cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for n >= 0, with a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

Examples

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...

References

  • M. Agronomof, Sur une suite récurrente, Mathesis (Series 4), Vol. 4 (1914), pp. 125-126.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

Programs

  • GAP
    a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018
  • Haskell
    a000073 n = a000073_list !! n
a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                          (zipWith (+) a000073_list $ tail a000073_list))
-- Reinhard Zumkeller, Dec 12 2011
  • Magma
    [n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
# second Maple program:
A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018
  • Mathematica
    CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)
  • Maxima
    A000073[0]:0$
A000073[1]:0$
A000073[2]:1$
A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
  makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */
  • PARI
    {a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */
  • PARI
    my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016
  • PARI
    a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018
  • Python
    def a(n, adict={0:0, 1:0, 2:1}):
    if n in adict:
        return adict[n]
    adict[n]=a(n-1)+a(n-2)+a(n-3)
    return adict[n] # David Nacin, Mar 07 2012
from functools import cache
@cache
def A000073(n: int) -> int:
    if n <= 1: return 0
    if n == 2: return 1
    return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022

Formula

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265. a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Extensions

Minor edits by M. F. Hasler, Apr 18 2018
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A007800 From a problem in AI planning: a(n) = 4+a(n-1)+a(n-2)+a(n-3)+a(n-4)-a(n-5)-a(n-6)-a(n-7), n>7.

Original entry on oeis.org

1, 2, 4, 8, 16, 31, 59, 111, 207, 384, 710, 1310, 2414, 4445, 8181, 15053, 27693, 50942, 93704, 172356, 317020, 583099, 1072495, 1972635, 3628251, 6673404, 12274314, 22575994, 41523738, 76374073, 140473833, 258371673, 475219609, 874065146
Offset: 1

Views

Author

Peter Jonsson [ petej(AT)ida.liu.se ]

Keywords

Comments

The number of length n binary words with fewer than 3 zeros between any pair of consecutive ones. - Jeffrey Liese, Dec 23 2010

Links

Crossrefs

Cf. A062544.

Programs

  • Maple
    for n from 1 to 5 do a[n]:= [1,2,4,8,16][n] od:
for n from 6 to 100 do a[n]:= 3*a[n-1]-2*a[n-2]-a[n-4]+a[n-5] od:
seq(a[n],n=1..100); # Robert Israel, Aug 19 2014
  • Mathematica
    LinearRecurrence[{3,-2,0,-1,1},{1,2,4,8,16},40] (* Harvey P. Dale, Apr 24 2013 *)
  • PARI
    Vec(-x*(x^4-x+1)/((x-1)^2*(x^3+x^2+x-1)) + O(x^100)) \\ Colin Barker, Aug 18 2014

Formula

a(1)=1, a(2)=2, a(3)=4, a(4)=8, a(5)=16, a(n)=3*a(n-1)-2*a(n-2)+0*a(n-3)- a(n-4)+ a(n-5). - Harvey P. Dale, Apr 24 2013
G.f.: -x*(x^4-x+1) / ((x-1)^2*(x^3+x^2+x-1)). - Colin Barker, Aug 18 2014
2*a(n) = A001590(n+4)-n. - R. J. Mathar, Aug 16 2017

A119407 Number of nonempty subsets of {1,2,...,n} with no gap of length greater than 4 (a set S has a gap of length d if a and b are in S but no x with a < x < b is in S, where b-a=d).

Original entry on oeis.org

1, 3, 7, 15, 31, 62, 122, 238, 462, 894, 1727, 3333, 6429, 12397, 23901, 46076, 88820, 171212, 330028, 636156, 1226237, 2363655, 4556099, 8782171, 16928187, 32630138, 62896622, 121237146, 233692122, 450456058, 868281979, 1673667337, 3226097529, 6218502937, 11986549817, 23104817656
Offset: 1

Views

Author

John W. Layman, Jul 25 2006

Keywords

Comments

The numbers of subsets of {1,2,...,n} with no gap of length greater than d, for d=1,2 and 3, seem to be given in A000217, A001924 and A062544, respectively.

Examples

			G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 62*x^6 + 122*x^7 + 238*x^8 + 462*x^9 + ...

