A001789 -id:A001789 - cp's OEIS frontend

A133156 Irregular triangle read by rows: coefficients of U(n,x), Chebyshev polynomials of the second kind with exponents in decreasing order.

1, 2, 4, -1, 8, -4, 16, -12, 1, 32, -32, 6, 64, -80, 24, -1, 128, -192, 80, -8, 256, -448, 240, -40, 1, 512, -1024, 672, -160, 10, 1024, -2304, 1792, -560, 60, -1, 2048, -5120, 4608, -1792, 280, -12, 4096, -11264, 11520, -5376, 1120, -84, 1

Offset: 0

Gary W. Adamson, Dec 16 2007

The Chebyshev polynomials of the second kind are defined by the recurrence relation: U(0,x) = 1; U(1,x) = 2x; U(n+1,x) = 2x*U(n,x) - U(n-1,x).
From Gary W. Adamson, Nov 28 2008: (Start)
Triangle read by rows, unsigned = A000012 * A028297.
Row sums of absolute values give the Pell series, A000129.
(End)
The row sums are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...}.
Triangle, with zeros omitted, given by (2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 27 2011
Coefficients in the expansion of sin((n+1)*x)/sin(x) in descending powers of cos(x). The length of the n-th row is A008619(n). - Jianing Song, Nov 02 2018

			The first few Chebyshev polynomials of the second kind are
    1;
    2x;
    4x^2 -    1;
    8x^3 -    4x;
   16x^4 -   12x^2 +   1;
   32x^5 -   32x^3 +   6x;
   64x^6 -   80x^4 +  24x^2 -   1;
  128x^7 -  192x^5 +  80x^3 -   8x;
  256x^8 -  448x^6 + 240x^4 -  40x^2 +  1;
  512x^9 - 1024x^7 + 672x^5 - 160x^3 + 10x;
  ...
From _Roger L. Bagula_ and _Gary W. Adamson_: (Start)
     1;
     2;
     4,    -1;
     8,    -4;
    16,   -12,    1;
    32,   -32,    6;
    64,   -80,   24,   -1;
   128,  -192,   80,   -8;
   256,  -448,  240,  -40,  1;
   512, -1024,  672, -160, 10;
  1024, -2304, 1792, -560, 60, -1; (End)
From  _Philippe Deléham_, Dec 27 2011: (Start)
Triangle (2, 0, 0, 0, 0, ...) DELTA (0, -1/2, 1/2, 0, 0, 0, 0, 0, ...) begins:
   1;
   2,   0;
   4,  -1,  0;
   8,  -4,  0,  0;
  16, -12,  1,  0,  0;
  32, -32,  6,  0,  0,  0;
  64, -80, 24, -1,  0,  0,  0; (End)

Tracale Austin, Hans Bantilan, Isao Jonas and Paul Kory, The Pfaffian Transformation, Journal of Integer Sequences, Vol. 12 (2009), page 25
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014. - From _Tom Copeland_, Oct 11 2014
Pantelis A. Damianou, A Beautiful Sine Formula, Amer. Math. Monthly 121 (2014), no. 2, 120-135. MR3149030
Caglar Koca and Ozgur B. Akan, Modelling 1D Partially Absorbing Boundaries for Brownian Molecular Communication Channels, arXiv:2402.15888 [q-bio.MN], 2024. See p. 9.
Wikipedia, Chebyshev polynomials

Cf. A038207, A053117.
Cf. A018297, A000129. - Gary W. Adamson, Nov 28 2008
Cf. A000079, A001787, A001788, A001789, A003472, A054849, A002409, A054851. - Philippe Deléham, Sep 12 2009
Cf. A091894, A097610, A099089, A207538.

t[n_, m_] = (-1)^m*Binomial[n - m, m]*2^(n - 2*m);
Table[Table[t[n, m], {m, 0, Floor[n/2]}], {n, 0, 10}];
Flatten[%] (* Roger L. Bagula, Dec 19 2008 *)

