A113300 -id:A113300 - cp's OEIS frontend

A085697 a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.

0, 0, 1, 1, 4, 16, 49, 169, 576, 1936, 6561, 22201, 75076, 254016, 859329, 2907025, 9834496, 33269824, 112550881, 380757169, 1288092100, 4357584144, 14741602225, 49870482489, 168710633536, 570743986576, 1930813074369, 6531893843049

Offset: 0

Emanuele Munarini, Jul 18 2003

In general, squaring the terms of a third-order linear recurrence with signature (x,y,z) will result in a sixth-order recurrence with signature (x^2 + y, x^2*y + z*x + y^2, x^3*z + 4*x*y*z - y^3 + 2*z^2, x^2*z^2 - x*y^2*z - z^2*y, z^2*y^2 - z^3*x, -z^4). - Gary Detlefs, Jan 10 2023

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,6,-1,0,-1).

Cf. A000073, A057597, A099463, A107239, A113300.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1-3*x-x^2-x^3)*(1+x+x^2-x^3)) )); // G. C. Greubel, Nov 20 2021

LinearRecurrence[{2,3,6,-1,0,-1},{0,0,1,1,4,16},30] (* Harvey P. Dale, Oct 26 2020 *)

t[0]:0$  t[1]:0$  t[2]:1$
t[n]:=t[n-1]+t[n-2]+t[n-3]$
makelist(t[n]^2,n,0,40); /* Emanuele Munarini, Mar 01 2011 */

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A085697(n): return T(n)^2
[A085697(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

G.f.: x^2*( 1-x-x^2-x^3 )/( (1-3*x-x^2-x^3)*(1+x+x^2-x^3) ).
a(n+6) = 2*a(n+5) + 3*a(n+4) + 6*a(n+3) - a(n+2) - a(n).
a(n) = (-A057597(n-2) + 3*A057597(n-1) + 6*A057597(n) + 5*A113300(n-1) - A099463(n-2))/11. - R. J. Mathar, Aug 19 2008

Offset corrected to match A000073 by N. J. A. Sloane, Sep 12 2020

Name corrected to match corrected offset by Michael A. Allen, Jun 10 2021

A089068 a(n) = a(n-1)+a(n-2)+a(n-3)+2 with a(0)=0, a(1)=0 and a(2)=1.

Original entry on oeis.org

0, 0, 1, 3, 6, 12, 23, 43, 80, 148, 273, 503, 926, 1704, 3135, 5767, 10608, 19512, 35889, 66011, 121414, 223316, 410743, 755475, 1389536, 2555756, 4700769, 8646063, 15902590, 29249424, 53798079, 98950095, 181997600, 334745776, 615693473

Offset: 0

Roger L. Bagula, Dec 03 2003

The a(n+2) represent the Kn12 and Kn22 sums of the square array of Delannoy numbers A008288. See A180662 for the definition of these knight and other chess sums. - Johannes W. Meijer, Sep 21 2010

Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Cf. A000931, A000073, A077939, A113300, A001057, A006054, A033505.

Cf. A000073 (Kn11 & Kn21), A089068 (Kn12 & Kn22), A180668 (Kn13 & Kn23), A180669 (Kn14 & Kn24), A180670 (Kn15 & Kn25). - Johannes W. Meijer, Sep 21 2010

Join[{a=0,b=0,c=1},Table[d=a+b+c+2;a=b;b=c;c=d,{n,50}]] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)
RecurrenceTable[{a[0]==a[1]==0,a[2]==1,a[n]==a[n-1]+a[n-2]+a[n-3]+2}, a[n],{n,40}] (* or *) LinearRecurrence[{2,0,0,-1},{0,0,1,3},40] (* Harvey P. Dale, Sep 19 2011 *)

a(n) = A008937(n-2)+A008937(n-1). - Johannes W. Meijer, Sep 21 2010
a(n) = A018921(n-5)+A018921(n-4), n>4. - Johannes W. Meijer, Sep 21 2010
a(n) = A000073(n+2)-1. - R. J. Mathar, Sep 22 2010
From Johannes W. Meijer, Sep 22 2010: (Start)
a(n) = a(n-1)+A001590(n+1).
a(n) = Sum_{m=0..n} A040000(m)*A000073(n-m).
a(n+2) = Sum_{k=0..floor(n/2)} A008288(n-k+1,k+1).
G.f. = x^2*(1+x)/((1-x)*(1-x-x^2-x^3)). (End)
a(n) = 2*a(n-1)-a(n-4), a(0)=0, a(1)=0, a(2)=1, a(3)=3. - Bruno Berselli, Sep 23 2010

Corrected and information added by Johannes W. Meijer, Sep 22 2010, Oct 22 2010

Definition based on arbitrarily set floating-point precision removed by R. J. Mathar, Sep 30 2010

A099463 Bisection of tribonacci numbers.