Links

Crossrefs

Cf. A000217, A001924, A062544, A107066.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x/((1-x)*(1-2*x+x^5)) )); // G. C. Greubel, Jun 05 2019
  • Mathematica
    Rest@CoefficientList[Series[x/((1-x)*(1-2*x+x^5)), {x, 0, 40}], x] (* G. C. Greubel, Jun 05 2019 *)
LinearRecurrence[{3,-2,0,0,-1,1},{1,3,7,15,31,62},40] (* Harvey P. Dale, Dec 04 2019 *)
  • PARI
    {a(n) = if( n<0, n = -n; polcoeff( x^5 / ((1 - x)^2 * (1 + x + x^2 + x^3 - x^4)) + x * O(x^n), n), polcoeff( x / ((1 - x)^2 * (1 - x - x^2 - x^3 - x^4)) + x * O(x^n), n))} /* Michael Somos, Dec 28 2012 */
  • PARI
    my(x='x+O('x^40)); Vec(x/((1-x)*(1-2*x+x^5))) \\ G. C. Greubel, Jun 05 2019
  • Sage
    a=(x/((1-x)*(1-2*x+x^5))).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Jun 05 2019

Formula

G.f. for number of nonempty subsets of {1,2,...,n} with no gap of length greater than d is x/((1-x)*(1-2*x+x^(d+1))). - Vladeta Jovovic, Apr 27 2008
From Michael Somos, Dec 28 2012: (Start)
G.f.: x/((1-x)^2*(1-x-x^2-x^3-x^4)) = x/((1-x)*(1-2*x+x^5)).
First difference is A107066. (End)
a(n-3) = Sum_{k=0..n} (n-k)*A000078(k) for n>3. - Greg Dresden, Jan 01 2021

Extensions

Terms a(25) onward added by G. C. Greubel, Jun 05 2019

A356619 a(n) = number of k-tuples (u(1), u(2), ..., u(k)) with 1 <= u(1) < u(2) < ... < u(k) <= n such that u(i) - u(i-1) <= 3 for i = 2,...,k.

Original entry on oeis.org

0, 1, 4, 11, 25, 52, 103, 198, 374, 699, 1298, 2401, 4431, 8166, 15037, 27676, 50924, 93685, 172336, 316999, 583077, 1072472, 1972611, 3628226, 6673378, 12274287, 22575966, 41523709, 76374043, 140473802, 258371641, 475219576, 874065112, 1607656425
Offset: 0

Views

Author

Clark Kimberling, Aug 24 2022

Keywords

Links

Crossrefs

Cf. A001891, A062544, A221949, A356620, A356621.

Programs

  • Mathematica
    maxDiff = 3;
t = Map[Length[Select[Map[{#, Max[Differences[#]]} &,
     Drop[Subsets[Range[#]], # + 1]], #[[2]] <= maxDiff &]] &, Range[16]]
FindGeneratingFunction[%, x]
FindLinearRecurrence[t]
LinearRecurrence[{3, -2, 0, -1, 1}, {0, 1, 4, 11, 25}, 45]

Formula

G.f.: x*(1 + x + x^2)/((-1 + x)^2*(1 - x - x^2 - x^3)).
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-4) + a(n-5).
a(n) = A221949(n+2)-1 for n >= 0.

A325473 Number of compositions of n with no part divisible by 3 and an even number of parts congruent to 4 or 5 modulo 6.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 22, 38, 67, 120, 217, 395, 722, 1323, 2428, 4460, 8197, 15070, 27711, 50961, 93724, 172377, 317042, 583122, 1072519, 1972660, 3628277, 6673431, 12274342, 22576023, 41523768, 76374104, 140473865, 258371706, 475219643, 874065181, 1607656496
Offset: 0

Views

Author

Brian Hopkins, Sep 06 2019

Keywords

Examples

			a(4) counts (1,1,1,1), (1,1,2), (1,2,1), (2,1,1), (2,2), but not (1,3) or (3,1) since they contain 3, neither (4) since that has an odd number of parts congruent to 4 or 5 mod 6.

Links

Crossrefs

Cf. A001590, A062544.

Formula

a(n) = (A001590(n+2) + n)/2, see Moser & Whitley reference, Theorem 3.
a(n) = A062544(n-3) + n for n >= 3 (also for n = 1 and 2 with A062544(-2) = A062544(-1) = 0), Moser & Whitney.
G.f.: (x^5-x^4+x^3-x^2+2*x-1)/((x^3+x^2+x-1)*(x-1)^2). - Alois P. Heinz, Sep 06 2019

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

Views

Author

Jonathan Vos Post, Apr 13 2008

Keywords

Comments

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

Examples

			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.|
========================================

Crossrefs

Columns n=4..6 are A000124, A000125, A055795.
Cf. A000045, A000073, A008937, A062544, A136431, A137176.

Programs

  • PARI
    \\ 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) }

Formula

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,...

Extensions

Examples corrected by R. J. Mathar, Apr 21 2008
Showing 1-6 of 6 results.

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