A073941 -id:A073941 - 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)

A081848 Number of numbers whose base-3/2 expansion (see A024629) has n digits.

Original entry on oeis.org

3, 3, 3, 6, 9, 12, 18, 27, 42, 63, 93, 141, 210, 315, 474, 711, 1065, 1599, 2397, 3597, 5394, 8091, 12138, 18207, 27309, 40965, 61446, 92169, 138255, 207381, 311073, 466608, 699912, 1049868, 1574802, 2362203, 3543306, 5314959, 7972437, 11958657

Offset: 1

N. J. A. Sloane, Apr 13 2003

Run lengths in A246435. - Reinhard Zumkeller, Sep 05 2014

			a(1) = 3 because 0, 1 and 2 each have 1 digit.

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.
J. S. Tanton, A collection of research problems.

Cf. A024629, A070885, A073941, A246435.

a081848 n = a081848_list !! (n-1)
a081848_list = 3 : tail (zipWith (-) (tail a070885_list) a070885_list)
-- Reinhard Zumkeller, Sep 05 2014

from itertools import islice
def A081848_gen(): # generator of terms
    yield (a:=3)
    while True:
        yield (b:=(a+1>>1)+(a&1))
        a += b
A081848_list = list(islice(A081848_gen(),70)) # Chai Wah Wu, Sep 20 2022

For n > 1, a(n) = A070885(n+1) - A070885(n). - Tom Edgar, Jun 25 2014

a(n) = 3*A073941(n). - Tom Edgar, Jul 21 2014

More terms from David Wasserman, Jun 28 2004

Edited by Charles R Greathouse IV, Aug 02 2010

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

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

A120160 a(n) = ceiling(Sum_{i=1..n-1} a(i)/4) for n >= 2 starting with a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 8, 10, 12, 15, 19, 24, 30, 37, 47, 58, 73, 91, 114, 142, 178, 222, 278, 347, 434, 543, 678, 848, 1060, 1325, 1656, 2070, 2588, 3235, 4043, 5054, 6318, 7897, 9871, 12339, 15424, 19280, 24100, 30125, 37656, 47070, 58838, 73547

Offset: 1

Graeme McRae, Jun 10 2006

nonn

From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture 1: a(n) equals the number of multiples of 4 whose representation in base 5/4 (see A024634) has n-1 digits. For example, a(8) = 3 because there are three multiples of 4 with n-1 = 7 digits in their representation in base 5/4: 36 = 4321031, 40 = 4321420, and 44 = 4321424.
Conjecture 2: a(n) equals 1/5 times the number of nonnegative integers with the property that their 5/4-expansion has n digits (assuming that the 5/4-expansion of 0 has 1 digit). For example, a(7)*5 = 10 because the following 10 numbers have 5/4 expansions with n = 7 digits: 35 = 4321030, 36 = 4321031, 37 = 4321032, 38 = 4321033, 39 = 4321034, 40 = 4321420, 41 = 4321421, 42 = 4321422, 43 = 4321423, and 44 = 4321424. (End)
From Petros Hadjicostas, Jul 23 2020: (Start)
Starting at a(11) = 5, we conjecture that this sequence gives all those numbers m for which when we place m persons on a circle, label them 1 through m, start counting at person number 1, and remove every 5th person, the last survivors have numbers in {1, 2, 3, 4}.
When m = 6, 12, 15, 37, 58, 142, 222, 347, ... the last survivor is person 1.
When m = 5, 19, 91, 434, 1656, 2070, 5054, ... the last survivor is person 2.
When m = 8, 10, 24, 30, 73, 114, 178, 278, ... the last survivor is person 3.
When m = 47, 543, 2588, 3235, 6318, 58838, ... the last survivor is person 4. (End)

