A008937 -id:A008937 - cp's OEIS frontend

A320452 Number of possible states when placing n tokens of 2 alternating types on 2 piles.

1, 2, 4, 8, 15, 28, 52, 96, 177, 326, 600, 1104, 2030, 3732, 6858, 12600, 23144, 42504, 78048, 143296, 263068, 482904, 886392, 1626912, 2985943, 5480012, 10056946, 18456056, 33868851, 62151788, 114050884, 209284710, 384034660, 704690938, 1293071688, 2372700708

Offset: 0

Bert Dobbelaere, Oct 20 2018

nonn

Piles start empty and have no height limit. A token can only be placed on top of a pile. The starting token is fixed.

Up to a(11) the terms are matching A008937(n+1).

			With alternating symbols A and B on two piles (starting with A), the following states emerge after placing 4 symbols in all 2^4 possible ways:
  B                                                            B
  A   A   B       B            B  B            B       B   A   A
  B   B   B   BB  A   AB  BA   A  A   AB  BA   A  BB   B   B   B
  A_  AB  AA  AA  AB  AB  AB  AB  BA  BA  BA  BA  AA  AA  BA  _A
All states are different, except the 13th state is a duplicate of the 4th.
Hence a(4)=15.

Martin Fuller, Table of n, a(n) for n = 0..40

For 2 token types on 3 piles, see A320731.

def fill(patterns, state_in, ply_nr, n_plies, n_players, n_stacks):
    if ply_nr>=n_plies:
        patterns.add(tuple(state_in))
    else:
        symbol=chr(ord('A')+ply_nr%n_players)
        for st in range(n_stacks):
            state_out=list(state_in)
            state_out[st]+=symbol
            fill(patterns, state_out, ply_nr+1, n_plies, n_players, n_stacks)
def A320452(n):
    n_plies, n_players, n_stacks = n, 2, 2
    patterns=set()
    state=[""]*n_stacks
    fill(patterns, state, 0, n_plies, n_players, n_stacks)
    return len(patterns)

a(33) onwards from Martin Fuller, Apr 09 2025

A141435 a(1) = 1, a(2) = 2; a(n) = a(n-a(1)) + a(n-a(2)) + a(n-a(3)) + a(n-a(4)) + ...

Original entry on oeis.org

1, 2, 3, 6, 11, 20, 38, 71, 132, 247, 461, 861, 1609, 3005, 5613, 10485, 19584, 36581, 68330, 127632, 238404, 445314, 831798, 1553712, 2902170, 5420945, 10125754, 18913838, 35329048, 65990929, 123264078, 230244265, 430071949, 803328933

Offset: 1

Raes Tom (tommy1729(AT)hotmail.com), Aug 06 2008

Thus we get a self-reference sequence that grows exponentially. a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-6) + a(n-11) + a(n-20) + ...
A Fibonacci-like sequence, even closer to the tribonacci numbers.
Lim n-> oo log (a(n))/n converges.

			a(6) = 20 because 20 = a(5) + a(4) + a(3) = 11 + 6 + 3
a(8) = 71 because 71 = a(7) + a(6) + a(5) + a(2) = 38 + 20 + 11 + 2

Cf. A058265, A008937, A001590, A000213, A000073, A096436, A102575, A111129.

A141435 := proc(n) option remember; local a,i; if n <= 3 then RETURN(n); else a :=0 ; for i from 1 to n-1 do if n-procname(i) < 1 then RETURN(a); else a := a+procname(n-procname(i)) ; fi; od; RETURN(a); fi; end: for n from 1 to 80 do printf("%d,",A141435(n)) ; od: # R. J. Mathar, Nov 03 2008

def A141435(terms):
    seq = [1, 2]
    for n in range(3, terms):
        s = 0
        for m in seq:
            if (n - m) > 0:
                s += seq[n - m - 1] #fix for python indexing
        seq.append(s)
    return seq
print(A141435(40)) # Andres Cruz y Corro A, Jun 19 2019

More terms from R. J. Mathar, Nov 03 2008

A202012 Expansion of (1-x+x^2)/((1-x)(1-x-x^2-x^3)).

