A081277 -id:A081277 - cp's OEIS frontend

A099193 a(n) = n(4n^6 + 70n^4 + 196n^2 + 45)/315.

0, 1, 14, 99, 476, 1765, 5418, 14407, 34232, 74313, 149830, 284075, 511380, 880685, 1459810, 2340495, 3644272, 5529233, 8197758, 11905267, 16970060, 23784309, 32826266, 44673751, 60018984, 79684825, 104642486, 136030779, 175176964, 223619261, 283131090, 355747103

Offset: 0

Jonathan Vos Post, Nov 16 2004

Kim asserts that every nonnegative integer can be represented by the sum of no more than 21 of these numbers.

Starting with 1 = binomial transform of [1, 13, 72, 220, 400, 432, 256, 0, 0, 0, ...], where (1, 13, 72, 220, 400, 432, 256) = row 7 of the Chebyshev triangle A081277. Also = row 7 of the array in A142978. - Gary W. Adamson, Jul 19 2008

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Index entries for linear recurrences with constant coefficients, signature (8, -28, 56, -70, 56, -28, 8, -1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099195 (m=8), A099196 (m=9), A099197 (m=10).
Cf. A000332.
Cf. A142978, A081277.

Table[SeriesCoefficient[x (1 + x)^6/(1 - x)^8, {x, 0, n}], {n, 0, 31}] (* Michael De Vlieger, Dec 14 2015 *)

concat(0, Vec(x*(1+x)^6/(1-x)^8 + O(x^40))) \\ Michel Marcus, Dec 14 2015

a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.
G.f.: x*(1+x)^6/(1-x)^8. - R. J. Mathar, Jul 18 2009
a(n) = 14*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

More terms from Michel Marcus, Dec 14 2015

A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.

Paul D. Hanna, Rows 0..45 of triangle, flattened.
Florian Luca, Attila Pethő, and László Szalay, Duplications in the k-generalized Fibonacci sequences, New York J. Math. (2021) Vol. 27, 1115-1133.

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Cf. A135387, A209149, A000012, A097805, A130595, A167374, A200139, A239473, A321620.

(* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
z = 8; a = 1; b = 1; c = 1; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
g[n_] := CoefficientList[w[n, -x], x];
TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
(* Clark Kimberling, Aug 04 2011 *)
T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)

{T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

/* Chebyshev Polynomials as Antidiagonals: */
{T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

# uses[riordan_square from A321620]
# Computes the unsigned triangle.
riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + ....  Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 19, 8, 1, 6, 25, 44, 33, 10, 1, 7, 36, 85, 96, 51, 12, 1, 8, 49, 146, 225, 180, 73, 14, 1, 9, 64, 231, 456, 501, 304, 99, 16, 1, 10, 81, 344, 833, 1182, 985, 476, 129, 18, 1, 11, 100, 489, 1408, 2471, 2668, 1765, 704, 163, 20, 1, 12

Offset: 0

Gary W. Adamson, Mar 19 2005

The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - Gary W. Adamson, Jul 17 2008
This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - Clark Kimberling, Mar 19 2012
This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - Wolfdieter Lang, May 13 2025

			The Riordan triangle T begins:
  n\k  0   1   2    3    4    5    6   7   8  9 10 ...
  ----------------------------------------------------
  0:   1
  1:   2   1
  2:   3   4   1
  3:   4   9   6    1
  4:   5  16  19    8    1
  5:   6  25  44   33   10    1
  6:   7  36  85   96   51   12    1
  7:   8  49 146  225  180   73   14   1
  8:   9  64 231  456  501  304   99  16   1
  9:  10  81 344  833 1182  985  476 129  18  1
  10: 11 100 489 1408 2471 2668 1765 704 163 20  1
  ... reformatted and extended by _Wolfdieter Lang_, May 13 2025
From _Wolfdieter Lang_, May 13 2025: (Start)
Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)

Reinhard Zumkeller, First 100 rows of triangle, flattened

Diagonal sums are A008937(n+1).
Cf. A048739 (row sums), A008288, A005900 (column 3), A014820 (column 4)
Cf. A081277, A142978 by antidiagonals, A119328, A110271 (matrix inverse).
Cf. A001100, A002464, A010673, A040000, A112478, A133872, A383857.

a104698 n k = a104698_tabl !! (n-1) !! (k-1)
a104698_row n = a104698_tabl !! (n-1)
a104698_tabl = [1] : [2,1] : f [1] [2,1] where
   f us vs = ws : f vs ws where
     ws = zipWith (+) ([0] ++ us ++ [0]) $
          zipWith (+) ([1] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Jul 17 2015

