A127893 -id:A127893 - cp's OEIS frontend

A095263 a(n+3) = 3a(n+2) - 2a(n+1) + a(n).

1, 3, 7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681, 145547525, 338356945, 786584466, 1828587033, 4250949112, 9882257736, 22973462017, 53406819691

Offset: 1

Gary W. Adamson, May 31 2004

a(n+1) = number of n-tuples over {0,1,2} without consecutive digits. For the general case see A096261.
Diagonal sums of Riordan array (1/(1-x)^3, x/(1-x^3)), A127893. - Paul Barry, Jan 07 2008
The signed variant (-1)^(n+1)*a(n+1) is the bottom right entry of the n-th power of the matrix [[0,1,0],[0,0,1],[-1,-2,-3]]. - Roger L. Bagula, Jul 01 2007
a(n) is the number of generalized compositions of n+1 when there are i^2/2-i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.1) gives a bijection between colored compositions of n, where each part k has one of binomial(k,2) colors, and 0,1,2 strings of length n-2 without sequential digits (i.e., avoiding 01 and 12). Cf. A052529. - Peter Bala, Sep 17 2013
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^2 - S^3; see A291000. - Clark Kimberling, Aug 24 2017
For n>1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 2 blocks (i.e., subintervals) from each interval.  For example, for n=6, a(5)=37 since the number of ways to split [6] into intervals and then select 2 blocks from each interval is C(6,2) + C(4,2)*C(2,2) + C(3,2)*C(3,2) + C(2,2)*C(4,2) + C(2,2)*C(2,2)*C(2,2). - Enrique Navarrete, May 20 2022

			a(9) = 1081 = 3*465 - 2*200 + 86.
M^9 * [1 0 0] = [a(7) a(8) a(9)] = [200 465 1081].
G.f. = x + 3*x^2 + 7*x^3 + 16*x^4 + 37*x^5 + 86*x^6 + 200*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, Jack N. Averitt College of Graduate Studies, Georgia Southern University, 2012.
Index entries for linear recurrences with constant coefficients, signature (3,-2,1).

Cf. A000931, A010912, A034943, A052529, A052530, A097550, A137531.
Cf. A052921 (first differences), A137229 (partial sums).
Column k=3 of A277666.

I:=[1,3,7]; [n le 3 select I[n] else 3*Self(n-1) -2*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 12 2021

A:= gfun:-rectoproc({a(n+3)=3*a(n+2)-2*a(n+1)+a(n),a(1)=1,a(2)=3,a(3)=7},a(n),remember):
seq(A(n),n=1..100); # Robert Israel, Sep 15 2014

a[1]=1; a[2]=3; a[3]=7; a[n_]:= a[n]= 3a[n-1] -2a[n-2] +a[n-3]; Table[a[n], {n, 22}] (* Or *)
a[n_]:= (MatrixPower[{{0,1,2,3}, {1,2,3,0}, {2,3,0,1}, {3,0,1,2}}, n].{{1}, {0}, {0}, {0}})[[2, 1]]; Table[ a[n], {n, 22}] (* Robert G. Wilson v, Jun 16 2004 *)
RecurrenceTable[{a[1]==1,a[2]==3,a[3]==7,a[n+3]==3a[n+2]-2a[n+1]+a[n]},a,{n,30}] (* Harvey P. Dale, Sep 17 2022 *)

[sum( binomial(n+k+1,3*k+2) for k in (0..(n-1)//2)) for n in (1..30)] # G. C. Greubel, Apr 12 2021

