A054143 -id:A054143 - cp's OEIS frontend

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset: 0

Wouter Meeussen, Oct 01 2002

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
  A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
  B) = [St2]*[dP]*[St2]^(-1)
  C) = [St1]^(-1)*[dP]*[St1],
  where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Tom Copeland, Appell polynomials, cumulants, noncrossing partitions, Dyck lattice paths, and inversion, 2014.
Tom Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015.
Sela Fried, The expected degree of noninvertibility of compositions of functions and a related combinatorial identity, arXiv:2202.13061 [math.CO], 2022.
J. R. Griggs, The Cycling of Partitions and Compositions under Repeated Shifts, Advances in Applied Mathematics, Volume 21, Issue 2, August 1998, Pages 205-227.
P. Olver, The canonical contact form p. 7.
D. P. Walsh, A short note on increasing acyclic functions

Cf. A007318, A181971, A228196, A228576.
Row sums are A000225, diagonal sums are A052952.
The number of acyclic functions is A058127.
Cf. A008292, A090582, A019538, A049019, A133314, A135278, A133437, A111785, A126216, A134685, A133932, A248727, A033282, A134264.
Cf. A000027, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582.
Cf. A325191.

Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018

a074909 n k = a074909_tabl !! n !! k
a074909_row n = a074909_tabl !! n
a074909_tabl = iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
-- Reinhard Zumkeller, Feb 25 2012

/* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018

A074909 := proc(n,k)
    if k > n or k < 0 then
        0;
    else
        binomial(n+1,k) ;
    end if;
end proc: # Zerinvary Lajos, Nov 09 2006

Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]

print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013

from math import comb, isqrt
def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003

Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 2014

A193842 Triangular array: the fission of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence ((x+2)^n: n >= 0). (Fission is defined at Comments.)

Original entry on oeis.org

1, 1, 4, 1, 7, 13, 1, 10, 34, 40, 1, 13, 64, 142, 121, 1, 16, 103, 334, 547, 364, 1, 19, 151, 643, 1549, 2005, 1093, 1, 22, 208, 1096, 3478, 6652, 7108, 3280, 1, 25, 274, 1720, 6766, 17086, 27064, 24604, 9841, 1, 28, 349, 2542, 11926, 37384, 78322, 105796