Original entry on oeis.org

1, 1, 3, 6, 11, 21, 39, 72, 133, 245, 451, 830, 1527, 2809, 5167, 9504, 17481, 32153, 59139, 108774, 200067, 367981, 676823, 1244872, 2289677, 4211373, 7745923, 14246974, 26204271, 48197169, 88648415, 163049856

Offset: 0

Philippe Deléham, Dec 08 2011

Antidiagonal sums of triangle T(n,k) = A104040(n,k)*(-1)^floor(k/2). - Philippe Deléham, Dec 11 2011

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Cf. A008937, A081172, A104040.

CoefficientList[Series[(1-x+x^2)/((1-x)(1-x-x^2-x^3)),{x,0,40}],x] (* or *) LinearRecurrence[{2,0,0,-1},{1,1,3,6},40] (* Harvey P. Dale, Apr 21 2014 *)

a(n) = 2*a(n-1) - a(n-4), n>3.
a(n) = A008937(n-1) - A008937(n) + A008937(n+1). - R. J. Mathar, Dec 10 2011
a(n+1)-a(n) = A081172(n+2). - Philippe Deléham, Dec 11 2011

A136337 Triangular sequence from both a quartic expansion polynomial and a four deep polynomial recursion: Expansion polynomial: f(x,t)=1/(1 - 2xt + t^4); Recursion polynomials: p(x, n) = 2xp(x, n - 1) - p(x, n - 4);.

Original entry on oeis.org

1, 0, 2, 0, 0, 4, 0, 0, 0, 8, -1, 0, 0, 0, 16, 0, -4, 0, 0, 0, 32, 0, 0, -12, 0, 0, 0, 64, 0, 0, 0, -32, 0, 0, 0, 128, 1, 0, 0, 0, -80, 0, 0, 0, 256, 0, 6, 0, 0, 0, -192, 0, 0, 0, 512, 0, 0, 24, 0, 0, 0, -448, 0, 0, 0, 1024

Offset: 1

Roger L. Bagula, Apr 12 2008

Row sums are A008937.
This sequence was a designed experiment in Umbral Calculus
using a quartic polynomial for the expansion base.
I was testing the recursion form in the polynomial as:
f(x,t)=1/(1-p(x,1)*t +t^m): m the depth of the recursion.
as a recursion of the form:
p(x,n)=p(x,1)*p(x,n-1)-p(x,n-m).

Cf. A008937.

(*expansion polynomial*) Clear[p, a] p[t_] = 1/(1 - 2*x*t + t^4) g = Table[ ExpandAll[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[ CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* recursion polynomial*) Clear[p] p[x,-1]=0;p[x, 0] = 1; p[x, 1] = 2x; p[x, 2] = 4x^2; p[x_, n_] := p[x, n] = 2*x*p[x, n - 1] - p[x, n - 4]; Table[ExpandAll[p[x, n]], {n, 0, Length[g] - 1}]; Flatten[Table[CoefficientList[p[x, n], x], {n, 0, Length[g] - 1}]]

f(x,t)=1/(1 - 2*x*t + t^4); f(x,t)=Sum[q(x,n)*t^n,{n,1,Infinity}]; p(x,-1)=0;p(x,0)=1;p(x,1)=2*x;p(x,2)=4*x^2; p(x, n) = 2*x*p(x, n - 1) - p(x, n - 4);

A136438 Hypertribonacci number array read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 1, 3, 4, 4, 0, 0, 1, 4, 7, 8, 7, 0, 0, 1, 5, 11, 15, 15, 13, 0, 0, 1, 6, 16, 26, 30, 28, 24, 0, 0, 1, 7, 22, 42, 56, 58, 52, 44, 0, 0, 1, 8, 29, 64, 98, 114, 110, 96, 81, 0, 0, 1, 9, 37, 93, 162, 212, 224, 206, 177, 149

Offset: 1

Jonathan Vos Post, Apr 13 2008

The hypertribonacci numbers are to the hyperfibonacci array of A136431 as the tribonacci numbers A000073 are to the Fibonacci numbers A000045.