A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
seq(seq(A104698(n, k), k=0..n), n=0..15); # R. J. Mathar, Sep 04 2011
T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # Peter Luschny, May 13 2025

u[1, ] = 1; v[1, ] = 1;
u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* Jean-François Alcover, Mar 10 2019, after Clark Kimberling *)

T(n,k)=sum(j=0,n-k,binomial(k,j)*binomial(n-j+1,k+1)) \\ Charles R Greathouse IV, Jan 16 2012

The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
From Paul Barry, Jul 18 2005: (Start)
Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - R. J. Mathar, Sep 04 2011
T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - Reinhard Zumkeller, Jul 17 2015
From Wolfdieter Lang, May 13 2025: (Start)
The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k){n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)

A167591 A triangle related to the a(n) formulas of the rows of the ED4 array A167584.

Original entry on oeis.org

1, 4, -2, 12, -8, 9, 32, -16, 120, -60, 80, 0, 952, -768, 525, 192, 160, 5664, -5008, 12396, -5670, 448, 896, 27888, -20672, 162740, -133128, 72765, 1024, 3584, 120064, -46720, 1537216, -1562464, 2557296, -1081080, 2304, 12288, 467712, 76800

Offset: 1

Johannes W. Meijer, Nov 10 2009

The a(n) formulas given below correspond to the first ten rows of the ED4 array A167584.

The recurrence relations of the a(n) formulas for the left hand triangle columns, see the cross-references below, lead to the sequences A013609, A003148, A081277 and A079628.

			Row 1: a(n) = 1.
Row 2: a(n) = 4*n - 2.
Row 3: a(n) = 12*n^2 - 8*n + 9.
Row 4: a(n) = 32*n^3 - 16*n^2 + 120*n - 60.
Row 5: a(n) = 80*n^4 + 0*n^3 + 952*n^2 - 768*n + 525.
Row 6: a(n) = 192*n^5 + 160*n^4 + 5664*n^3 - 5008*n^2 + 12396*n - 5670.
Row 7: a(n) = 448*n^6 + 896*n^5 + 27888*n^4 - 20672*n^3 + 162740*n^2 - 133128*n + 72765.
Row 8: a(n) = 1024*n^7 + 3584*n^6 + 120064*n^5 - 46720*n^4 + 1537216*n^3 - 1562464*n^2 + 2557296*n - 1081080.
Row 9: a(n) = 2304*n^8 + 12288*n^7 + 467712*n^6 + 76800*n^5 + 11589216*n^4 - 12058368*n^3 + 47963568*n^2 - 38278080*n + 18243225.
Row 10: a(n) = 5120*n^9 + 38400*n^8 + 1686528*n^7 + 1540608*n^6 + 73898880*n^5 - 66179520*n^4 + 631348672*n^3 - 669559008*n^2 + 869709780*n - 344594250.

A167584 is the ED4 array.
A000012, A016825, A167585, A167586 and A167587 equal the first five rows of the ED4 array.
A001787, A167592, A167593, A168307 and A168308 equal the first five left hand triangle columns.
A001193 equals the first right hand triangle column.
A024199 equals the row sums.
Cf. A013609, A003148, A079628 and A081277.

Comment and formulas added by Johannes W. Meijer, Nov 23 2009

A167580 A triangle related to the a(n) formulas of the rows of the ED3 array A167572.

Original entry on oeis.org

1, 6, -1, 20, 0, 3, 56, 28, 98, -15, 144, 192, 1080, -48, 105, 352, 880, 7568, 2024, 6534, -945, 832, 3328, 40976, 31616, 132444, -8112, 10395, 1920, 11200, 187488, 274480, 1593960, 286900, 972162, -135135, 4352, 34816, 761600, 1784320, 13962848

Offset: 1

Johannes W. Meijer, Nov 10 2009

The a(n) formulas given below correspond to the first ten rows of the ED3 array A167572.

The recurrence relations of the a(n) formulas for the left hand triangle columns, see the cross-references below, lead to the sequences A013609, A003148, A081277 and A079628.

			Row 1: a(n) = 1.
Row 2: a(n) = 6*n - 1.
Row 3: a(n) = 20*n^2 + 0*n + 3.
Row 4: a(n) = 56*n^3 + 28*n^2 + 98*n - 15.
Row 5: a(n) = 144*n^4 + 192*n^3 + 1080*n^2 - 48*n + 105.
Row 6: a(n) = 352*n^5 + 880*n^4 + 7568*n^3 + 2024*n^2 + 6534*n - 945.
Row 7: a(n) = 832*n^6 + 3328*n^5 + 40976*n^4 + 31616*n^3 + 132444*n^2 - 8112*n + 10395.
Row 8: a(n) = 1920*n^7 + 11200*n^6 + 187488*n^5 + 274480*n^4 + 1593960*n^3 + 286900*n^2 + 972162*n - 135135.
Row 9: a(n) = 4352*n^8 + 34816*n^7 + 761600*n^6 + 1784320*n^5 + 13962848*n^4 + 7874944*n^3 + 29641200*n^2 - 2080800*n + 2027025.
Row 10: a(n) = 9728*n^9 + 102144*n^8 + 2830848*n^7 + 9645312*n^6 + 98382912*n^5 + 106720416*n^4 + 522283552*n^3 + 69265488*n^2 + 255468870*n - 34459425.