Offset: 0

Clark Kimberling, Aug 07 2011

Suppose that p = p(n)*x^n + p(n-1)*x^(n-1) + ... + p(1)*x + p(0) is a polynomial and that Q is a sequence of polynomials:
...
q(k,x) = t(k,0)*x^k + t(k,1)*x^(k-1) + ... + t(k,k-1)*x + t(k,k),
...
for k = 0, 1, 2, ...  The Q-downstep of p is the polynomial given by
...
D(p) = p(n)*q(n-1,x) + p(n-1)*q(n-2,x) + ... + p(1)*q(0,x). (Note that p(0) does not appear. "Q-downstep" as just defined differs slightly from "Q-downstep" as defined for a different purpose at A193649.)
...
Now suppose that P = (p(n,x): n >= 0) and Q = (q(n,x): n >= 0) are sequences of polynomials, where n indicates degree.  The fission of P by Q, denoted by P^^Q, is introduced here as the sequence W = (w(n,x): n >= 0) of polynomials defined by w(0,x) = 1 and w(n,x) = D(p(n+1,x)).
...
Strictly speaking, ^^ is an operation on sequences of polynomials.  However, if P and Q are regarded as numerical triangles (of coefficients of polynomials), then ^^ can be regarded as an operation on numerical triangles.  In this case, row n of P^^Q, for n > 0, is given by the matrix product P(n+1)*QQ(n), where P(n+1) =(p(n+1,n+1), p(n+1,n), ..., p(n+1,2), p(n+1,1)) and QQ(n) is the (n+1)-by-(n+1) matrix given by
...
q(n,0) .. q(n,1)............. q(n,n-1) .... q(n,n)
0 ....... q(n-1,0)........... q(n-1,n-2)... q(n-1,n-1)
0 ....... 0.................. q(n-2,n-3) .. q(n-2,n-2)
...
0 ....... 0.................. q(1,0) ...... q(1,1)
0 ....... 0 ................. 0 ........... q(0,0).
Here, the polynomial q(k,x) is taken to be
q(k,0)*x^k + q(k,1)x^(k-1) + ... + q(k,k)*x + q(k,k);
i.e., "q" is used instead of "t".
...
Example:  Let p(n,x) = (x+1)^n and q(n,x) = (x+2)^n.  Then
...
w(0,x) = 1 by the definition of W,
w(1,x) = D(p(2,x)) = 1*(x+2) + 2*1 = x + 4,
w(2,x) = D(p(3,x)) = 1*(x^2+4*x+4) + 3*(x+2) + 3*1 = x^2 + 7*x + 13,
w(3,x) = D(p(4,x)) = 1*(x^3+6*x^2+12*x+8) + 4*(x^2+4x+4) + 6*(x+2) + 4*1 = x^3 + 10*x^2 + 34*x + 40.
...
From these first 4 polynomials in the sequence P^^Q, we can write the first 4 rows of P^^Q when P, Q, and P^^Q are regarded as triangles:
1
1...4
1...7....13
1...10...34...40
...
In the following examples, r(P^^Q) is the mirror of P^^Q, obtained by reversing the rows of P^^Q.  Let u denote the polynomial x^n + x^(n-1) + ... + x + 1.
...
..P........Q...........P^^Q........r(P^^Q)
(x+1)^n....(x+2)^n.....A193842.....A193843
(x+1)^n....(x+1)^n.....A193844.....A193845
(x+2)^n....(x+1)^n.....A193846.....A193847
(2x+1)^n...(x+1)^n.....A193856.....A193857
(x+1)^n....(2x+1)^n....A193858.....A193859
(x+1)^n.......u........A054143.....A104709
..u........(x+1)^n.....A074909.....A074909
..u...........u........A002260.....A004736
(x+2)^n.......u........A193850.....A193851
..u.........(x+2)^n....A193844.....A193845
(2x+1)^n......u........A193860.....A193861
..u.........(2x+1)^n...A115068.....A193862
...
Regarding A193842,
col 1 ...... A000012
col 2 ...... A016777
col 3 ...... A081271
w(n,n) ..... A003462
w(n,n-1) ... A014915

			First six rows, for 0 <= k <= n and 0 <= n <= 5:
  1
  1...4
  1...7....13
  1...10...34....40
  1...13...64....142...121
  1...16...103...334...547...364

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Digital Library of Mathematical Functions, Hypergeometric function, analytic properties.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A193722 (fusion of P by Q), A193649 (Q-residue), A193843 (mirror of A193842).

[ (&+[3^(k-j)*Binomial(n-j,k-j): j in [0..k]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

fission := proc(p, q, n) local d, k;
p(n+1,0)*q(n,x)+add(coeff(p(n+1,x),x^k)*q(n-k,x), k=1..n);
seq(coeff(%,x,n-k), k=0..n) end:
A193842_row := n -> fission((n,x) -> (x+1)^n, (n,x) -> (x+2)^n, n);
for n from 0 to 5 do A193842_row(n) od; # Peter Luschny, Jul 23 2014
# Alternatively:
p := (n,x) -> add(x^k*(1+3*x)^(n-k),k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od; # Peter Luschny, Jun 18 2017

(* First program *)
z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 2)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193842 *)
TableForm[Table[h[n], {n, 0, z}]]  (* A193843 *)
Flatten[Table[h[n], {n, -1, z}]]
(* Second program *)
Table[SeriesCoefficient[((x+3)^(n+1) -1)/(x+2), {x,0,n-k}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

T(n,k) = sum(j=0,k, 3^(k-j)*binomial(n-j,k-j)); \\ G. C. Greubel, Feb 18 2020

from mpmath import mp, hyp2f1
mp.dps = 100; mp.pretty = True
def T(n,k):
    return 3^k*binomial(n,k)*hyp2f1(1,-k,-n,1/3)-0^(n-k)//2
for n in range(7):
    print([int(T(n,k)) for k in (0..n)]) # Peter Luschny, Jul 23 2014

# Second program using the 'fission' operation.
def fission(p, q, n):
    F = p(n+1,0)*q(n,x)+add(expand(p(n+1,x)).coefficient(x,k)*q(n-k,x) for k in (1..n))
    return [expand(F).coefficient(x,n-k) for k in (0..n)]
A193842_row = lambda k: fission(lambda n,x: (x+1)^n, lambda n,x: (x+2)^n, k)
for n in range(7): A193842_row(n) # Peter Luschny, Jul 23 2014