Let M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 -2 3]; then M^n *[1 0 0] = [a(n-2) a(n-1) a(n)].
a(n)/a(n-1) tends to 2.3247179572..., an eigenvalue of M and a root of the characteristic polynomial. [Is that constant equal to 1 + A060006? - Michel Marcus, Oct 11 2014] [Yes, the limit is the root of the equation -1 + 2*x - 3*x^2 + x^3 = 0, after substitution x = y + 1 we have the equation for y: -1 - y + y^3 = 0, y = A060006. - Vaclav Kotesovec, Jan 27 2015]
Related to the Padovan sequence A000931 as follows : a(n)=A000931(3n+4). Also the binomial transform of A000931(n+4).
From Paul Barry, Jul 06 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, n-2*k+1).
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, 3*k-1). (End)
From Paul Barry, Jan 07 2008: (Start)
G.f.: x/(1 -3*x +2*x^2 -x^3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k+2,3*k+2).
a(n) = Sum_{k=0..n} binomial(n,k) * Sum_{j=0..floor((k+4)/2)} binomial(j,k-2j+4). (End)
If p[i]=i(i-1)/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>=2, a(n-1)=det A. - Milan Janjic, May 02 2010
a(n) = A000931(3*n + 4). - Michael Somos, Sep 18 2012

Edited by Paul Barry, Jul 06 2004

Corrected and extended by Robert G. Wilson v, Jun 05 2004

A052529 Expansion of (1-x)^3/(1 - 4x + 3x^2 - x^3).

Original entry on oeis.org

1, 1, 4, 13, 41, 129, 406, 1278, 4023, 12664, 39865, 125491, 395033, 1243524, 3914488, 12322413, 38789712, 122106097, 384377665, 1209982081, 3808901426, 11990037126, 37743426307, 118812495276, 374009739309, 1177344897715

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

a(n+1) is the number of distinct matrix products in (A+B+C+D)^n where commutator [A,B] = [A,C] = [B,C] = 0 but D does not commute with A, B, or C. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Starting (1, 4, 13, ...) = INVERT transform of the triangular series, (1, 3, 6, 10, ...). Example: a(5) = 129 = termwise products of (1, 1, 4, 13, 41) and (15, 10, 6, 3, 1) = (15 + 10 + 24 + 39 + 41). - Gary W. Adamson, Apr 10 2009
a(n) is the number of generalized compositions of n when there are i^2/2+i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.2) gives a bijection between colored compositions of n, where each part k has one of binomial(k+1,2) colors, and 0,1,2,3 strings of length n-1 avoiding 10, 20 and 21. Cf. A095263. For a refinement of this sequence counting binomial(k+1,2)-colored compositions by the number of parts see A127893. - Peter Bala, Sep 17 2013

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 80.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3 , example 14.
C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, Jack N. Averitt College of Graduate Studies, Georgia Southern University, 2012.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 459
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Index entries for linear recurrences with constant coefficients, signature (4,-3,1).

Cf. A001906, A052530, A055991, A095263.
Trisection of A000930.
First differences of A052544.
Row sums of triangle A127893.

I:=[1, 1, 4, 13, 41, 129]; [n le 6 select I[n] else 4*Self(n-1) -3*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012

spec := [S,{S=Sequence(Prod(Z,Sequence(Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
f:= gfun:-rectoproc({a(n+4)-4*a(n+3)+3*a(n+2)-a(n+1), a(0) = 1, a(1) = 1, a(2) = 4, a(3) = 13},a(n),`remember`):
seq(f(n),n=0..40); # Robert Israel, Dec 19 2014

CoefficientList[Series[(-1+x)^3/(-1+4*x-3*x^2+x^3),{x,0,40}],x] (* Vincenzo Librandi, Jun 22 2012 *)
LinearRecurrence[{4,-3,1},{1,1,4,13},30] (* Harvey P. Dale, Oct 04 2015 *)

my(x='x+O('x^30)); Vec((1-x)^3/(1-4*x+3*x^2-x^3)) \\ G. C. Greubel, May 12 2019

((1-x)^3/(1-4*x+3*x^2-x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 12 2019