A167572 is the ED3 array.
A000012, A016969, A167573, A167574 and A167575 equal the first five rows of the ED3 array.
A014480, A167581, A167582, A168305 and A168306 equal the first five left hand triangle columns.
A001147 equals the first right hand triangle column.
A167576 equals the row sums.
Cf. A013609, A003148, A079628 and A081277.

Comment and links added by Johannes W. Meijer, Nov 23 2009

A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 16, 48, 56, 32, 9, 1, 32, 112, 160, 120, 50, 11, 1, 64, 256, 432, 400, 220, 72, 13, 1, 128, 576, 1120, 1232, 840, 364, 98, 15, 1, 256, 1280, 2816, 3584, 2912, 1568, 560, 128, 17, 1, 512, 2816, 6912, 9984, 9408, 6048, 2688, 816, 162, 19, 1

Offset: 0

Philippe Deléham, Nov 13 2011

Riordan array ((1-x)/(1-2x),x/(1-2x)).
Product A097805*A007318 as infinite lower triangular arrays.
Product A193723*A130595 as infinite lower triangular arrays.
T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - Geoffrey Critzer, Nov 18 2012
Apparently, rows of this array are unsigned diagonals of A028297. - Tom Copeland, Oct 11 2014
Unsigned A118800, so my conjecture above is true. - Tom Copeland, Nov 14 2016

			Triangle begins:
   1
   1,   1
   2,   3,   1
   4,   8,   5,   1
   8,  20,  18,   7,   1
  16,  48,  56,  32,   9,   1
  32, 112, 160, 120,  50,  11,   1

Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Cf. A007318, A028297, A059260, A097805, A118801, A130595.

nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n

T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.

G.f.: (1-x)/(1-(2+y)*x).

T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.

T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - Tom Copeland, Nov 15 2016

T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - Tom Copeland, Nov 17 2016

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 24 2018

A079628 Array of coefficients of P(n,x) = det (M(n,x)) where M(n,x) is the n X n matrix m(i,j)=x if i>j m(i,j)=1-x if i<=j.

Original entry on oeis.org

1, 1, -1, 1, -3, 2, 1, -5, 8, -4, 1, -7, 18, -20, 8, 1, -9, 32, -56, 48, -16, 1, -11, 50, -120, 160, -112, 32, 1, -13, 72, -220, 400, -432, 256, -64, 1, -15, 98, -364, 840, -1232, 1120, -576, 128, 1, -17, 128, -560, 1568, -2912, 3584, -2816, 1280, -256, 1, -19, 162, -816, 2688, -6048, 9408, -9984, 6912, -2816, 512, 1

Offset: 0

Benoit Cloitre, Jan 30 2003

Formatted as a triangular array, this is [1, 0, 0, 0, 0, 0, 0, ...] DELTA [ -1, -1, 0, 0, 0, 0, 0, 0, ...] (see construction in A084938). - Philippe Deléham, Aug 09 2005

			det(M(4,x))=1-7x+18x^2-20x^3+8x^4.
1;
1,-1;
1,-3,2;
1,-5,8,-4;
1,-7,18,-20,8;
1,-9,32,-56,48,-16;
1,-11,50,-120,160,-112,32;
1,-13,72,-220,400,-432,256,-64;
1,-15,98,-364,840,-1232,1120,-576,128;
1,-17,128,-560,1568,-2912,3584,-2816,1280,-256,

Cf. A081277.

Cf. A167580 and A167591. - Johannes W. Meijer, Nov 23 2009

A079628 := proc(n,k) local x; expand((-1)^n* (x-1)*(2*x-1)^(n-1)) ;coeftayl(%,x=0,k) ; end proc: # R. J. Mathar, Nov 04 2011

P(n, x)= (-1)^n*(x-1)*(2*x-1)^(n-1).

G.f.: (1+x*y)/(1+2*x*y-x). - R. J. Mathar, Aug 11 2015

Sign added to formula. - R. J. Mathar, Nov 04 2011

A081281 Square array of binomial transforms of Chebyshev polynomial coefficients.