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = Sum_{i = 0..k} 3^(k-i)*binomial(n-i,k-i).
O.g.f.: 1/((1 - x*t)*(1 - (1 + 3*x)*t)) = 1 + (1 + 4*x)*t + (1 + 7*x + 13*x^2)*t^2 + ....
The n-th row polynomial is R(n,x) = (1/(2*x + 1))*((3*x + 1)^(n+1) - x^(n+1)). (End)
T(n,k) = T(n-1,k) + 4*T(n-1,k-1) - T(n-2,k-1) - 3*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 4, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 17 2014
T(n,k) = 3^k * C(n,k) * hyp2F1(1, -k, -n, 1/3) with or without the additional term -0^(n-k)/2 depending on the exact definition of the hypergeometric function used. Compare formulas 15.2.5 and 15.2.6 in the DLMF reference. - Peter Luschny, Jul 23 2014

Name and Comments edited by Petros Hadjicostas, Jun 05 2020

A005672 a(n) = Fibonacci(n+1) - 2^floor(n/2).

Original entry on oeis.org

0, 0, 0, 1, 1, 4, 5, 13, 18, 39, 57, 112, 169, 313, 482, 859, 1341, 2328, 3669, 6253, 9922, 16687, 26609, 44320, 70929, 117297, 188226, 309619, 497845, 815656, 1313501, 2145541

Offset: 0

N. J. A. Sloane

R. K. Guy, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. K. Guy, Letter to N. J. A. Sloane, 1987
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992

Gives diagonal sums of triangle A054143.

A005672:=z**3/(z**2+z-1)/(-1+2*z**2); # conjectured by Simon Plouffe in his 1992 dissertation
with (combinat):a[ -1]:=0:a[1]:=0:a[2]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-2] od: seq(a[n-1], n=0..31); # Zerinvary Lajos, Mar 17 2008

a(n) = Fibonacci(n-1) + 2*a(n-2), a(-1)=0, a(1)=0, a(2)=1. - Zerinvary Lajos, Mar 17 2008

A204201 Triangle based on (0,1/3,1) averaging array.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 15, 22, 1, 7, 21, 37, 46, 1, 8, 28, 58, 83, 94, 1, 9, 36, 86, 141, 177, 190, 1, 10, 45, 122, 227, 318, 367, 382, 1, 11, 55, 167, 349, 545, 685, 749, 766, 1, 12, 66, 222, 516, 894, 1230, 1434, 1515, 1534, 1, 13, 78, 288, 738, 1410

Offset: 1

Clark Kimberling, Jan 12 2012

For a1, let
t(n,1)=[a+t(n-1,1)]/2,
t(n,n)=[b+t(n-1,n-1)]/2,
t(n,k)=[t(n-1,k-1)+t(n-1,k)]/2 for 2<=k<=n-1.
We call (t(n,k)) the (a,r,b) averaging array.  If a and b
are integers and r is a rational number, then multiplying
row n of (t(n,k)) by the LCM of its denominators yields a
triangle of integers; A204201 arises in this manner from
(a,r,b)=(0,1/3,1).
...
Guide to related arrays:
(a,r,b).........triangle
(0,1/2,1).......A054143
(0,1/3,1).......A204201
(0,2/3,1).......A204202
(0,1/4,1).......A204203
(0,3/4,1).......A204204
(0,1/5,1).......A204205
(1,3/2,2).......A204206
(1,2,3).........A204207

			The (0,1/3,1) averaging array has these first four rows:
1/3
1/6....2/3
1/12...5/12...5/6
1/24...1/4....5/8...11/12.
Multiplying those rows by 3,6,12,24, respectively:
1
1...4
1...5...10
1...6...15...22
The first nine rows:
1
1...4
1...5...10
1...6...15...22
1...7...21...37...46
1...8...28...58...83...94
1...9...36...86...141..177..190
1...10..45...122..227..318..367..382
1...11..55...167..349..545..685..749..766

Cf. A204202.

