A083286 -id:A083286 - cp's OEIS frontend

A073941 a(n) = ceiling((Sum_{k=1..n-1} a(k)) / 2) for n >= 2 starting with a(1) = 1.

1, 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

Offset: 1

Reinhard Zumkeller, Nov 20 2002

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
From Petros Hadjicostas, Jul 20 2020: (Start)
We describe Schuh's counting-off game (pp. 373-375 and 377-379). Assume m people are standing on a circle and they are labeled 1 through m (say clockwise). We start with the person labeled 1 and every 3rd person drops out (in a variation of the famous Josephus problem). The process is repeated until only one person is left.
This sequence describes those numbers m for which either the person labeled 1 or the person labeled 2 is the last survivor.
From a(4) = 2 to a(53) = 775795914 (see T. D. Noe's b-file), the values agree with those in Schuh (1968, p. 374) and Burde (1987, p. 207). a(54) = 1163693871 while both Schuh and Burde have 1063693871 (a difference in the 2nd digit starting on the left). a(55) = 1745540806 while both Schuch and Burde have 1595540806.
Schuh (1968) obtains the numbers in the following way. Suppose we know a(n) and the corresponding number i(n) of the last survivor (i(n) = 1 or 2). We multiply a(n) by 3/2 (cf. Burde's use of fractional bases).
If the product is an integer, that is a(n+1) and the corresponding last survivor is the same.
If the product is not an integer, then a(n+1) = floor(a(n)*3/2) if the last survivor i(n) = 2 (and the new last survivor is i(n+1) = 1), and a(n+1) = ceiling(a(n)*3/2) if the last survivor is i(n) = 1 (and the new last survivor is i(n+1) = 2).
Note that a(53) = 775795914 and a(54) = (3/2)*a(53) = 1163693871 (not 1063693871), so it seems Schuh did a mistake and Burde copied it. Also (3/2)*1163693871 = 1745540806.5. Since a(53) = 775795914 corresponds to number 2, we round down, i.e., a(54) = 1745540806 (and move to number 1). If, however, we multiply the incorrect 1063693871 by 3/2 and round down, we get Schuh and Burde's incorrect value 1595540806 for a(54).
Numbers a(n) that correspond to last survivors being number 1 are tabulated in A081614 while numbers a(n) that correspond to last survivors being number 2 are tabulated in A081615. (End)
a(n) is the number of times (n-1) appears in A061420. - Chinmaya Dash, Aug 19 2020

Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968. [See Table 18, p. 374. Only the terms from a(6) = 4 forward are shown in the table. The table is definitely related to this sequence.]

T. D. Noe, Table of n, a(n) for n = 1..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. [The author deals with the representation of n in fractional bases k/(k-1) and its relation to counting-off games. Here k = 3. See the table on p. 207. See also the review in MathSciNet (MR0889384) by R. G. Stoneham.]
B. Chen, R. Chen, J. Guo, S. Lee et al, On Base 3/2 and its sequences, arXiv:1808.04304 [math.NT], 2018.
Tom Edgar, Hailey Olafson, and James Van Alstine, Approximating the Fibonacci Sequence, Integers 16 (2016), #A63.

Same as log_2(A082125(n)), for n > 2. - Ralf Stephan, Apr 16 2002
Apart from initial term, same as A005428, which has further information.
a(n+4) = A079719(n)+2. Cf. A082416.
Partial sums for various start indices are in A006999, A061419, A061418. - Ralf Stephan, Apr 17 2003
Is this the same as A081848/3?
The constant c is (2/9)*K(3) (see A083286). - Ralf Stephan, May 29 2003
Cf. A081614, A081615.