A generating function for U(n) is 1/(1 - 2tx + t^2). Given A038207, shift down columns to allow for (1, 1, 2, 2, 3, 3, ...) terms in each row, then insert alternate signs.
T(n,m) = (-1)^m*binomial(n - m, m)*2^(n - 2*m). - Roger L. Bagula and Gary W. Adamson, Dec 19 2008
From Tom Copeland, Feb 11 2016: (Start)
Shifted o.g.f.: G(x,t) = x/(1 - 2x + tx^2).
A053117 is a reflected, aerated version of this entry; A207538, an unsigned version; and A099089, a reflected, shifted version.
The compositional inverse of G(x,t) is Ginv(x,t) = ((1 + 2x) - sqrt((1 + 2x)^2 - 4tx^2))/(2tx) = x - 2x^2 + (4 + t)x^3 - (8 + 6t)x^4 + ..., a shifted o.g.f. for A091894 (mod signs with A091894(0,0) = 0.). Cf. A097610 with h_1 = -2 and h_2 = t. (End)

More terms from Philippe Deléham, Sep 12 2009

A054851 a(n) = 2^(n-7)*binomial(n,7). Number of 7D hypercubes in an n-dimensional hypercube.

Original entry on oeis.org

1, 16, 144, 960, 5280, 25344, 109824, 439296, 1647360, 5857280, 19914752, 65175552, 206389248, 635043840, 1905131520, 5588385792, 16066609152, 45364543488, 126012620800, 344876646400, 931166945280, 2483111854080

Offset: 7

Henry Bottomley, Apr 14 2000

If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n>6, a(n) is equal to the number of (n+7)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007

G. C. Greubel, Table of n, a(n) for n = 7..1000
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (16,-112,448,-1120,1792,-1792, 1024,-256).

Cf. A000079, A001787, A001788, A001789, A003472, A054849, A002409.

a(n) = A038207(n,7).

List([7..30], n-> 2^(n-7)*Binomial(n,7)); # G. C. Greubel, Aug 27 2019

[2^(n-7)*Binomial(n,7): n in [7..30]]; // G. C. Greubel, Aug 27 2019

seq(binomial(n+7,7)*2^n,n=0..21); # Zerinvary Lajos, Jun 23 2008

Table[2^(n-7)*Binomial[n,7], {n,7,30}] (* G. C. Greubel, Aug 27 2019 *)

vector(23, n, 2^(n-1)*binomial(n+6, 7)) \\ G. C. Greubel, Aug 27 2019

a(n) = 2*a(n-1) + A002409(n-1).
a(n+8) = A082141(n+1)/2.
G.f.: x^7/(1-2*x)^8. - Colin Barker, Sep 04 2012
a(n) = Sum_{i=7..n} binomial(i,7)*binomial(n,i). Example: for n=11, a(11) = 1*330 + 8*165 + 36*55 + 120*11 + 330*1 = 5280. - Bruno Berselli, Mar 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=7} 1/a(n) = 14*log(2) - 259/30.
Sum_{n>=7} (-1)^(n+1)/a(n) = 10206*log(3/2) - 124117/30. (End)

More terms from James Sellers, Apr 15 2000

A027608 Expansion of 1/((1-x)(1-2x)^4).

Original entry on oeis.org

1, 9, 49, 209, 769, 2561, 7937, 23297, 65537, 178177, 471041, 1216513, 3080193, 7667713, 18808833, 45547521, 109051905, 258473985, 607125505, 1414529025, 3271557121, 7516192769, 17163091969

Offset: 0

N. J. A. Sloane

Michael De Vlieger, Table of n, a(n) for n = 0..3288
Michał Adamaszek and Henry Adams, On Vietoris-Rips complexes of hypercube graphs, arXiv:2103.01040 [math.CO], 2021.
M. H. Albert, M. D. Atkinson, and R. Brignall, The enumeration of three pattern classes using monotone grid classes, E. J. Combinat. 19 (3) (2012) P20. Chapter 5.5.1 (with leading zeros).
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
Santiago López de Medrano, On the genera of moment-angle manifolds associated to dual-neighborly polytopes, combinatorial formulas and sequences, arXiv:2003.07508 [math.GT], 2020.
Index entries for linear recurrences with constant coefficients, signature (9,-32,56,-48,16).

Cf. A001789 (first differences).