a = 0; r = 1/3; b = 1;
t[1, 1] = r;
t[n_, 1] := (a + t[n - 1, 1])/2;
t[n_, n_] := (b + t[n - 1, n - 1])/2;
t[n_, k_] := (t[n - 1, k - 1] + t[n - 1, k])/2;
u[n_] := Table[t[n, k], {k, 1, n}]
Table[u[n], {n, 1, 5}]   (* averaging array *)
u = Table[(1/2) (1/r) 2^n*u[n], {n, 1, 12}];
TableForm[u]             (* A204102 triangle *)
Flatten[u]               (* A204201 sequence *)

From Philippe Deléham, Dec 24 2013: (Start)
T(n,n) = A033484(n-1).
Sum{k=1..n} T(n,k) = A053220(n).
T(n,k) = T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-2*T(n-2,k-2), T(1,1)=1, T(2,1)=1, T(2,2)=4, T(n,k)=0 if k<1 or if k>n. (End)

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 15, 11, 5, 1, 31, 26, 16, 6, 1, 63, 57, 42, 22, 7, 1, 127, 120, 99, 64, 29, 8, 1, 255, 247, 219, 163, 93, 37, 9, 1, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1, 2047, 2036, 1981, 1816, 1486, 1024, 562, 232, 67

Offset: 0

Gary W. Adamson, Mar 19 2005

This array (A104709) is the mirror of the fission, A054143, of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence (q(n,x): n >= 0) given by q(n,x) = x^n + x^(n-1) + ... + x + 1.  See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
The elements of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A110813(n,k) assuming the same offset in both triangles. - R. J. Mathar, Mar 15 2013
From Paul Curtz, Jun 12 2019: (Start)
Numerators of the triangle [Curtz, page 15, triangle (E)]:
     1/2;
     3/4,  1/4;
     7/8,  4/8,    1/8;
   15/16, 11/16,  5/16,  1/16;
   31/32, 26/31, 16/32,  6/32, 1/32;
   63/64, 57/64, 42/64, 22/64, 7/64, 1/64;
   ...
Denominators - Numerators: Triangle A054143.
  1;
  1, 3;
  1, 4, 7;
  1, 5, 11, 15;
  ...
(E) is a transform which accelerates the convergence of series.
For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have
  1*(1/2) = 1/2,
  1*(3/4) - (1/2)*(1/4) = 5/8,
  1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,
  1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,
  ...
This is A068566/A068565. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   3,  1;
   7,  4,  1;
  15, 11,  5,  1;
  31, 26, 16,  6,  1;
  63, 57, 42, 22,  7,  1;
  ...

Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.

A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

z = 10;
p[n_, x_] := (x + 1)^n;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A054143 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
(* Clark Kimberling, Aug 07 2011 *)

Begin with A055248 as a triangle, delete leftmost column.
The Riordan array factors as (1/(1-2*x), x)*(1/(1-x), x/(1-x)) - the sequence array for 2^n times Pascal's triangle. - Paul Barry, Aug 05 2005
T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - Paul Barry, Jan 12 2006
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
Working with an offset of 0, we have exp(x) * (e.g.f. for row n) = (e.g.f. for diagonal n). For example, for n = 3 we have exp(x)*(15 + 11*x + 5*x^2/2! + x^3/3!) = 15 + 26*x + 42*x^2/2! + 64*x^3/3! + 93*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), x/(1 - x)). - Peter Bala, Dec 21 2014
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: A(x,y) = Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - 3*x - x*y + 2*x^2 + 2*x^2*y) = 1/((1 - 2*x)*(1 - x*(y+1))).
The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).
Let B(x,y) be the bivariate o.g.f. of triangular array A054143. Because A054143 is the mirror image of the current array, we have A(x,y) = B(x*y, 1/y) and B(x,y) = A(x*y, 1/y). This makes it easy to identify lower diagonals of the array.
For example, if we want to identify the second lower diagonal of the array (i.e., 7, 11, 16, 22, ...), we take the 2nd derivative of B(x,y) with respect to y, set y = 0, and divide by 2!. (Note that columns in A054143 start at k = 0.) We get the g.f. x^2*(7 - 10*x + 4*x^2)/(1 - x)^3.
It is then easy to derive that T(n,n-2) = A000124(n+1) = (n+1)*(n+2)/2 + 1 for n >= 2 (by ignoring the first three terms of A000124). Of course, in the current case, it is much easier to use the formula for T(n,k) to find T(n,n-2). (End)
T(n,0) = 2^(n+1) - 1 for n >= 0; T(n,k) = T(n-1,k) + T(n-1,k-1) for 1 <= k <= n. - Peter Bala, Jan 30 2023
T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - Bernard Schott, Feb 22 2023

Name edited and offset changed by Petros Hadjicostas, Jun 04 2020

A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 2, 3, 1, 0, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 0, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 0, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 0, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 0

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(0,2n)=T(1,n) by T(0,2n)=T(m,n) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058394, A058395, A057884, (and effectively A007318).