Original entry on oeis.org

1, 1, 2, 1, 4, 5, 1, 6, 15, 14, 1, 8, 29, 54, 41, 1, 10, 47, 126, 189, 122, 1, 12, 69, 238, 513, 648, 365, 1, 14, 95, 398, 1101, 1998, 2187, 1094, 1, 16, 125, 614, 2057, 4788, 7533, 7290, 3281, 1, 18, 159, 894, 3501, 9858, 19899, 27702, 24057, 9842, 1, 20, 197, 1246

Offset: 0

Paul Barry, Mar 16 2003

Square array of binomial transforms of rows of A081277. Rows include A007051, A006234, A081278, A081279, A081280.

			Read by antidiagonals, the array appears as {1}, {1,2},{1,4,5}, {1,6,15,14}, {1,8,29,54,41},...

Rows have g.f. (1-2x)(1-x)^(n-1)/(1-3x)^(n+1).

A136523 Triangle T(n,k) = A053120(n,k) + A053120(n-1,k), read by rows.

Original entry on oeis.org

1, 1, 1, -1, 1, 2, -1, -3, 2, 4, 1, -3, -8, 4, 8, 1, 5, -8, -20, 8, 16, -1, 5, 18, -20, -48, 16, 32, -1, -7, 18, 56, -48, -112, 32, 64, 1, -7, -32, 56, 160, -112, -256, 64, 128, 1, 9, -32, -120, 160, 432, -256, -576, 128, 256, -1, 9, 50, -120, -400, 432, 1120, -576, -1280, 256, 512

Offset: 0

Roger L. Bagula, Mar 23 2008

			Triangle begins as:
   1;
   1,  1;
  -1,  1,   2;
  -1, -3,   2,    4;
   1, -3,  -8,    4,    8;
   1,  5,  -8,  -20,    8,   16;
  -1,  5,  18,  -20,  -48,   16,   32;
  -1, -7,  18,   56,  -48, -112,   32,   64;
   1, -7, -32,   56,  160, -112, -256,   64,   128;
   1,  9, -32, -120,  160,  432, -256, -576,   128, 256;
  -1,  9,  50, -120, -400,  432, 1120, -576, -1280, 256, 512;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000007, A001792, A001793, A001794, A000330, A008794, A011782.
Cf. A025192, A040000 (row sums), A053120, A057077, A081277, A109613.
Cf. A124182.

function A053120(n,k)
  if ((n+k) mod 2) eq 1 then return 0;
  elif n eq 0 then return 1;
  else return (-1)^Floor((n-k)/2)*(n/(n+k))*Binomial(Floor((n+k)/2), k)*2^k;
  end if;
end function;
A136523:= func< n,k | A053120(n,k) + A053120(n-1,k) >;
[A136523(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 26 2023

A053120[n_, k_]:= Coefficient[ChebyshevT[n,x], x, k];
T[n_, k_]:= T[n, k]= A053120[n,k] + A053120[n-1,k];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten

def A053120(n,k):
    if (n+k)%2==1: return 0
    elif n==0: return 1
    else: return floor((-1)^((n-k)//2)*(n/(n+k))*binomial((n+k)//2, k)*2^k)
def A136523(n,k): return A053120(n,k) + A053120(n-1,k)
flatten([[A136523(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 26 2023

T(n, k) = A053120(n,k) + A053120(n-1,k).
Sum_{k=0..n} T(n, k) = A040000(n).
From G. C. Greubel, Jul 26 2023: (Start)
T(n, 0) = A057077(n).
T(n, 1) = (-1)^floor((n-1)/2) * A109613(n-1).
T(n, 2) = (-1)^floor((n-2)/2) * A008794(n-1).
T(n, 3) = (-1)^floor((n+1)/2) * A000330(n-1).
T(n, n) = A011782(n).
T(n, n-1) = A011782(n-1).
T(n, n-2) = -A001792(n-2).
T(n, n-4) = A001793(n-3).
T(n, n-6) = -A001794(n-6).
Sum_{k=0..n} (-1)^k*T(n,k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A000007(n) + [n=1].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (-1)^floor(n/2)*A025192(floor(n/2)). (End)

Edited by G. C. Greubel, Jul 26 2023

Previous Showing 11-19 of 19 results.

cp's OEIS Frontend

A099193 a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.

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A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

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A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

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A167591 A triangle related to the a(n) formulas of the rows of the ED4 array A167584.

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A167580 A triangle related to the a(n) formulas of the rows of the ED3 array A167572.

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A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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A079628 Array of coefficients of P(n,x) = det (M(n,x)) where M(n,x) is the n X n matrix m(i,j)=x if i>j m(i,j)=1-x if i<=j.

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A081281 Square array of binomial transforms of Chebyshev polynomial coefficients.

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A099193 a(n) = n(4n^6 + 70n^4 + 196n^2 + 45)/315.

A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.