Original entry on oeis.org

0, 1, 2, 7, 24, 81, 274, 927, 3136, 10609, 35890, 121415, 410744, 1389537, 4700770, 15902591, 53798080, 181997601, 615693474, 2082876103, 7046319384, 23837527729, 80641778674, 272809183135, 922906855808, 3122171529233

Offset: 0

Paul Barry, Oct 16 2004

Binomial transform of A099462.
From Paul Barry, Feb 07 2006: (Start)
a(n+1) gives row sums of number triangle A114123 or A184883.
Partial sums are A113300. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 24.
Meng-Han Wu, Henryk A. Witek, and Rafał Podeszwa, Clar Covers and Zhang-Zhang Polynomials of Zigzag and Armchair Carbon Nanotubes, MATCH Commun. Math. Comput. Chem. (2025) Vol. 93, 415-462. See p. 437.
Index entries for linear recurrences with constant coefficients, signature (3,1,1).

Cf. A000073, A058265, A099462, A113300, A114123, A184883.

[n le 3 select (n-1) else 3*Self(n-1) +Self(n-2) +Self(n-3): n in [1..31]]; // G. C. Greubel, Nov 20 2021

LinearRecurrence[{3,1,1},{0,1,2},30] (* or *) Join[{0},Mean/@ Partition[ LinearRecurrence[ {1,1,1},{1,1,1},60],2]] (* Harvey P. Dale, Apr 02 2012 *)

def A184883(n, k): return simplify( hypergeometric([-k, 2*(k-n)], [1], 2) )
def A099463(n): return sum( A184883(n, k) for k in (0..n) )
[0]+[A099463(n-1) for n in (1..40)] # G. C. Greubel, Nov 20 2021

G.f.: x*(1-x)/(1-3*x-x^2-x^3).
a(n) = Sum_{k=0..n} binomial(n, k)*Sum_{j=0..floor((k-1)/2)} binomial(j, k-2*j-1)*4^j.
From Paul Barry, Feb 07 2006: (Start)
a(n) = 3*a(n-1) + a(n-2) + a(n-3).
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(2*k, n-k-j)*C(n-k, j)*2^(n-k-j). (End)
a(n)/a(n-1) tends to 3.38297576..., the square of the tribonacci constant A058265. - Gary W. Adamson, Feb 28 2006
If p[1]=2, p[2]=3, p[i]=4, (i>2), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n+1) = det A. - Milan Janjic, May 02 2010

A073717 a(n) = T(2n+1), where T(n) are the tribonacci numbers A000073.

Original entry on oeis.org

0, 1, 4, 13, 44, 149, 504, 1705, 5768, 19513, 66012, 223317, 755476, 2555757, 8646064, 29249425, 98950096, 334745777, 1132436852, 3831006429, 12960201916, 43844049029, 148323355432, 501774317241, 1697490356184, 5742568741225

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 05 2002

In general, the bisection of a third-order linear recurrence with signature (x,y,z) will result in a third-order recurrence with signature (x^2 + 2*y, 2*z*x - y^2, z^2). - Gary Detlefs, May 29 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Meng-Han Wu, Henryk A. Witek, Rafał Podeszwa, Clar Covers and Zhang-Zhang Polynomials of Zigzag and Armchair Carbon Nanotubes, MATCH Commun. Math. Comput. Chem. (2025) Vol. 93, 415-462. See p. 437.
Index entries for linear recurrences with constant coefficients, signature (3,1,1).

Cf. A000073, A099463, A113300.

Row sums of A216182.

[n le 3 select (n-1)^2 else 3*Self(n-1) +Self(n-2) +Self(n-3): n in [1..31]]; // G. C. Greubel, Nov 19 2021

CoefficientList[Series[(x+x^2)/(1-3x-x^2-x^3), {x, 0, 30}], x]
LinearRecurrence[{3,1,1},{0,1,4},30] (* Harvey P. Dale, Sep 07 2015 *)

def A073717_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x)/(1-3*x-x^2-x^3) ).list()
A073717_list(30) # G. C. Greubel, Nov 19 2021

a(n) = 3*a(n-1) + a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=4.
G.f.: x*(1+x)/(1-3*x-x^2-x^3).
a(n+1) = Sum_{k=0..n} A216182(n,k). - Philippe Deléham, Mar 11 2013
a(n) = A113300(n-1) + A113300(n). - R. J. Mathar, Jul 04 2019

A154948 Riordan array ((1+x)/(1-x^2)^2, x(1+x)/(1-x)).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 2, 6, 5, 1, 3, 10, 14, 7, 1, 3, 15, 30, 26, 9, 1, 4, 21, 55, 70, 42, 11, 1, 4, 28, 91, 155, 138, 62, 13, 1, 5, 36, 140, 301, 363, 242, 86, 15, 1, 5, 45, 204, 532, 819, 743, 390, 114, 17, 1, 6, 55, 285, 876, 1652, 1925, 1375, 590, 146, 19, 1

Offset: 0

Paul Barry, Jan 17 2009

Row sums are A113300(n+1). Diagonal sums are A154949.