			Rows are (1,0,1,0,1,0,1,...), (1,1,1,1,1,1,...), (1,2,2,2,2,2,...), (1,3,4,4,4,...) etc.

Rows are A000035 (A000012 with zeros), A000012, A040000 etc. Columns are A000012, A001477, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863 etc. Diagonals include A000079, A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, etc. The triangles A008949, A054143 and A055248 also appear in the half of the array which is not powers of 2.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(1, 1)=1, T(0, 2n)=T(1, n) and T(0, 2n+1)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2).

A193820 Triangular array: the fusion of polynomial sequences P and Q given by p(n,x)=(x+1)^n and q(n,x)=x^n+x^(n-1)+...+x+1.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 7, 8, 8, 1, 5, 11, 15, 16, 16, 1, 6, 16, 26, 31, 32, 32, 1, 7, 22, 42, 57, 63, 64, 64, 1, 8, 29, 64, 99, 120, 127, 128, 128, 1, 9, 37, 93, 163, 219, 247, 255, 256, 256, 1, 10, 46, 130, 256, 382, 466, 502, 511, 512, 512, 1, 11, 56

Offset: 0

Clark Kimberling, Aug 06 2011

See A193722 for the definition of fusion of two sequences of polynomials or triangular arrays.

Variant of A054143 and A008949. - R. J. Mathar, Mar 03 2013

			First six rows:
  1
  1....1
  1....2....2
  1....3....4....4
  1....4....7....8....8
  1....5....11...15...16...16

Cf. A193722, A128175, A193823, A045623 (row sums), A009766.

Cf. A054143, A008949.

A193820 := (n,k) -> `if`(k=0 or n=0,1, A193820(n-1,k-1)+A193820(n-1,k));
seq(print(seq(A193820(n,k),k=0..n+1)),n=0..10); # Peter Luschny, Jan 22 2012

z = 10; a = 1; b = 1;
p[n_, x_] := (a*x + b)^n
q[0, x_] := 1
q[n_, x_] := x*q[n - 1, x] + 1; q[n_, 0] := q[n, x] /. x -> 0;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]]   (* A193820 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]  (* A128175 *)

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = sum {i = 0..k} binomial(n-1,k-i) for 0 <= k <= n.
O.g.f.: (1 - x*t)^2/( (1 - 2*x*t)*(1 - (1 + x)*t) ) = 1 + (1 + x)*t + (1 + 2*x + 2*x^2)*t^2 + ....
The n-th row polynomial R(n,x) for n >= 1 is given by R(n,x) = 1/(1 - x)*( (x + 1)^(n-1) - 2^(n-1)*x^(n+1) ). Cf. A193823. (End)

A117670 Triangle read by rows: partial sums of the Pascal triangle minus 1.

Original entry on oeis.org

1, 2, 3, 3, 6, 7, 4, 10, 14, 15, 5, 15, 25, 30, 31, 6, 21, 41, 56, 62, 63, 7, 28, 63, 98, 119, 126, 127, 8, 36, 92, 162, 218, 246, 254, 255, 9, 45, 129, 255, 381, 465, 501, 510, 511, 10, 55, 175, 385, 637, 847, 967, 1012, 1022, 1023