a073941 n = a073941_list !! (n-1)
a073941_list = 1 : f [1] where
   f xs = x' : f (x':xs) where x' = (1 + sum xs) `div` 2
-- Reinhard Zumkeller, Oct 26 2011

f[s_] := Append[s, Ceiling[Plus @@ s/2]]; Nest[f, {1}, 41] (* Robert G. Wilson v, Jul 07 2006 *)

v=vector(100);s=v[1]=1;for(i=2,#v,s+=(v[i]=(s+1)\2));v \\ Charles R Greathouse IV, Feb 11 2011

from itertools import islice
def A073941_gen(): # generator of terms
    a, c = 1, 0
    yield 1
    while True:
        yield (a:=(c:=c+a)+1>>1)
A073941_list = list(islice(A073941_gen(),70)) # Chai Wah Wu, Sep 20 2022

a(n) = ceiling(c*(3/2)^n-1/2) where c = 0.3605045561966149591015446628665... - Benoit Cloitre, Nov 22 2002
If 2^m divides a(i) then 2^(m-1)*3^1 divides a(i+1) and so on... until finally, 3^m divides a(i+m). - Ralf Stephan, Apr 20 2003
a(n) = A081848(n)/3. - Tom Edgar, Jul 21 2014
a(n) = A005428(n-2). - Tanya Khovanova and PRIMES STEP Senior group, Jun 03 2018

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

Original entry on oeis.org

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)

A061419 a(n) = ceiling(a(n-1)*3/2) with a(1) = 1.

Original entry on oeis.org

1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315, 473, 710, 1065, 1598, 2397, 3596, 5394, 8091, 12137, 18206, 27309, 40964, 61446, 92169, 138254, 207381, 311072, 466608, 699912, 1049868, 1574802, 2362203, 3543305, 5314958, 7972437, 11958656

Offset: 1

Henry Bottomley, May 02 2001

nonn

It appears that this sequence is the (L)-sieve transform of {3,6,9,12,...,3n,...} = A008585. (See A152009 for the definition of the (L)-sieve transform.) - John W. Layman, Jan 06 2009

			a(6) = ceiling(8*3/2) = 12.

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.30.1, p. 196.

Harry J. Smith, Table of n, a(n) for n = 1..500
Zakir Deniz, Topology of acyclic complexes of tournaments and coloring, Applicable Algebra in Engineering, Communication, March 2015, Volume 26, Issue 1-2, pp. 213-226.
A. Dubickas, On integer sequences generated by linear maps, Glasg. Math. J. 51(2) (2009), 243-252.
Don Knuth, Ambidextrous Numbers, Preprint, September 2022.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33(2) (1991), 235-240.
Eric Weisstein's World of Mathematics, Power Ceilings.
Index entries for sequences related to the Josephus Problem

Cf. A002379, A003312, A034082, A061418, A061420, A070885, A083286.

First differences are in A073941.

[ n eq 1 select 1 else Ceiling(Self(n-1)*3/2): n in [1..40] ]; // Klaus Brockhaus, Nov 14 2008

a:=proc(n) option remember: if n=1 then 1 else ceil(procname(n-1)*3/2) fi; end; seq(a(n),n=1..40); # Muniru A Asiru, Jun 07 2018

a=1;a=Table[a=Ceiling[a*3/2],{n,0,4!}] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
NestList[Ceiling[3#/2]&,1,39] (* Stefano Spezia, Dec 08 2024 *)

{ a=2/3; for (n=1, 500, write("b061419.txt", n, " ", a=ceil(a*3/2)) ) } \\ Harry J. Smith, Jul 22 2009

from itertools import islice
def A061419_gen(): # generator of terms
    a = 2
    while True:
        yield a-1
        a += a>>1
A061419_list = list(islice(A061419_gen(),70)) # Chai Wah Wu, Sep 20 2022

a(n) = A061418(n) - 1 = floor(K*(3/2)^n) where K = 1.08151366859...
The constant K is (2/3)*K(3) (see A083286). - Ralf Stephan, May 29 2003
a(1) = 1, a(n) = A070885(n)/3. - Benoit Cloitre, Aug 18 2002
a(n) = ceiling((a(n-1) + a(n-2))*9/10) - Franklin T. Adams-Watters, May 01 2006

A061418 a(n) = floor(a(n-1)*3/2) with a(1) = 2.

Original entry on oeis.org

2, 3, 4, 6, 9, 13, 19, 28, 42, 63, 94, 141, 211, 316, 474, 711, 1066, 1599, 2398, 3597, 5395, 8092, 12138, 18207, 27310, 40965, 61447, 92170, 138255, 207382, 311073, 466609, 699913, 1049869, 1574803, 2362204, 3543306, 5314959, 7972438

Offset: 1

Henry Bottomley, May 02 2001

Can be stated as the number of animals starting from a single pair if any pair of animals can produce a single offspring (as in the game Minecraft, if the player allows offspring to fully grow before breeding again). - Denis Moskowitz, Dec 05 2012

Maximum number of partial products that can be added in a Wallace tree multiplier with n-1 full adder stages. - Chinmaya Dash, Aug 19 2020

			a(6) = floor(9*3/2) = 13.

