A002426 -id:A002426 - cp's OEIS frontend

A111808 Left half of trinomial triangle (A027907), triangle read by rows.

1, 1, 1, 1, 2, 3, 1, 3, 6, 7, 1, 4, 10, 16, 19, 1, 5, 15, 30, 45, 51, 1, 6, 21, 50, 90, 126, 141, 1, 7, 28, 77, 161, 266, 357, 393, 1, 8, 36, 112, 266, 504, 784, 1016, 1107, 1, 9, 45, 156, 414, 882, 1554, 2304, 2907, 3139, 1, 10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350

Offset: 1

Reinhard Zumkeller, Aug 17 2005

Consider a doubly infinite chessboard with squares labeled (n,k), ranks or rows n in Z, files or columns k in Z (Z denotes ...,-2,-1,0,1,2,... ); number of king-paths of length n from (0,0) to (n,k), 0 <= k <= n, is T(n,n-k). - Harrie Grondijs, May 27 2005. Cf. A026300, A114929, A114972.

Triangle of numbers C^(2)(n-1,k), n>=1, of combinations with repetitions from elements {1,2,...,n} over k, such that every element i, i=1,...,n, appears in a k-combination either 0 or 1 or 2 times (cf. also A213742-A213745). - Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

Harrie Grondijs, Neverending Quest of Type C, Volume B - the endgame study-as-struggle.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Eric Weisstein's World of Mathematics, Trinomial Triangle
Eric Weisstein's World of Mathematics, Trinomial Coefficient

Row sums give A027914; central terms give A027908;
T(n, 0) = 0;
T(n, 1) = n for n>1;
T(n, 2) = A000217(n) for n>1;
T(n, 3) = A005581(n) for n>2;
T(n, 4) = A005712(n) for n>3;
T(n, 5) = A000574(n) for n>4;
T(n, 6) = A005714(n) for n>5;
T(n, 7) = A005715(n) for n>6;
T(n, 8) = A005716(n) for n>7;
T(n, 9) = A064054(n-5) for n>8;
T(n, n-5) = A098470(n) for n>4;
T(n, n-4) = A014533(n-3) for n>3;
T(n, n-3) = A014532(n-2) for n>2;
T(n, n-2) = A014531(n-1) for n>1;
T(n, n-1) = A005717(n) for n>0;
T(n, n) = central terms of A027907 = A002426(n).

T := (n,k) -> simplify(GegenbauerC(k, -n, -1/2)):
for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 09 2016

Table[GegenbauerC[k, -n, -1/2], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

(1 + x + x^2)^n = Sum(T(n,k)*x^k: 0<=k<=n) + Sum(T(n,k)*x^(2*n-k): 0<=k

T(n, k) = A027907(n, k) = Sum_{i=0,..,(k/2)} binomial(n, n-k+2*i) * binomial(n-k+2*i, i), 0<=k<=n.

T(n, k) = GegenbauerC(k, -n, -1/2). - Peter Luschny, May 09 2016

Corrected and edited by Johannes W. Meijer, Oct 05 2010

A077042 Square array read by falling antidiagonals of central polynomial coefficients: largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^k = ((1-x^n)/(1-x))^k, i.e., the coefficient of x^floor(k(n-1)/2) and of x^ceiling(k(n-1)/2); also number of compositions of floor(k*(n+1)/2) into exactly k positive integers each no more than n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 6, 7, 4, 1, 1, 0, 1, 10, 19, 12, 5, 1, 1, 0, 1, 20, 51, 44, 19, 6, 1, 1, 0, 1, 35, 141, 155, 85, 27, 7, 1, 1, 0, 1, 70, 393, 580, 381, 146, 37, 8, 1, 1, 0, 1, 126, 1107, 2128, 1751, 780, 231, 48, 9, 1, 1, 0, 1, 252, 3139

Offset: 0

Henry Bottomley, Oct 22 2002

From Michel Marcus, Dec 01 2012: (Start)
A pair of numbers written in base n are said to be comparable if all digits of the first number are at least as big as the corresponding digit of the second number, or vice versa. Otherwise, this pair will be defined as uncomparable. A set of pairwise uncomparable integers will be called anti-hierarchic.
T(n,k) is the size of the maximal anti-hierarchic set of integers written with k digits in base n.
For example, for base n=2 and k=4 digits:
- 0 (0000) and 15 (1111) are comparable, while 6 (0110) and 9 (1001) are uncomparable,
- the maximal antihierarchic set is {3 (0011), 5 (0101), 6 (0110), 9 (1001), 10 (1010), 12 (1100)} with 6 elements that are all pairwise uncomparable. (End)

