops out, either it is person #1 or #2 who is left at the end. A081614 and A081615 give the subsequences where it is person #1 (respectively #2) who is left.
The state changes just when a(n) is odd: it therefore indicates whether the sum of a(0) to a(n) is odd (1 means no, 2 means yes).
The sum a(0) to a(n) is never divisible by 3 (for n >= 0); it is 1 mod 3 precisely when the sum a(0) to a(n-1) is odd and thus indicates the state at the previous step. - Franklin T. Adams-Watters, May 14 2008
The number of nodes at level n of a planted binary tree with alternating branching and non-branching nodes. - Joseph P. Shoulak, Aug 26 2012
Take Sum_{k=1..n} a(k) objects, and partition them into 3 parts. It is always possible to generate those parts using addends once each from the initial n terms, and this is the fastest growing sequence with this property. For example, taking 1+1+2+3+4+6+9 = 26 objects, if we partition them [10,9,7], we can generate these sizes as 10 = 9+1, 9 = 6+3, 7 = 4+2+1. The corresponding sequence partitioning into 2 parts is the powers of 2, A000079. In general, to handle partitioning into k parts, replace the division by 2 in the definition with division by k-1. - Franklin T. Adams-Watters, Nov 07 2015
a(n) is the number of even integers that have n+1 digits when written in base 3/2. For example, there are 2 even integers that use three digits in base 3/2: 6 and 8: they are written as 210 and 212, respectively. - Tanya Khovanova and PRIMES STEP Senior group, Jun 03 2018

			n........0...1...2...3...4...5...6...7...8...9..10..11..12..13..14.
state=1......1...........4...6...9..........31.....70..105.........
state=2..1.......2...3..............14..21......47.........158..237

F. Schuh, The Master Book of Mathematical Recreations. Dover, NY, 1968, page, 374, Table 18, union of columns 1 and 2 (which are A081614 and A081615).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
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.
B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its sequences, arXiv:1808.04304 [math.NT], 2018.
L. Halbeisen and N. Hungerbuehler, The Josephus Problem, J. Théor. Nombres Bordeaux 9(2) (1997), 303-318.
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. A005427, A073941, A082416. Union of A081614 and A081615.
First differences of D_3(n) (A061419) in the terminology of Odlyzko and Wilf. - Ralf Stephan, Apr 23 2002
Same as log_2(A082125(n+3)). - Ralf Stephan, Apr 16 2002
Apart from initial terms, same as A073941, which has further information.
a(n) is the number of positive even k for which A024629(k) has n+1 digits. - Glen Whitney, Jul 09 2017
Partial sums are in A061419, A061418, A006999.
Cf. A000079, A072493, A120160, A120170, A120178, A120186, A120194, A120202.

a005428 n = a005428_list !! n
a005428_list = (iterate j (1, 1)) where
   j (a, s) = (a', (s + a') `mod` 2) where
     a' = (3 * a + (1 - s) * a `mod` 2) `div` 2
-- Reinhard Zumkeller, May 10 2015 (fixed), Oct 26 2011

f[s_] := Append[s, Ceiling[(1 + Plus @@ s)/2]]; Nest[f, {1}, 40] (* Robert G. Wilson v, Jul 07 2006 *)
nxt[{t_,a_}]:=Module[{c=Ceiling[(1+t)/2]},{t+c,c}]; NestList[nxt,{1,1},50][[All,2]] (* Harvey P. Dale, Nov 05 2017 *)

{ a=1; s=2; for(k=1,50, print1(a,", "); a=(3*a+s-1)\2; s=(s+a)%3; ) } \\ Max Alekseyev, Mar 28 2009

s=0;vector(50,n,-s+s+=s\2+1)  \\ M. F. Hasler, Oct 14 2012

from itertools import islice
def A005428_gen(): # generator of terms
    a, c = 1, 0
    yield 1
    while True:
        yield (a:=1+((c:=c+a)>>1))
A005428_list = list(islice(A005428_gen(),30)) # Chai Wah Wu, Sep 21 2022

a(0) = 1; a(n) = ceiling((1 + Sum_{k=0..n-1} a(k)) / 2). - Don Reble, Apr 23 2003
a(1) = 1, s(1) = 2, and for n > 1, a(n) = floor((3*a(n-1) + s(n-1) - 1) / 2), s(n) = (s(n-1) + a(n)) mod 3. - Max Alekseyev, Mar 28 2009
a(n) = floor(1 + (sum of preceding terms)/2). - M. F. Hasler, Oct 14 2012

More terms from Hans Havermann, Apr 23 2003

Definition replaced with a simpler formula due to Don Reble, by M. F. Hasler, Oct 14 2012

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.

Original entry on oeis.org

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

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

A178853 "Josephus problem": n persons stand in a circle and eliminate every seventh person counting clockwise until only person a(n) is remaining.

Original entry on oeis.org

1, 2, 3, 2, 4, 5, 5, 4, 2, 9, 5, 12, 6, 13, 5, 12, 2, 9, 16, 3, 10, 17, 1, 8, 15, 22, 2, 9, 16, 23, 30, 5, 12, 19, 26, 33, 3, 10, 17, 24, 31, 38, 2, 9, 16, 23, 30, 37, 44, 1, 8, 15, 22, 29, 36, 43, 50, 57, 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 6, 13, 20, 27, 34, 41, 48, 55, 62, 69, 76

Offset: 1

Roland Schroeder (florola(AT)gmx.de), Jun 18 2010

nonn

Several other versions of this sequence are already in the OEIS. - N. J. A. Sloane, Jun 24 2010

Leonhard Euler, Observationes circa novum et singulare progressionum genus. In: Novi Comment. Akadem. Petropol. Vol.20 (1775).
Eric Weisstein's World of Mathematics, Josephus Problem.
Index entries for sequences related to the Josephus Problem.

Cf. A005427, A006257.

Needs["Combinatorica`"]
a[n_] := Last@ InversePermutation@ Josephus[n, 7]; Array[a, 79] (* Robert G. Wilson v, Jul 31 2010 *)

a(29) onwards from Robert G. Wilson v, Jul 31 2010

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).

Showing 1-6 of 6 results.

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A072493 a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.

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A005428 a(n) = ceiling((1 + sum of preceding terms) / 2) starting with a(0) = 1.

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A178853 "Josephus problem": n persons stand in a circle and eliminate every seventh person counting clockwise until only person a(n) is remaining.

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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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