a(n) = Sum_{a=0..n} (Sum_{b=0..n} (Sum_{c=0..n} C(n-b-c,a)*C(n-a-c,b)*C(n-a-b,c))).
G.f.: (1 - x)^3/(1 - 4*x + 3*x^2 - x^3).
a(n) = 4*a(n-1) - 3*a(n-2) + a(n-3) for n>=4.
a(n) = Sum_{alpha = RootOf(-1+4*x-3*x^2+x^3)} (1/31)*(6 - 5*alpha - 3*alpha^2) * alpha^(-1-n).
For n>0, a(n) = Sum_{k=0..n-1} Sum_{i=0..k} Sum_{j=0..i} a(j). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=0..n} binomial(n+2*k-1, n-k). - Vladeta Jovovic, Mar 23 2003
If p[i]=i(i+1)/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)=det A. - Milan Janjic, May 02 2010
Recurrence equation: a(n) = Sum_{k = 1..n} 1/2*k*(k+1)*a(n-k) with a(0) = 1. - Peter Bala, Sep 19 2013
a(n) = Sum_{i=0..n} (n-i)*A052544(i) = A052544(n) - A052544(n-1) for n>=1. - Areebah Mahdia, Jul 07 2020

A122432 Riordan array (1/(1+x)^3,x).

Original entry on oeis.org

1, -3, 1, 6, -3, 1, -10, 6, -3, 1, 15, -10, 6, -3, 1, -21, 15, -10, 6, -3, 1, 28, -21, 15, -10, 6, -3, 1, -36, 28, -21, 15, -10, 6, -3, 1, 45, -36, 28, -21, 15, -10, 6, -3, 1, -55, 45, -36, 28, -21, 15, -10

Offset: 0

Paul Barry, Sep 04 2006

Sequence array for (-1)^n*C(n+2,2). Inverse of A122431. Row sums are -A083392(n+1). Antidiagonal sums are (-1)^n*A002623(n).
Call the unsigned version of this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A127893. - Peter Bala, Jul 22 2014
From Wolfdieter Lang, Apr 05 2020: (Start)
Triangle T(n, k) has the k=0 column (-1)^n*A000217(n+1) = (-1)^n*binomial(n+2, 2), then repeated and down-shifted.
The unsigned triangle, i.e., Tup(n, k) := (-1)^(n-k)*T(n-1,k-1) = binomial(n-k+2, 2) with n >= 1, k = 1..n, gives the number of triangles of length k (in some units), for k = 1..n, in the matchstick arrangement (or tower of cards, with n cards as basis) with an enclosing triangle of length n, but only triangles with orientation (up) like the enclosing triangle are counted. The total number of matchsticks (cards) is 3*A000217(n). (See the comment by Andrew Howroyd in A085691). Recurrence: Tup(n, k) = 0 for n < k, Tup(1, 1) = 1, and Tup(n, k) = Tup(n-1, k) + n - k + 1, for n >= 2, k = 1..n. Row sums give A000292(n). (End)

			The triangle T(n, k) begins:
n\k  0   1   2   3   4   5   6  7  8  9 ...
-------------------------------------------
0:   1
1  :-3   1
2:   6  -3   1
3: -10   6  -3   1
4:  15 -10   6  -3   1
5; -21  15 -10   6  -3   1
6:  28 -21  15 -10   6  -3   1
7: -36  28 -21  15 -10   6  -3  1
8:  45 -36  28 -21  15 -10   6 -3  1
9: -55  45 -36  28 -21  15 -10  6 -3  1
... reformattet by - _Wolfdieter Lang_, Apr 05 2020

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000217, A000292, A083392, A085691, A122431, A127893.

/* As triangle */ [[(-1)^(n-k)*Binomial(n-k+2, 2): k in [1..n]]: n in [1..10]]; // G. C. Greubel, Oct 29 2017

Table[(-1)^(n - k)*Binomial[n - k + 2, 2], {n, 0, 49}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 29 2017 *)

for(n=0,10, for(k=0,n, print1((-1)^(n-k)*binomial(n-k+2,2), ", "))) \\ G. C. Greubel, Oct 29 2017

Number triangle T(n, k) = [k<=n]*(-1)^(n-k)*binomial(n-k+2, 2).