			Rows of square array start:
  1,    0,    0,    0,    0,    0,    0, ...
  1,    1,    1,    1,    1,    1,    1, ...
  1,    1,    2,    3,    6,   10,   20, ...
  1,    1,    3,    7,   19,   51,  141, ...
  1,    1,    4,   12,   44,  155,  580, ...
  1,    1,    5,   19,   85,  381, 1751, ...
  ...
Read by antidiagonals:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 2, 1, 1;
  0, 1, 3, 3, 1, 1;
  0, 1, 6, 7, 4, 1, 1;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Denis Bouyssou, Thierry Marchant, Marc Pirlot, The size of the largest antichains in products of linear orders, arXiv:1903.07569 [math.CO], 2019.
J. W. Sander, On maximal antihierarchic sets of integers, Discrete Mathematics, Volume 113, Issues 1-3, 5 April 1993, Pages 179-189.
Index entries for sequences related to compositions

Rows include A000007, A000012, A001405, A002426, A005190, A005191, A018901, A025012, A025013, A025014, A025015, A201549, A225779, A201550. Columns include A000012, A000012, A001477, A077043, A005900, A077044, A071816, A133458.
Central diagonal is A077045, with A077046 and A077047 either side. Cf. A067059, A270918, A201552.
Cf. A273975.

t[n_, k_] := Max[ CoefficientList[ Series[ ((1-x^n) / (1-x))^k, {x, 0, k*(n-1)}], x]]; t[0, 0] = 1; t[0, ] = 0; Flatten[ Table[ t[n-k, k], {n, 0, 12}, {k, n, 0, -1}]] (* _Jean-François Alcover, Nov 04 2011 *)

T(n,k)=if(n<1 || k<1,k==0,vecmax(Vec(((1-x^n)/(1-x))^k)))

By the central limit theorem, T(n,k) is roughly n^(k-1)*sqrt(6/(Pi*k)).

T(n,k) = Sum{j=0,h/n} (-1)^j*binomial(k,j)*binomial(k-1+h-n*j,k-1) with h=floor(k*(n-1)/2), k>0. - Michel Marcus, Dec 01 2012

A094061 Number of n-moves paths of a king starting and ending at the origin of an infinite chessboard.

Original entry on oeis.org

1, 0, 8, 24, 216, 1200, 8840, 58800, 423640, 3000480, 21824208, 158964960, 1171230984, 8668531872, 64574844048, 483114856224, 3630440899800, 27379154692032, 207172490054816, 1572194644061184, 11962847247681616, 91242602561647680, 697438669619791008

Offset: 0

Matthijs Coster, Apr 29 2004

The chessboard here is the full four-quadrant board Z X Z.
This is an analog of A054474 for walks on a square grid where the steps can be made diagonally as well.
a(n) is the constant term in the expansion of ((x + 1/x) * (y + 1/y) + x^2 + 1/x^2 + y^2 + 1/y^2)^n. - Seiichi Manyama, Nov 03 2019

D. Joyner, "Adventures in Group Theory: Rubik's Cube, Merlin's Machine and Other Mathematical Toys", Johns Hopkins University Press, 2002, pp. 79

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Peter Bala, A note on A094061

Row 2 of A327751.

Cf. A002426, A098070, A126869, A253974, A254129, A254459, A329024.

a:=array(0..30):a[0]:=1:a[1]:=0:a[2]:=8:a[3]:=24:for n from 3 to 29 do a[n+1]:= (n*(5*n+1)*a[n]+2*(15*n^2+6*n-5)*a[n-1]-8*(5*n^2-23*n+21)*a[n-2]-64*(n-2)^2*a[n-3])/(n+1)^2: print(n+1,a[n+1]) od:
# second Maple program
a:= proc(n) option remember; `if`(n<3, (n-1)*(9*n-2)/2,
      ((n-1)*(3*n-1)*(3*n-4) *a(n-1)
      +(108*n^3-396*n^2+452*n-152) *a(n-2)
      +32*(3*n-2)*(n-2)^2 *a(n-3))/ (n^2*(3*n-5)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 02 2012