			The array a(k,n) begins:
========================================
n=0..|.1.|.2.|...3.|..4.|...5.|....6.|...7..|.....8.|.....9.|....10.|
========================================
k=0..|.0.|.0.|...1.|..2.|...4.|....7.|..13..|....24.|....44.|....81.| A000073
k=1..|.0.|.0.|...2.|..4.|...8.|...15.|..28..|....52.|....96.|...177.| A008937
k=2..|.0.|.0.|...3.|..7.|..15.|...30.|..58..|...110.|...206.|...383.| A062544
k=3..|.0.|.0.|...4.|.11.|..26.|...56.|..114.|...224.|...430.|...813.|
k=4..|.0.|.0.|...5.|.16.|..42.|...98.|..212.|...436.|...866.|..1679.|
k=5..|.0.|.0.|...6.|.22.|..64.|..162.|..374.|...810.|..1676.|..3355.|
k=6..|.0.|.0.|...7.|.29.|..93.|..255.|..629.|..1439.|..3115.|..6470.|
k=7..|.0.|.0.|...8.|.37.|.130.|..385.|.1014.|..2453.|..5568.|.12038.|
k=8..|.0.|.0.|...9.|.46.|.176.|..561.|.1575.|..4028.|..9596.|.21634.|
k=9..|.0.|.0.|..10.|.56.|.232.|..793.|.2368.|..6396.|.15992.|.37626.|
k=10.|.0.|.0.|..11.|.67.|.299.|.1092.|.3460.|..9856.|.25848.|.63474.|
========================================

Columns n=4..6 are A000124, A000125, A055795.

Cf. A000045, A000073, A008937, A062544, A136431, A137176.

\\ create the n X n matrix of nonzero values
hypertribo(n)={ local(M=matrix(n,n)); M[1,]=Vec(1/(1-x-x^2-x^3)+O(x^n));
M[,1]=vector(n,i,1)~; for(i=2,n, for(j=2,n, M[i,j]=M[i-1,j]+M[i,j-1])); M}
{ hypertribo(10) }

a(k,n) = apply partial sum operator k times to tribonacci numbers A000073.

M. F. Hasler notes that the 8th column = vector(25,n,binomial(n+5,6)+binomial(n+5,4)+2*binomial(n+3,1)). R. J. Mathar points out that the repeated partial sums are quickly computed from their o.g.f.s (-1)^(k+1)*x^2/(-1+x+x^2+x^3)/(-1+x)^k, k=1,2,3,...

Examples corrected by R. J. Mathar, Apr 21 2008

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A320452 Number of possible states when placing n tokens of 2 alternating types on 2 piles.

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A141435 a(1) = 1, a(2) = 2; a(n) = a(n-a(1)) + a(n-a(2)) + a(n-a(3)) + a(n-a(4)) + ...

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A202012 Expansion of (1-x+x^2)/((1-x)(1-x-x^2-x^3)).

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A136337 Triangular sequence from both a quartic expansion polynomial and a four deep polynomial recursion: Expansion polynomial: f(x,t)=1/(1 - 2*x*t + t^4); Recursion polynomials: p(x, n) = 2*x*p(x, n - 1) - p(x, n - 4);.

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A136438 Hypertribonacci number array read by antidiagonals.

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Extensions

A136337 Triangular sequence from both a quartic expansion polynomial and a four deep polynomial recursion: Expansion polynomial: f(x,t)=1/(1 - 2xt + t^4); Recursion polynomials: p(x, n) = 2xp(x, n - 1) - p(x, n - 4);.