Iain Fox, Table of n, a(n) for n = 1..1000 (first 500 terms from Harry J. Smith)
Don Knuth, Ambidextrous Numbers, Preprint, September 2022.
M. van de Vel, Determination of msd(L^n), J. Algebraic Combin. 9(2) (1999), 161-171. See Table 5. - _N. J. A. Sloane_, Mar 26 2012
Mark van Wijk, The Quest for the Best Thread-Safe Java List, Univ. of Twente (Netherlands 2022).
Wikipedia, Wallace tree.

Cf. A002379, A003312, A034082, A061419, A083286.

First differences are in A073941.

[ n eq 1 select 2 else Floor(Self(n-1)*(3/2)): n in [1..39] ]; // Klaus Brockhaus, Nov 14 2008

{ a=4/3; for (n=1, 500, a=a*3\2; write("b061418.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 22 2009

first(n) = my(v=vector(n)); v[1]=2; for(i=2, n, v[i]=v[i-1]*3\2); v \\ Iain Fox, Jul 15 2022

from itertools import islice
def A061418_gen(): # generator of terms
    a = 2
    while True:
        yield a
        a += a>>1
A061418_list = list(islice(A061418_gen(),70)) # Chai Wah Wu, Sep 20 2022

a(n) = A061419(n) + 1 = ceiling(K*(3/2)^n) where K = 1.08151366859...

The constant K is (2/3)*K(3) (see A083286). - Ralf Stephan, May 29 2003

A006999 Partitioning integers to avoid arithmetic progressions of length 3.

Original entry on oeis.org

0, 1, 2, 4, 7, 11, 17, 26, 40, 61, 92, 139, 209, 314, 472, 709, 1064, 1597, 2396, 3595, 5393, 8090, 12136, 18205, 27308, 40963, 61445, 92168, 138253, 207380, 311071, 466607, 699911, 1049867, 1574801, 2362202, 3543304, 5314957, 7972436

Offset: 0

N. J. A. Sloane, D. R. Hofstadter, and James Propp, Jul 15 1977

a(n) = A006997(3^n-1).
It appears that, aside from the first term, this is the (L)-sieve transform of A016789 ={2,5,8,11,...,3n+2....}. This has been verified up to a(30)=311071. See A152009 for the definition of the (L)-sieve transform. - John W. Layman, Nov 20 2008
a(n) is also the largest-index square reachable in n jumps if we start at square 0 of the Infinite Sidewalk. - Jose Villegas, Mar 27 2023

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
B. Chen, R. Chen, J. Guo, S. Lee et al, On Base 3/2 and its sequences, arXiv:1808.04304 [math.NT], 2018.
Joseph Gerver, James Propp and Jamie Simpson, Greedily partitioning the natural numbers into sets free of arithmetic progressions Proc. Amer. Math. Soc. 102 (1988), no. 3, 765-772.
D. R. Hofstadter, Eta-Lore [Cached copy, with permission]
D. R. Hofstadter, Pi-Mu Sequences [Cached copy, with permission]
D. R. Hofstadter and N. J. A. Sloane, Correspondence, 1977 and 1991
James Propp and N. J. A. Sloane, Email, March 1994
Jane Street, Traversing the Infinite Sidewalk (2023).

Cf. A061419, A061418, A005428 (first differences), A083286.
Cf. A006997, A016789, A152009.
Cf. A003312.

a006999 n = a006999_list !! n
a006999_list = 0 : map ((`div` 2) . (+ 2) . (* 3)) a006999_list
-- Reinhard Zumkeller, Oct 26 2011

a[0] = 0; a[n_] := a[n] = Floor[(3 a[n-1] + 2)/2];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Aug 01 2018 *)