Product of A154950 and A007318.

			Triangle begins
  1;
  1,  1;
  2,  3,  1;
  2,  6,  5,  1;
  3, 10, 14,  7,  1;
  3, 15, 30, 26,  9,  1;
  4, 21, 55, 70, 42, 11, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

[ (&+[Binomial(k-1, j)*Binomial(n-j+1, k+1): j in [0..n+1]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

seq(seq( add(binomial(k-1, j)*binomial(n-j+1, k+1), j=0..n+1), k=0..n), n=0..10); # G. C. Greubel, Feb 18 2020

Table[Binomial[n+1, k+1]*Hypergeometric2F1[-n+k, -k+1, -n-1, -1], {n, 0, 5}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

[[ sum(binomial(k-1, j)*binomial(n-j+1, k+1) for j in (0..n+1)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 18 2020

Number triangle T(n,k) = Sum_{j=0..n+1} C(n+1-j,k+1)*C(k-1,j).

T(n, k) = binomial(n+1,k+1)*2F1(-(n-k), -(k-1); -(n+1); -1). - G. C. Greubel, Feb 18 2020

a(45)=0 removed by Georg Fischer, Feb 18 2020

A113301 Sum of odd-indexed terms of tribonacci numbers.

Original entry on oeis.org

0, 1, 5, 18, 62, 211, 715, 2420, 8188, 27701, 93713, 317030, 1072506, 3628263, 12274327, 41523752, 140473848, 475219625, 1607656477, 5438662906, 18398864822, 62242913851, 210566269283, 712340586524, 2409830942708, 8152399683933, 27579370581033, 93300342369742

Offset: 0

Jonathan Vos Post, Oct 24 2005

A000073 is the tribonacci numbers. A113300 is the sum of even-indexed terms of tribonacci numbers. A099463 is the bisection of the tribonacci numbers. A113300(n) + A113301(n) = cumulative sum of tribonacci numbers = A008937(n). Primes in A113300 include a(2) = 5, a(5) = 211, a(9) = 27701, .... A113300 is semiprime for n = 4, 10, 14, ...

			a(0) = 0 = A000073(1);
a(1) = 0+1 = A000073(1) + A000073(3) = 1;
a(2) = 0+1+4 = A000073(1) + A000073(3) + A000073(5) = 5, prime;
a(3) = 0+1+4+13 = A000073(1) + A000073(3) + A000073(5) + A000073(7) = 18;
a(4) = 0+1+4+13+44 = A000073(1) + A000073(3) + A000073(5) + A000073(7) + A000073(9) = 62 = 2 * 31, semiprime;
a(5) = 0+1+4+13+44+149 = A000073(1) + A000073(3) + A000073(5) + A000073(7) + A000073(9) + A000073(11) = 211, prime.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,0,-1).

Cf. A000073, A008937, A099463, A113300.

I:=[0,1,5,18]; [n le 4 select I[n] else 4*Self(n-1) - 2*Self(n-2) -Self(n-4): n in [1..41]]; // G. C. Greubel, Nov 20 2021

Accumulate[Take[LinearRecurrence[{1,1,1},{0,1,1},40],{1,-1,2}]] (* or *) LinearRecurrence[{4,-2,0,-1},{0,1,5,18},30] (* Harvey P. Dale, Apr 12 2013 *)

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A113301(n): return sum(T(2*j+1) for j in (0..n))
[A113301(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

a(n) = Sum_{j=0..n} A000073(2*j+1).
a(n) + A113300(n) = A008937(n).
a(n) = 4*a(n-1) - 2*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=18. - Harvey P. Dale, Apr 12 2013
G.f.: x*(1+x) / ((1-x)*(1-3*x-x^2-x^3)). - Colin Barker, May 06 2013

More terms from Colin Barker, May 06 2013

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A085697 a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.

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A089068 a(n) = a(n-1)+a(n-2)+a(n-3)+2 with a(0)=0, a(1)=0 and a(2)=1.

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A099463 Bisection of tribonacci numbers.

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A073717 a(n) = T(2n+1), where T(n) are the tribonacci numbers A000073.

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A154948 Riordan array ((1+x)/(1-x^2)^2, x(1+x)/(1-x)).

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A113301 Sum of odd-indexed terms of tribonacci numbers.

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