Recurrence: T(n, k) = - T(n-1, k) + (-1)^(n-k)*(n-k+1), for n >= 0, and k = 0..n. - Wolfdieter Lang, Apr 06 2020

A127895 Riordan array (1/(1+x)^3, x/(1+x)^3).

Original entry on oeis.org

1, -3, 1, 6, -6, 1, -10, 21, -9, 1, 15, -56, 45, -12, 1, -21, 126, -165, 78, -15, 1, 28, -252, 495, -364, 120, -18, 1, -36, 462, -1287, 1365, -680, 171, -21, 1, 45, -792, 3003, -4368, 3060, -1140, 231, -24, 1, -55, 1287, -6435, 12376, -11628, 5985, -1771, 300, -27, 1

Offset: 0

Paul Barry, Feb 04 2007

The matrix inverse of the convolution triangle of A001764 (number of ternary trees). - Peter Luschny, Oct 09 2022

			Triangle begins
    1;
   -3,     1;
    6,    -6,     1;
  -10,    21,    -9,      1;
   15,   -56,    45,    -12,      1;
  -21,   126,  -165,     78,    -15,      1;
   28,  -252,   495,   -364,    120,    -18,     1;
  -36,   462, -1287,   1365,   -680,    171,   -21,     1;
   45,  -792,  3003,  -4368,   3060,  -1140,   231,   -24,   1;
  -55,  1287, -6435,  12376, -11628,   5985, -1771,   300, -27,   1;
   66, -2002, 12870, -31824,  38760, -26334, 10626, -2600, 378, -30, 1;

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

Inverse is A127898.
Alternating sign version of A127893.
Cf. A001764, A095263, A127896.

[(-1)^(n-k)*Binomial(n+2*k+2, n-k): k in [0..n], n in [0..10]]; // G. C. Greubel, Apr 29 2018

# Uses function InvPMatrix from A357585. Adds column 1, 0, 0, ... to the left.
InvPMatrix(10, n -> binomial(3*n, n)/(2*n+1)); # Peter Luschny, Oct 09 2022

Table[(-1)^(n-k)*Binomial[n+2*k+2, n-k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 29 2018 *)

for(n=0, 10, for(k=0,n, print1((-1)^(n-k)*binomial(n+2*k+2, n-k), ", "))) \\ G. C. Greubel, Apr 29 2018

flatten([[(-1)^(n-k)*binomial(n+2*k+2, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 16 2021

T(n, k) = (-1)^(n-k)*binomial(n +2*k +2, n-k).
Sum_{k=0..n} T(n, k) = A127896(n) (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^n*A095263(n) (diagonal sums).

Terms a(50) onward added by G. C. Greubel, Apr 29 2018

A127894 Inverse of Riordan array (1/(1-x)^3, x/(1-x)^3).

Original entry on oeis.org

1, -3, 1, 12, -6, 1, -55, 33, -9, 1, 273, -182, 63, -12, 1, -1428, 1020, -408, 102, -15, 1, 7752, -5814, 2565, -760, 150, -18, 1, -43263, 33649, -15939, 5313, -1265, 207, -21, 1, 246675, -197340, 98670, -35880, 9750, -1950, 273, -24, 1

Offset: 0

Paul Barry, Feb 04 2007

First column is (-1)^n*A001764(n+1). Row sums are (-1)^n*A006013(n). Inverse of A127893.

			Triangle begins
1,
-3, 1,
12, -6, 1,
-55, 33, -9, 1,
273, -182, 63, -12, 1,
-1428, 1020, -408, 102, -15, 1,
7752, -5814, 2565, -760, 150, -18, 1,
-43263, 33649, -15939, 5313, -1265, 207, -21, 1,
246675, -197340, 98670, -35880, 9750, -1950, 273, -24, 1,
-1430715, 1170585, -610740, 237510, -71253, 16443, -2842, 348, -27, 1,
8414640, -7012200, 3786588, -1553472, 503440, -129456, 26040, -3968, 432, -30, 1

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

Cf. A001764, A006013, A127893.