PARI
```
a(n)=if(n<1,0,floor((3*a(n-1)+2)/2))
```

a(n) = A061419(n) - 1.
a(n) = A061418(n) - 2.
a(n) = floor((3a(n-1)+2)/2).
a(n) = -1 + floor(c*(3/2)^n) where c=1.0815136... - Benoit Cloitre, Jan 10 2002; this constant c is 2/3*K(3) (see A083286). - Ralf Stephan, May 29 2003
a(n+1) = (3*a(n))/2+1 if a(n) is even. a(n+1) = (3*a(n)+1)/2 if a(n) is odd. - Miquel Cerda, Jun 15 2019

More terms from James Sellers, Feb 06 2000

A070885 a(n) = (3/2)a(n-1) if a(n-1) is even; (3/2)(a(n-1)+1) if a(n-1) is odd.

Original entry on oeis.org

1, 3, 6, 9, 15, 24, 36, 54, 81, 123, 186, 279, 420, 630, 945, 1419, 2130, 3195, 4794, 7191, 10788, 16182, 24273, 36411, 54618, 81927, 122892, 184338, 276507, 414762, 622143, 933216, 1399824, 2099736, 3149604, 4724406, 7086609, 10629915

Offset: 1

Eric W. Weisstein, May 14 2002

nonn

The smallest positive number such that A024629(a(n)) has n digits, per page 9 of the Tanton reference in Links. - Glen Whitney, Sep 17 2017

Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, 2002, p. 123.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
B. Chen, R. Chen, J. Guo, S. Lee et al., On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018.
James Tanton, Exploding Dots, Chapter 9
Eric Weisstein's World of Mathematics, Wolfram Sequences

The constant K is 2/3*K(3) (see A083286). - Ralf Stephan, May 29 2003
Cf. A003312.
Cf. A081848.
Cf. A205083 (parity of terms).

a070885 n = a070885_list !! (n-1)
a070885_list = 1 : map (flip (*) 3 . flip div 2 . (+ 1)) a070885_list
-- Reinhard Zumkeller, Sep 05 2014

A070885 := proc(n)
    option remember;
    if n = 1 then
        return 1;
    elif type(procname(n-1),'even') then
        procname(n-1) ;
    else
        procname(n-1)+1 ;
    end if;
    %*3/2 ;
end proc:
seq(A070885(n),n=1..80) ; # R. J. Mathar, Jun 18 2018

NestList[If[EvenQ[#],3/2 #,3/2 (#+1)]&,1,40] (* Harvey P. Dale, May 18 2018 *)

from itertools import islice
def A070885_gen(): # generator of terms
    a = 1
    while True:
        yield a
        a += (a+1>>1)+(a&1)
A070885_list = list(islice(A070885_gen(),70)) # Chai Wah Wu, Sep 20 2022

For n > 1, a(n) = 3*A061419(n) = 3*floor(K*(3/2)^n) where K=1.08151366859... - Benoit Cloitre, Aug 18 2002
a(n) = 3*ceiling(a(n-1)/2). - Benoit Cloitre, Apr 25 2003
a(n+1) = a(n) + A081848(n), for n > 1. - Reinhard Zumkeller, Sep 05 2014

A214986 Power ceiling array for the golden ratio, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 5, 1, 1, 12, 21, 22, 7, 1, 1, 20, 55, 94, 48, 12, 1, 1, 33, 144, 399, 329, 134, 18, 1, 1, 54, 377, 1691, 2255, 1487, 323, 30, 1, 1, 88, 987, 7164, 15456, 16492, 5796, 872, 47, 1, 1, 143, 2584, 30348, 105937, 182900