a[n_]:=Module[{f=(x+x^-1+y+y^-1+x y+x^-1y+x^-1y^-1+x y^-1)^n,s}, s=Series[f,{x,0,0},{y,0,0}]; SeriesCoefficient[s,{0,0}]]; Table[a[n], {n,1,22}] (* Armin Vollmer (Armin.Vollmer(AT)kabelleipzig.de), May 01 2006 *)
CoefficientList[Series[1/(1+4*x)*LegendreP[-1/2,1-32*x*(1+x)/(1+4*x)^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Aug 16 2013 *)

a[0]:1$
a[1]:0$
a[2]:8$
a[3]:24$
a[n]:=((n-1)*(3*n-1)*(3*n-4) *a[n-1]
      +(108*n^3-396*n^2+452*n-152) *a[n-2]
      +32*(3*n-2)*(n-2)^2 *a[n-3])/ (n^2*(3*n-5))$
A094061(n):=a[n]$
makelist(A094061(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

{a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*polcoef((1+x+1/x)^k, 0)^2)} \\ Seiichi Manyama, Oct 29 2019

{a(n) = polcoef(polcoef(((x+1/x)*(y+1/y)+x^2+1/x^2+y^2+1/y^2)^n, 0), 0)} \\ Seiichi Manyama, Nov 03 2019

D-finite with recurrence (n+1)^2*a(n+1) = n*(5*n+1)*a(n) + 2*(15*n^2+6*n-5)*a(n-1) - 8*(5*n^2-23*n+21) *a(n-2) - 64*(n-2)^2*a(n-3).
G.f.: (2/(Pi*(1+4*x))) * EllipticK(4*sqrt(x*(1+x))/(1+4*x)) = 1/(1+4*x) * hypergeom([1/2,1/2], [1], 16*(x*(1+x))/(1+4*x)^2). - Sergey Perepechko, Jan 15 2011
a(n) ~ 2^(3*n+1)/(3*Pi*n). - Vaclav Kotesovec, Aug 16 2013
a(n) = (1/Pi^2) * Integral_{y = 0..Pi} Integral_{x = 0..Pi} (2*cos(x) + 2*cos(y) + 4*cos(x)*cos(y))^n dx dy. - Peter Bala, Feb 14 2017
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A002426(k)^2. - Seiichi Manyama, Oct 29 2019
From Peter Bala, Feb 08 2022: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 5 and positive integers n and k. (End)
a(n) = Sum_{j = 0..n} Sum_{k = 0..j} binomial(2*j,j)^2*binomial(j,k)* binomial(n+j-k,2*j)*(-4)^(n-j-k). - Peter Bala, Mar 19 2022

More terms from and entry improved by Sergey Perepechko, Sep 06 2004

A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 19, 19, 1, 9, 37, 85, 51, 1, 11, 61, 231, 381, 141, 1, 13, 91, 489, 1451, 1751, 393, 1, 15, 127, 891, 3951, 9331, 8135, 1107, 1, 17, 169, 1469, 8801, 32661, 60691, 38165, 3139, 1, 19, 217, 2255, 17151, 88913, 273127, 398567, 180325, 8953, 1

Offset: 1

R. H. Hardin, Dec 02 2011

Equivalently, the number of compositions of n*(k + 1) into n parts with maximum part size 2*k+1. - Andrew Howroyd, Oct 14 2017

			Some solutions for n=7, k=3:
..1...-2....1...-1....1...-3....0....0....1....2....3...-3....0....2....1....0
.-1....2...-2....2....2....2...-1....0....2....2...-2...-1...-2...-1....2...-1
.-3...-1....1...-3....2....1....0....1....3....0....2....0...-1....2...-2...-1
..0....3....3....3...-2...-2....3....3...-3...-3....0...-1...-1...-1....0....3
..2...-1...-1...-1...-3....0...-3...-2....1...-1...-1....1....1....0....3...-1
..2...-1...-3....0....2....3....0....1...-2....1....1....1....3...-2...-3...-3
.-1....0....1....0...-2...-1....1...-3...-2...-1...-3....3....0....0...-1....3
Table starts:
.   1,      1,       1,        1,        1,         1,...
.   3,      5,       7,        9,       11,        13,...
.   7,     19,      37,       61,       91,       127,...
.  19,     85,     231,      489,      891,      1469,...
.  51,    381,    1451,     3951,     8801,     17151,...
. 141,   1751,    9331,    32661,    88913,    204763,...
. 393,   8135,   60691,   273127,   908755,   2473325,...
.1107,  38165,  398567,  2306025,  9377467,  30162301,...
.3139, 180325, 2636263, 19610233, 97464799, 370487485,...