[(n/3)*(n^2+3*n+8)*2^n +1: n in [0..40]]; // G. C. Greubel, Aug 24 2022

CoefficientList[Series[1/((1-x)*(1-2x)^4), {x, 0, 22}], x] (* Michael De Vlieger, Jun 23 2020 *)
LinearRecurrence[{9,-32,56,-48,16},{1,9,49,209,769},30] (* Harvey P. Dale, Apr 09 2021 *)

Vec(1/((1-x)*(1-2*x)^4)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

[(n/3)*(n^2+3*n+8)*2^n + 1 for n in (0..40)] # G. C. Greubel, Aug 24 2022

a(n-1) = 1 + (n-1)*2^(n+1) + ((n^3 - 7*n + 6)*2^(n-1))/3, n >= 1. - Roger Voles, Dec 07 2004, index corrected by R. J. Mathar, Mar 14 2011
a(n) = A119258(n+4,n). - Reinhard Zumkeller, May 11 2006
a(n) = 1 + n*2^(n+2) + (((n+1)^3 - 7*(n+1) + 6)*2^n)/3 = (n/3)*(n^2 + 3*n + 8)*2^n + 1, n >= 0. - Daniel Forgues, Nov 01 2012
E.g.f.: exp(x) + (8/3)*x*(3 + 3*x + x^2)*exp(2*x). - G. C. Greubel, Aug 24 2022

A099089 Riordan array (1, 2+x).

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 0, 4, 8, 0, 0, 1, 12, 16, 0, 0, 0, 6, 32, 32, 0, 0, 0, 1, 24, 80, 64, 0, 0, 0, 0, 8, 80, 192, 128, 0, 0, 0, 0, 1, 40, 240, 448, 256, 0, 0, 0, 0, 0, 10, 160, 672, 1024, 512, 0, 0, 0, 0, 0, 1, 60, 560, 1792, 2304, 1024, 0, 0, 0, 0, 0, 0, 12, 280, 1792, 4608, 5120, 2048

Offset: 0

Paul Barry, Sep 25 2004

Row sums are A000129. Diagonal sums are A008346. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).
Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/2, -1/2, 0, 0, 0, 0, ...] DELTA [2, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008
As an upper right triangle (in the example), table rows give number of points, edges, faces, cubes, 4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009

			Triangle begins:
  1;
  0,  2;
  0,  1,  4;
  0,  0,  4,  8;
  0,  0,  1, 12, 16;
  0,  0,  0,  6, 32, 32;
  0,  0,  0,  1, 24, 80, 64;
The entries can also be interpreted as the antidiagonal reading of the following array:
  1,    2,    4,    8,   16,   32,   64,  128,  256,  512, 1024,... A000079
  0,    1,    4,   12,   32,   80,  192,  448, 1024, 2304, 5120,... A001787
  0,    0,    1,    6,   24,   80,  240,  672, 1792, 4608,11520,... A001788
  0,    0,    0,    1,    8,   40,  160,  560, 1792, 5376,15360,... A001789
  0,    0,    0,    0,    1,   10,   60,  280, 1120, 4032,13440,...
  0,    0,    0,    0,    0,    1,   12,   84,  448, 2016, 8064,...
  0,    0,    0,    0,    0,    0,    1,   14,  112,  672, 3360,...
  0,    0,    0,    0,    0,    0,    0,    1,   16,  144,  960,...
  0,    0,    0,    0,    0,    0,    0,    0,    1,   18,  180,...
  0,    0,    0,    0,    0,    0,    0,    0,    0,    1,   20,...
  0,    0,    0,    0,    0,    0,    0,    0,    0,    0,    1,...

H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

Eric W. Weisstein's Mathworld, Hypercube.

Cf. A053118, A008312, A062715, A038207.

Number triangle T(n,k) = binomial(k, n-k)*2^k*(1/2)^(n-k); columns have g.f. (2*x+x^2)^k.
G.f.: 1/(1-2y*x-y*x^2). - Philippe Deléham, Nov 20 2011
Sum_ {k=0..n} T(n,k)*x^k = A000007(n), A000129(n+1), A090017(n+1), A090018(n), A190510(n+1), A190955(n+1) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Nov 20 2011
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1), T(0,0) = 1, T(1,0) = T(2,0) = 0, T(1,1) = 2, T(2,1) = 1, T(2,2) = 4, T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Oct 30 2013

A081143 5th binomial transform of (0,0,0,1,0,0,0,0,......).