Table[If[k == 0, (-1)^(n-1)*Binomial[3*n, n-k]/(2*n+1), (-1)^(n-k-1)*((k + 1)/(n))*Binomial[3*n, n-k-1]], {n, 1, 100}, {k, 0, n-1}] // Flatten (* G. C. Greubel, Apr 29 2018 *)

for(n=1,10, for(k=0,n-1, print1(if(k==0, (-1)^(n-1)*binomial(3*n, n-k)/(2*n+1), (-1)^(n-k-1)*((k+1)/n)*binomial(3*n, n-k-1)), ", "))) \\ G. C. Greubel, Apr 29 2018

Terms a(39) onward added by G. C. Greubel, Apr 29 2018

A206294 Riordan array (1, x/(1-x)^3).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 6, 6, 1, 0, 10, 21, 9, 1, 0, 15, 56, 45, 12, 1, 0, 21, 126, 165, 78, 15, 1, 0, 28, 252, 495, 364, 120, 18, 1, 0, 36, 462, 1287, 1365, 680, 171, 21, 1, 0, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1

Offset: 0

Philippe Deléham, Feb 05 2012

The convolution triangle of the triangular numbers A000217. - Peter Luschny, Oct 07 2022

			Triangle begins:
1
0, 1
0, 3, 1
0, 6, 6, 1
0, 10, 21, 9, 1
0, 15, 56, 45, 12, 1
0, 21, 126, 165, 78, 15, 1
0, 28, 252, 495, 364, 120, 18, 1
0, 36, 462, 1287, 1365, 680, 171, 21, 1
0, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1
0, 55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1
0, 66, 2002, 12870, 31824, 38760, 26324, 10626, 2600, 378, 30, 1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. Columns: A000007, A000217 (triangular numbers), A000389, A000581, A001288, A010967..(+3)..A011000, A017714..(+3)..A017762.
Row sums are A052529.
Cf. A127893.

# Uses function PMatrix from A357368.
PMatrix(10, n -> n * (n + 1) / 2); # Peter Luschny, Oct 07 2022

Table[If[n == 0 && k == 0 , 1, Binomial[n - 1 + 2 k, n - k]], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 25 2017 *)

{T(n,k)=polcoeff(1/(1-x+x*O(x^(n-k)))^(3*k),n-k)}

{T(n,k)=polcoeff(polcoeff((1-x)^3/((1-x)^3-y*x +x*O(x^n)),n,x),k,y)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

Triangle T(n,k), read by rows, given by (0, 3, -1, 2/3, -1/6, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
T(n,0) = 0^n, T(n,k) = C(n-1+2k, n-k) for k > 0.
T(n,n) = 1, T(k+1,k) = 3*k = A008585(k), T(k+2,k) = A081266(k).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A052529(n), A052910(n) for x = 0, 1, 2 respectively.
G.f.: (1-x)^3/((1-x)^3-y*x).

cp's OEIS Frontend

A095263 a(n+3) = 3*a(n+2) - 2*a(n+1) + a(n).

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A052529 Expansion of (1-x)^3/(1 - 4*x + 3*x^2 - x^3).

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A122432 Riordan array (1/(1+x)^3,x).

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A127895 Riordan array (1/(1+x)^3, x/(1+x)^3).

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A127894 Inverse of Riordan array (1/(1-x)^3, x/(1-x)^3).

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A206294 Riordan array (1, x/(1-x)^3).

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A095263 a(n+3) = 3a(n+2) - 2a(n+1) + a(n).

A052529 Expansion of (1-x)^3/(1 - 4x + 3x^2 - x^3).