Offset: 1

Clark Kimberling, Oct 28 2012

row 0: A000012 ... row 6: A049660
row 1: A000071 ... row 8: A049668
row 2: A001906 ... col 0: A000012
row 3: A049652 ... col 1: A169986
row 4: A004187
For x>1, define c(x,0) = 1 and c(x,n) = ceiling(x*c(x,n-1)) for n>0.  Row m of A214986 is the sequence c(r^m,n), where r = golden ratio = (1 + sqrt(5))/2.  The name of the array corresponds to the power ceiling function f(x) = limit of c(x,n)/x^n as n increases without bound; f(x) generalizes the case for x = 3/2 as described under "Power Ceilings" at MathWorld.  For a graph of f(x), see the Mathematica program at A083286.
The term "power ceiling sequence" extends to sequences generated by recurrences P(n) = ceiling(x*P(n-1)) + g(n), and "power ceiling functions" f(x) to the limit of P(n)/x^n in case x>1 and g(n)/x^n -> 0.
Suppose that h is a nonnegative integer and g(n) is a constant.  If x is a positive integer power of the golden ratio r, then f(x), in many cases, lies in the field Q(sqrt(5)).  Examples matching rows of A214986, using g(n) = 0, follow:
...
x ... P ........ f(x)
r ... A000071 .. (5 + 2*sqrt(5))/2 = 1.8944... (A010532)
r^2 . A001906 .. (5 + 3*sqrt(5))/10 = 1.7082...(A176015)
r^3 . A049652 .. (25 + 11*sqrt(5))/40 = 1.2399...
r^4 . A004187 .. (15 + 7*sqrt(5))/10 = 1.0219...
...
If k is odd, then f(r^k) = r^k((b(k) + c(k))/d(k)), where
b(k) = L(j)^2 + L(j-1)^2, where j=[(k+1)/2], L=A000032 (Lucas numbers); c(k) = (L(k)+2)*sqrt(5); d(k) = 10*F(k)*L(k), where F=A000045 (Fibonacci numbers).  If k is even, then f(r^k) = r^k/(F(k)*sqrt(5)).

			Northwest corner:
1...1....1.....1......1.......1
1...2....4.....7......12......20
1...3....8.....21.....55......144
1...5....22....94.....399.....1691
1...7....48....329....2255....15456
1...19...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..35, flattened
Eric Weisstein's World of Mathematics, Power Ceilings

Cf. A000045, A214978, A214984, A214987.

r = GoldenRatio;
s[x_, 0] := 1; s[x_, n_] := Ceiling[x*s[x, n - 1]];
t = TableForm[Table[s[r^m, n], {m, 0, 10}, {n, 0, 10}]  ]
u = Flatten[Table[s[r^m, n - m], {n, 0, 10}, {m, 0, n}]]

The odd-numbered rows of A214986 are even-numbered rows of A213978; the even-numbered rows of A214986 are odd-numbered rows of A214984.

A061420 a(n) = a(ceiling((n-1)*2/3)) + 1 with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 0

Henry Bottomley, May 02 2001

nonn

Least k such that f^(k)(n) = 0 where f(x) = floor(2/3*x) and f^(k+1)(x) = f(f^(k)(x)). - Benoit Cloitre, May 26 2007

Number of 3:2 compressor stages in a Wallace tree multiplier starting with (n+2) partial products. - Chinmaya Dash, Aug 18 2020

			a(10) = a(ceiling(9*2/3)) + 1 = a(6) + 1 = 4 + 1 = 5.

Clark Kimberling, Table of n, a(n) for n = 0..2000
K. A. C. Bickerstaff, M. Schulte and E. E. Swartzlander, Reduced area multipliers, Proceedings of International Conference on Application Specific Array Processors (ASAP '93), Venice, Italy, 1993, pp. 478-489. See Table 1 p. 480.
William J. Gilbert, Radix Representations of Quadratic Fields, Journal of Mathematical Analysis and Applications 83 (1981) pp 264-274. Gilbert (page 273) cites Wang and Washburn (below) in connection with the length of the base 3/2 expansion of the even positive integers.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
E. T. H. Wang, Phillip C. Washburn, Problem E2604, American Mathematical Monthly 84 (1977) pp. 821-822.
Index entries for sequences related to the Josephus Problem

Cf. A029837, A061419, A083286 (the constant c).

[IsZero(n) select 0 else Self(Floor(2*n/3)+1)+1: n in [0..90]]; // Bruno Berselli, Oct 31 2012

a:= n-> `if`(n=0, 0, a(ceil((n-1)*2/3))+1):
seq(a(n), n=0..100);  # Alois P. Heinz, Oct 29 2012