Original entry on oeis.org

0, 0, 0, 1, 20, 250, 2500, 21875, 175000, 1312500, 9375000, 64453125, 429687500, 2792968750, 17773437500, 111083984375, 683593750000, 4150390625000, 24902343750000, 147857666015625, 869750976562500, 5073547363281250

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, four-fold convolution of A000351 (powers of 5).

With a different offset, number of n-permutations (n=4)of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly three u's. Example: a(4)=20 because we have uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu, xuuu, uuuy, uuyu, uyuu and yuuu. - Zerinvary Lajos, Jun 03 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (20,-150,500,-625).

Cf. A038846, A036216, A001789, A081144, A081135.

[5^(n-3) * Binomial(n, 3): n in [0..25]]; // Vincenzo Librandi, Aug 06 2013

seq(binomial(n,3)*5^(n-3), n=0..25); # Zerinvary Lajos, Jun 03 2008

CoefficientList[Series[x^3/(1-5x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
LinearRecurrence[{20,-150,500,-625}, {0,0,0,1}, 30] (* Harvey P. Dale, Dec 24 2015 *)

vector(31, n, my(m=n-1); 5^(m-3)*binomial(m,3)) \\ G. C. Greubel, Mar 05 2020

[lucas_number2(n, 5, 0)*binomial(n,3)/5^3 for n in range(0, 22)] # Zerinvary Lajos, Mar 12 2009

a(n) = 20*a(n-1) - 150*a(n-2) + 500*a(n-3) - 625*a(n-4), with a(0)=a(1)=a(2)=0, a(3)=1.
a(n) = 5^(n-3)*binomial(n,3).
G.f.: x^3/(1-5*x)^4.
E.g.f.: x^3*exp(x)/6. - G. C. Greubel, Mar 05 2020
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=3} 1/a(n) = 240*log(5/4) - 105/2.
Sum_{n>=3} (-1)^(n+1)/a(n) = 540*log(6/4) - 195/2. (End)

A130810 If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n) is equal to the number of 4-subsets of X containing none of X_i, (i=1,...,n).

Original entry on oeis.org

16, 80, 240, 560, 1120, 2016, 3360, 5280, 7920, 11440, 16016, 21840, 29120, 38080, 48960, 62016, 77520, 95760, 117040, 141680, 170016, 202400, 239200, 280800, 327600, 380016, 438480, 503440, 575360, 654720, 742016, 837760, 942480, 1056720

Offset: 4

Milan Janjic, Jul 16 2007

Number of n permutations (n>=4) of 3 objects u,v,z, with repetition allowed, containing n-4 u's. Example: if n=4 then n-4 =(0) zero u, a(1)=16 because we have vvvv zzzz vvvz zzzv vvzv zzvz vzvv zvzz zvvv vzzz vvzz zzvv vzvz zvzv zvvz vzzv. - Zerinvary Lajos, Aug 05 2008

a(n) is the number of 3-dimensional elements in an n-cross polytope where n>=4. - Patrick J. McNab, Jul 06 2015

Harlan J. Brothers, Pascal's Prism: Supplementary Material.
Milan Janjić, Two Enumerative Functions. [Wayback Machine link]
Eric Weisstein's World of Mathematics, Cross Polytope.

Cf. A000332, A038207, A000079, A001787, A001788, A001789, A003472, A054849, A002409, A054851, A140325, A140354, A046092, A130809. Equals twice A162668.