Offset: 1

Arie Bos, Jul 06 2008, Jul 08 2008

Imagine that you are in a building with floors starting at floor 1, the lowest floor and you have a large number of eggs. For each floor in the building, you want to know whether or not an egg dropped from that floor will break.
If an egg breaks when dropped from floor i, then all eggs are guaranteed to break when dropped from any floor j > i. Likewise, if an egg doesn't break when dropped from floor i, then all eggs are guaranteed to never break when dropped from any floor j <= i.
a(n,k) is the maximum number of floors where you can determine whether or not an egg will break when dropped from any floor, with the following restrictions: you may drop a maximum of n eggs (one at a time, from any floors of your choosing) and you may break a maximum of k eggs.
Each row of the triangle is the running sum of the corresponding row with the first 1 omitted of Pascal's triangle (A007318), see A008949, A054143, A193820.
The k-th entry in the n-th row is the number of possible combinations of on/off switches after k attempts to turn on a switch in a set of n distinguishable switches. An attempt to turn on the same switch twice does not result in a new combination. See example. - Sergei Viznyuk, Jun 24 2012
T(n,k) is the number of nonempty subsets of the n-set with at most k elements, see example. - Joerg Arndt, May 04 2014

			Triangle a(n,k) begins:
n\k  1   2    3    4    5    6    7     8     9    10 ...
1:   1
2:   2   3
3:   3   6    7
4:   4  10   14   15
5:   5  15   25   30   31
6:   6  21   41   56   62   63
7:   7  28   63   98  119  126  127
8:   8  36   92  162  218  246  254   255
9:   9  45  129  255  381  465  501   510   511
10: 10  55  175  385  637  847  967  1012  1022  1023
...  Reformatted and extended by _Wolfdieter Lang_, Feb 07 2013
From _Sergei Viznyuk_, Jun 24 2012: (Start)
For example, we have n=3 distinguishable switches A,B,C (third row above). We attempt k=2 times to turn on a switch at random. The possible resulting combinations are:
A=on, B=off, C=off (the same A switch was turned on 2 times)
A=off, B=on, C=off (the same B switch was turned on 2 times)
A=off, B=off, C=on (the same C switch was turned on 2 times)
A=on, B=on, C=off  (switches A and B were turned on)
A=on, B=off, C=on  (switches A and C were turned on)
A=off, B=on, C=on  (switches B and C were turned on)
Thus, we have 6 different combinations, which is the number 6 at row n=3 column k=2 in the sequence above.
(End)
From _Joerg Arndt_, May 04 2014: (Start)
There are T(4,2) = 10 subsets of {0, 1, 2, 3}:
01:    1...    { 0 }
02:    11..    { 0, 1 }
03:    111.    { 0, 1, 2 }
04:    11.1    { 0, 1, 3 }
05:    1.1.    { 0, 2 }
06:    1.11    { 0, 2, 3 }
07:    1..1    { 0, 3 }
08:    .1..    { 1 }
09:    .11.    { 1, 2 }
10:    .111    { 1, 2, 3 }
11:    .1.1    { 1, 3 }
12:    ..1.    { 2 }
13:    ..11    { 2, 3 }
14:    ...1    { 3 }
(End)

Susanne Wienand, Table of n, a(n) for n = 1..1830
Google code.jam, Problem C. Egg Drop
Sergei Viznyuk, C-Program
Sergei Viznyuk, Local copy of C-Program

Table[Sum[Binomial[n, m], {m, k}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 25 2015 *)

tabl(nrows) = {for (n=1, nrows, for (k=1, n, print1(sum(m=1,k,binomial(n,m)), ", ");); print(););} \\ Michel Marcus, May 21 2013

a(n,1) = n ; a(n,n) = 2^n-1; a(n+1,k+1) = 1 + a(n,k) + a(n,k-1), 0 < k < n.

a(n,k) = sum(binomial(n,m),m=1..k), 1 <= k <= n. (see the running sum comment above). - Wolfdieter Lang, Feb 07 2013

cp's OEIS Frontend

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Python

Formula

Extensions

A193842 Triangular array: the fission of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence ((x+2)^n: n >= 0). (Fission is defined at Comments.)

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Sage

Formula

Extensions

A005672 a(n) = Fibonacci(n+1) - 2^floor(n/2).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

Formula

A204201 Triangle based on (0,1/3,1) averaging array.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)*binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)*(1-2*x)), x/(1-x)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A193820 Triangular array: the fusion of polynomial sequences P and Q given by p(n,x)=(x+1)^n and q(n,x)=x^n+x^(n-1)+...+x+1.

Views

Author

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

A061929 Triangle with n >= k >= 0 where a(n,k) = 2^k3^(n-k)(C(n+1,0)+C(n+1,1)+...C(n+1,k)).