(* 1st program, using the alternative definition *)
a[0] = 0; a[n_] := a[Floor[2 n/3]] + 1;
Table[a[n], {n, 0, 120}]
(* 2nd program, using Cloitre's recurrence *)
f[x_] := Floor[2 x/3]; g[0, x_] := f[x];
g[k_, x_] := f[g[k - 1, x]];
u[n_] := Flatten[Table[g[k, n], {k, 0, 12}]]
v[n_] := First[Position[u[n], 0]];
Flatten[Table[v[n], {n, 1, 120}]]
(* 3rd program, using the constant c *)
f[n_] := -Floor[-Log[3/2, (n + 1)/1.62227050288476731595695098289932]]
Table[f[n], {n, 1, 120}]
(* Clark Kimberling, Oct 23 2012 *)

a(n) = if(n<0, 0, s=n; c=0; while(floor(s)>0, s=floor(2/3*s); c++); c) \\ Benoit Cloitre, May 26 2007

a(n) = a(n-1) + 1 if n is in A061419; a(n) = a(n-1) otherwise.
From Clark Kimberling, Oct 19 2012: (Start)
a(n) = a(floor(2*n/3)) + 1, where a(0) = 0 (alternative definition).
Washburn's solution of Problem E2604 (see References) shows that (for n>0), a(n) = -floor(-L((n+1)/c)), where L is the logarithm with base 3/2 and
  c = lim_{n->infinity} (2/3)^n*s(n) where s(n) = floor(3*s(n-1)/2) + 1 and s(0)=0.  The editors state that "It may be interesting to know whether c is irrational or even transcendental"; c = 1.62227050288476731595695098289932... .
Odlyzko and Wilf also discuss the defining recurrence, and they, after Washburn, give a formula for the sequence using c, as in the third Mathematica program below.
(End)

A083287 Continued fraction expansion of K(3), a constant related to the Josephus problem.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 5, 10, 19, 1, 4, 4, 4, 3, 10, 1, 42, 2, 23, 33, 1, 4, 7, 1, 12, 1, 1, 2, 9, 2, 11, 3, 4, 1, 1, 3, 2, 4, 25, 3, 1, 16, 5, 10, 1, 1, 1, 3, 1, 1, 1, 3, 2, 2, 1, 1, 1, 2, 3, 2, 1, 3, 4, 3, 1, 1, 117, 2, 1, 12, 4, 1, 4, 3, 3, 15, 1, 5, 16, 7, 2, 7, 21, 1, 3, 1, 2, 2, 2, 1, 1, 1, 1

Offset: 0

Ralf Stephan, Apr 23 2003

The constant K(3)=1.62227050288476731595695... is related to the Josephus problem with q=3 and the computation of A054995.

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. A054995, A083286 (decimal expansion).

For[p = 1; nn = 10^4; n = 1, n <= nn, n++, p = Ceiling[3/2*p]]; p/(3/2)^nn // ContinuedFraction[#, 93] & (* Jean-François Alcover, Jul 11 2013, after Pari *)

p=1; N=10^4; for(n=1, N, p=ceil(3/2*p)); c=(p/(3/2)^N)+0. \\ This gives K(3) not the sequence!

Offset changed by Andrew Howroyd, Aug 07 2024

A083198 a(n) = A061419(n) - A002379(n).

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 7, 11, 16, 24, 36, 54, 80, 121, 181, 271, 407, 610, 915, 1372, 2058, 3088, 4631, 6947, 10420, 15630, 23446, 35169, 52753, 79129, 118693, 178039, 267059, 400589, 600883, 901324, 1351986, 2027979, 3041968

Offset: 5

Ralf Stephan, Jun 01 2003

lim (a(n) * (2/3)^n) = 2/3 * A083286 - 1.

p=1; for(n=2, 100, p=p+ceil(p/2); print1(p-floor((3/2)^n)", "))

cp's OEIS Frontend

A073941 a(n) = ceiling((Sum_{k=1..n-1} a(k)) / 2) for n >= 2 starting with a(1) = 1.

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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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A061419 a(n) = ceiling(a(n-1)*3/2) with a(1) = 1.

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A061418 a(n) = floor(a(n-1)*3/2) with a(1) = 2.

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A006999 Partitioning integers to avoid arithmetic progressions of length 3.

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A070885 a(n) = (3/2)*a(n-1) if a(n-1) is even; (3/2)*(a(n-1)+1) if a(n-1) is odd.

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A070885 a(n) = (3/2)a(n-1) if a(n-1) is even; (3/2)(a(n-1)+1) if a(n-1) is odd.