a:= n-> binomial(2*n,4) +binomial(n,2) -n*binomial(2*n-2,2);
seq(binomial(n, n-4)*2^4, n=4..37); # Zerinvary Lajos, Dec 07 2007

a[n_] := 16 * Binomial[n, 4]; Array[a, 34, 4] (* Amiram Eldar, Jul 25 2025 *)

a(n) = 16 * binomial(n, 4); \\ Amiram Eldar, Jul 25 2025

a(n) = binomial(2*n,4) + binomial(n,2) - n*binomial(2*n-2,2).
a(n) = binomial(n,4)*16. - Zerinvary Lajos, Dec 07 2007
G.f.: 16*x^4/(1-x)^5. - Colin Barker, Apr 14 2012
a(n) = 2*n*(n-1)*(n-2)*(n-3)/3 = 2*A162668(n-3). - Robert Israel, Jul 06 2015
a(n) = 16 * A000332(n). - Alois P. Heinz, Oct 26 2020
E.g.f.: 2*exp(x)*x^4/3. - Stefano Spezia, Jul 17 2025
From Amiram Eldar, Jul 25 2025: (Start)
Sum_{n>=4} 1/a(n) = 1/12.
Sum_{n>=4} (-1)^n/a(n) = 2*log(2) - 4/3. (End)

A162007 Third left hand column of the EG1 triangle A162005.

Original entry on oeis.org

1, 270, 36096, 4766048, 704357760, 120536980224, 24060789342208, 5590122715250688, 1503080384197754880, 464520829174515630080, 163839204411117787938816, 65500849343294249018327040

Offset: 3

Johannes W. Meijer, Jun 27 2009

Third left hand column of the EG1 triangle A162005.
Other left hand columns are A000182 and A162006.
Related to A094665, A083061 and A156919.
A000079, A036289 and A100381 appear in the a(n, 3) formula.
A001789, A003472, A054849, A002409, A054851, A140325 and A140354 (scaled by 2^(m-1)) appear one by one in the a(n, m) formulas for m= 4 and higher .

nmax := 14; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1)*(x+1)*T1(i-1, x+1) - 2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1) * A156919(n-1, m-1) end do end do: m:=3; for n from m to nmax do a(n, m) := sum((-1)^(m-p1-1)*sum(2^(n-q-1)*binomial(n-q-1, m-p1-1) * A094665(n-1, q) * A156919(q, p1), q=1..n-m+p1), p1=0..m-1) od: seq(a(n, m), n = m..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

a(n) = sum((-1)^(m-p-1)*sum(2^(n-q-1)*binomial(n-q-1,m-p-1)*A094665(n-1,q)* A156919(q,p),q=1..n-m+p), p=0..m-1) with m = 3.

A055584 Triangle of partial row sums (prs) of triangle A055252.

Original entry on oeis.org

1, 5, 1, 19, 6, 1, 63, 25, 7, 1, 192, 88, 32, 8, 1, 552, 280, 120, 40, 9, 1, 1520, 832, 400, 160, 49, 10, 1, 4048, 2352, 1232, 560, 209, 59, 11, 1, 10496, 6400, 3584, 1792, 769, 268, 70, 12, 1, 26624, 16896, 9984, 5376, 2561, 1037, 338, 82, 13, 1, 66304, 43520

Offset: 0

Wolfdieter Lang, May 26 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is (((1-z)^3)/(1-2*z)^4)/(1-x*z/(1-z)).
This is the fourth member of the family of Riordan-type matrices obtained from A007318(n,m) (Pascal's triangle read as lower triangular matrix) by repeated application of the prs-procedure.
The column sequences appear as A049612(n+1), A055585, A001794, A001789(n+3), A027608, A055586 for m=0..5.

			[0] 1
[1] 5, 1
[2] 19, 6, 1
[3] 63, 25, 7, 1
[4] 192, 88, 32, 8, 1
[5] 552, 280, 120, 40, 9, 1
[6] 1520, 832, 400, 160, 49, 10, 1
[7] 4048, 2352, 1232, 560, 209, 59, 11, 1
Fourth row polynomial (n=3): p(3, x)= 63 + 25*x + 7*x^2 + x^3.

Cf. A007318, A055248, A055249, A055252. Row sums: A049600(n+1, 4).

T := (n, k) -> binomial(n, k)*hypergeom([4, k - n], [k + 1], -1):
for n from 0 to 7 do seq(simplify(T(n, k)), k = 0..n) od; # Peter Luschny, Sep 23 2024

a(n, m)=sum(A055252(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n

Column m recursion: a(n, m)= sum(a(j, m), j=m..n-1)+ A055252(n, m), n >= m >= 0, a(n, m) := 0 if n

G.f. for column m: (((1-x)^3)/(1-2*x)^4)*(x/(1-x))^m, m >= 0.