			From _Petros Hadjicostas_, Jul 23 2020: (Start)
We explain why 6 and 8 belong to the sequence related to the Josephus problem (for the case k = 5) while 7 does not.
When we place m = 6 people on a circle, label them 1 through 6, start counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 4 -> 6 -> 2 -> 3. Thus the last survivor is person 1, so 6 belongs to this sequence.
When we place m = 7 people on a circle, label them 1 through 7, start counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 3 -> 2 -> 4 -> 7 -> 1. Thus the last survivor is person 6 (not in {1, 2, 3, 4}), so 7 does not belong to this sequence.
When we place m = 8 people on a circle, label them 1 through 8, start the counting at person 1, and remove every 5th person, the list of people we remove is 5 -> 2 -> 8 -> 7 -> 1 -> 4 -> 6. Thus the last survivor is 3, so 8 belongs to this sequence. (End)

G. C. Greubel, 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 = 5. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]
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. A011782, A024634, A072493, A073941, A112088, A120170, A120178, A120186, A120194, A120202.

function f(n, a, b)
  t:=0;
    for k in [1..n-1] do
      t+:= a+Ceiling((b+t)/4);
    end for;
  return t;
end function;
g:= func< n, a, b | f(n+1, a, b)-f(n, a, b) >;
A120160:= func< n | n le 4 select 1 else g(n-4, 1, 0) >;
[A120160(n): n in [1..60]]; // G. C. Greubel, Sep 01 2023

f[s_]:= Append[s, Ceiling[Plus @@ s/4]]; Nest[f,1,53] (* Robert G. Wilson v *)
(* Second program *)
f[n_]:= f[n]= 1 + Quotient[Sum[f[k], {k,n-1}], 4];
A120160[n_]:= If[n==1, 1, f[n-1]];
Table[A120160[n], {n, 60}] (* G. C. Greubel, Sep 01 2023 *)

/* PARI program for the general case of Josephus problem. We use the Burde-Thériault algorithm. Here k = 5. 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

@CachedFunction
def A120160(n): return 1 + ceil( sum(A120160(k) for k in range(1,n))//4 )
[A120160(n) for n in range(1,61)] # G. C. Greubel, Sep 01 2023

Edited and extended by Robert G. Wilson v, Jul 07 2006

Appended the first missing term. - Tom Edgar, Jul 18 2014

A120134 a(n) = 4 + floor(Sum_{k=1..n-1} a(k) / 2).

Original entry on oeis.org

4, 6, 9, 13, 20, 30, 45, 67, 101, 151, 227, 340, 510, 765, 1148, 1722, 2583, 3874, 5811, 8717, 13075, 19613, 29419, 44129, 66193, 99290, 148935, 223402, 335103, 502655, 753982, 1130973, 1696460, 2544690, 3817035, 5725552, 8588328, 12882492

Offset: 1

Graeme McRae, Jun 10 2006

nonn

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

Cf. A073941, A072493, A112088, A120135 - A120209, A172161.

a := proc(n) option remember; 4 + iquo(add(a(k), k = 1..n-1), 2) end:
seq(a(n), n = 1..37); # Peter Luschny, May 07 2023

f[s_]:= Append[s, 4+Floor[Plus @@ s/2]]; Nest[f, {4}, 37] (* Robert G. Wilson v, Jun 10 2006 *)

@CachedFunction
def A120134(n): return 4 + sum(A120134(k) for k in range(1,n))//2
print([A120134(n) for n in range(1,41)])  # G. C. Greubel, May 07 2023

a(n) ~ c * (3/2)^n, where c = 3.9320886310283094862025842648411406926537121258867950257873967568454133192... - Vaclav Kotesovec, May 07 2023

Name edited by Peter Luschny, May 07 2023

cp's OEIS Frontend

A082416 Parity of A073941(n).

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A129272 a(n)=(1/2)*(2*b(n+1)-3*b(n)+(1-(-1)^b(n))/2) where b(n)=A073941(n).

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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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A081848 Number of numbers whose base-3/2 expansion (see A024629) has n digits.

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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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Magma

PARI

PARI

Python

Formula

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

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Haskell

A129272 a(n)=(1/2)(2b(n+1)-3*b(n)+(1-(-1)^b(n))/2) where b(n)=A073941(n).