T(n, k) = binomial(n, k)*hypergeom([4, k - n], [k + 1], -1). - Peter Luschny, Sep 23 2024

A118441 Triangle L, read by rows, equal to the matrix log of A118435, with the property that L^2 consists of a single diagonal (two rows down from the main diagonal).

Original entry on oeis.org

0, 1, 0, -4, 2, 0, -12, 12, 3, 0, 32, -48, -24, 4, 0, 80, -160, -120, 40, 5, 0, -192, 480, 480, -240, -60, 6, 0, -448, 1344, 1680, -1120, -420, 84, 7, 0, 1024, -3584, -5376, 4480, 2240, -672, -112, 8, 0, 2304, -9216, -16128, 16128, 10080, -4032, -1008, 144, 9, 0

Offset: 0

Paul D. Hanna, Apr 28 2006

L = log(A118435) = log(H*[C^-1]*H], where C=Pascal's triangle and H=A118433 where H^2 = I (identity matrix).

			The matrix log, L = log(H*[C^-1]*H], begins:
     0;
     1,     0;
    -4,     2,      0;
   -12,    12,      3,     0;
    32,   -48,    -24,     4,     0;
    80,  -160,   -120,    40,     5,     0;
  -192,   480,    480,  -240,   -60,     6,     0;
  -448,  1344,   1680, -1120,  -420,    84,     7,   0;
  1024, -3584,  -5376,  4480,  2240,  -672,  -112,   8,  0;
  2304, -9216, -16128, 16128, 10080, -4032, -1008, 144,  9,  0;
  ...
The matrix square, L^2, is a single diagonal:
  0;
  0, 0;
  2, 0,  0;
  0, 6,  0,  0;
  0, 0, 12,  0,  0;
  0, 0,  0, 20,  0,  0;
  0, 0,  0,  0, 30,  0,  0;
  ...
From _Peter Luschny_, Apr 23 2020: (Start)
In unsigned form and without the main diagonal, as computed by the Maple script:
  [0], [0]
  [1], [1]
  [2], [4,   2]
  [3], [12,  12,   3]
  [4], [32,  48,   24,   4]
  [5], [80,  160,  120,  40,   5]
  [6], [192, 480,  480,  240,  60,  6]
  [7], [448, 1344, 1680, 1120, 420, 84, 7] (End)

Cf. A118435 (exp(L)), A118442 (column 0), A118443 (row sums), A027471 (unsigned row sums); A118433 (self-inverse triangle), A001815 (column 1?), A001789 (third of column 2?).

# Generalized Worpitzky transform of the harmonic numbers.
CL := p -> PolynomialTools:-CoefficientList(expand(p), x):
H := n -> add(1/k, k=1..n):
Trow := proc(n) local k,v; if n=0 then return [0] fi;
add(add((-1)^(n-v)*binomial(k,v)*H(k)*(-x+v-1)^n, v=0..k), k=0..n); CL(%) end:
for n from 0 to 7 do Trow(n) od; # Peter Luschny, Apr 23 2020

nmax = 12;
h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
L = MatrixLog[H.Inverse[Cn].H ];
Table[L[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)

/* From definition of L as matrix log of H*C^-1*H: */
{L(n,k)=local(H=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2-(c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))),N=(H*C^-1*H)); Log=sum(p=1,n+1,-(N^0-N)^p/p);Log[n+1,k+1]}
for(n=0, 10, for(k=0, n, print1(L(n, k), ", ")); print(""))

/* The matrix power L^m is given by: */
{L(n,k,m)=if(m%2==0,if(n==k+m,n!/k!*2^(n-k-m)/(n-k-m)!), if(n>=k+m,n!/k!*2^(n-k-m)/(n-k-m)!*(-1)^(m\2+(n+1)\2-k\2+n-k)))}
for(n=0, 10, for(k=0, n, print1(L(n, k,1), ", ")); print(""))

For even exponents of L, L^(2m) is a single diagonal:
if n == k+2m, then [L^(2m)](n,k) = n!/k!*2^(n-k-2m)/(n-k-2m)!; else if n != k+2m: [L^(2m)](n,k) = 0.
For odd exponents of L:
if n >= k+2m+1, then [L^(2m+1)](n,k) = n!/k!*2^(n-k-2m-1)/(n-k-2m-1)!*(-1)^(m+[(n+1)/2]-[k/2]+n-k); else if n < k+2m+1: [L^(2m)](n,k) = 0.
Unsigned row sums equals A027471(n+1) = n*3^(n-1).

A155863 Triangle T(n,k) = n(n^2 - 1)binomial(n-2, k-1) for 1 <= k <= n-1, n >= 2, and T(n,0) = T(n,n) = 1 for n >= 0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 24, 24, 1, 1, 60, 120, 60, 1, 1, 120, 360, 360, 120, 1, 1, 210, 840, 1260, 840, 210, 1, 1, 336, 1680, 3360, 3360, 1680, 336, 1, 1, 504, 3024, 7560, 10080, 7560, 3024, 504, 1, 1, 720, 5040, 15120, 25200, 25200, 15120, 5040, 720, 1, 1, 990, 7920, 27720, 55440, 69300, 55440, 27720, 7920, 990, 1

Offset: 0

Roger L. Bagula, Jan 29 2009

			Triangle begins:
  1;
  1,   1;
  1,   6,    1;
  1,  24,   24,     1;
  1,  60,  120,    60,     1;
  1, 120,  360,   360,   120,     1;
  1, 210,  840,  1260,   840,   210,     1;
  1, 336, 1680,  3360,  3360,  1680,   336,     1;
  1, 504, 3024,  7560, 10080,  7560,  3024,   504,    1,
  1, 720, 5040, 15120, 25200, 25200, 15120,  5040,  720,   1;
  1, 990, 7920, 27720, 55440, 69300, 55440, 27720, 7920, 990, 1;
  ...

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

Cf. A001789, A052771, A155864, A155865.

A155863:= func< n,k | k eq 0 or k eq n select 1 else 6*Binomial(n+1, 3)*Binomial(n-2, k-1) >;
[A155863(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 04 2021

(* First program *)
p[n_, x_]:= p[n, x]= If[n==0, 1, 1 + x^n + x*D[(x+1)^(n+1), {x, 3}]];
Flatten[Table[CoefficientList[p[n,x], x], {n,0,12}]]
(* Second program *)
T[n_, k_]:= If[k==0 || k==n, 1, 6*Binomial[n+1, 3]*Binomial[n-2, k-1]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 04 2021 *)

T(n, k):= ratcoef(expand(x^n + n*(n^2 -1)*x*(x+1)^(n-2) + (1 + (-1)^(2^n))/2), x, k)$
create_list(T(n, k), n, 0, 12, k, 0, n); /* Franck Maminirina Ramaharo, Dec 03 2018 */

def A155863(n,k): return 1 if (k==0 or k==n) else 6*binomial(n+1, 3)*binomial(n-2, k-1)
flatten([[A155863(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 04 2021

T(n, k) = coefficients of p(n, x), where p(n, x) = 1 + x^n + x*((d/dx)^3 (x+1)^(n+1)) and T(0, 0) = 1.
From Franck Maminirina Ramaharo, Dec 03 2018: (Start)
T(n, k) = (n-1)*n*(n+1)*binomial(n-2, k-1) with T(n, 0) = T(n, n) = 1.
n-th row polynomial is x^n + n*(n^2 - 1)*x*(x+1)^(n-2) + (1 + (-1)^(2^n))/2.
G.f.: 1/(1 - y) + 1/(1 - x*y) + (6*x*y^2)/(1 - y - x*y)^4 - 1.
E.g.f.: exp(y) + exp(x*y) + (3*x*y^2 + (x + x^2)*y^3)*exp((1 + x)*y) - 1. (End)
Sum_{k=0..n} T(n, k) = 2 - [n=0] + 6*A001789(n+1) = 2 - [n=0] + A052771(n+1). - G. C. Greubel, Jun 04 2021

Edited and name clarified by Franck Maminirina Ramaharo, Dec 03 2018

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A155863 Triangle T(n,k) = n(n^2 - 1)binomial(n-2, k-1) for 1 <= k <= n-1, n >= 2, and T(n,0) = T(n,n) = 1 for n >= 0, read by rows.