R. H. Hardin, Table of n, a(n) for n = 1..9999
Michelle Rudolph-Lilith and Lyle E. Muller, On a link between Dirichlet kernels and central multinomial coefficients, Discrete Mathematics 338.9 (2015): 1567-1572.
Wikipedia, Dirichlet kernel.

Columns 1-10: A002426, A005191, A025012, A025014, A201549, A201550, A201551, A322538, A322539, A322540.
Rows 3-10: A003215, A063496(n+1), A083669, A201553, A201554, A322535, A322536, A322537.
Cf. A286928.

seq(print(seq(add((-1)^i*binomial(n, i)*binomial((k+1)*n-(2*k+1)*i-1, n-1), i = 0..floor((1/2)*n)), k = 1..10)), n = 1..10); # Peter Bala, Oct 16 2024

comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r - 1 - i*m, k - 1]*Binomial[k, i], {i, 0, Floor[(r - k)/m]}];  T[n_, k_] := comps[n*(k + 1), 2*k + 1, n]; Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)

comps(r, m, k)=sum(i=0, floor((r-k)/m), (-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
T(n,k) = comps(n*(k+1), 2*k+1, n); \\ Andrew Howroyd, Oct 14 2017

Empirical: T(n,k) = Sum_{i=0..floor(k*n/(2*k+1))} (-1)^i*binomial(n,i)* binomial((k+1)*n-(2*k+1)*i-1,n-1).
The above empirical formula is true and can be derived from the formula for the number of compositions with given number of parts and maximum part size. - Andrew Howroyd, Oct 14 2017
Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k + 1
T(3,k) = 3*k^2 + 3*k + 1
T(4,k) = (16/3)*k^3 + 8*k^2 + (14/3)*k + 1
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (185/12)*k^2 + (35/6)*k + 1
T(6,k) = (88/5)*k^5 + 44*k^4 + 46*k^3 + 25*k^2 + (37/5)*k + 1
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (2275/18)*k^4 + (357/4)*k^3 + (6643/180)*k^2 + (259/30)*k + 1
T(m,k) = (1/Pi)*integral_{x=0..Pi} (sin((k+1/2)x)/sin(x/2))^m dx; for the proof see Dirichlet Kernel link; so f(m,n) = (1/Pi)*integral_{x=0..Pi} (Sum_{k=-n..n} exp(I*k*x))^m dx = sum(integral(exp(I(k_1+...+k_m).x),x=0..Pi)/Pi,{k_1,...,k_m=-n..n}) = sum(delta_0(k1+...+k_m),{k_1,...,k_m=-n..n}) = number of arrays of m integers in -n..n with sum zero. - Yalcin Aktar, Dec 03 2011
T(n, k) = the constant term in the expansion of (x^(-k) + ... + x^(-1) + 1 + x + ... + x^k)^n = the coefficient of x^(k*n) (i.e., the central coefficient) in the expansion of (1 + x + ... + x^(2*k))^n = the coefficient of x^(k*n) in the expansion of ( (1 - x^(2*k+1))/(1 - x) )^n. Expanding the binomials and collecting terms gives the empirical formula above. - Peter Bala, Oct 16 2024

A071675 Array read by antidiagonals of trinomial coefficients.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 0, 3, 3, 1, 0, 0, 2, 6, 4, 1, 0, 0, 1, 7, 10, 5, 1, 0, 0, 0, 6, 16, 15, 6, 1, 0, 0, 0, 3, 19, 30, 21, 7, 1, 0, 0, 0, 1, 16, 45, 50, 28, 8, 1, 0, 0, 0, 0, 10, 51, 90, 77, 36, 9, 1, 0, 0, 0, 0, 4, 45, 126, 161, 112, 45, 10, 1, 0, 0, 0, 0, 1, 30, 141, 266, 266

Offset: 0

Henry Bottomley, May 30 2002

Read as a number triangle, this is the Riordan array (1, x(1+x+x^2)) with T(n,k) = Sum_{i=0..floor((n+k)/2)} C(k,2i+2k-n)*C(2i+2k-n,i). Rows start {1}, {0,1}, {0,1,1}, {0,1,2,1}, {0,0,3,3,1},... Row sums are then the trinomial numbers A000073(n+2). Diagonal sums are A013979. - Paul Barry, Feb 15 2005

Let {a_(k,i)}, k>=1, i=0,...,k, be the k-th antidiagonal of the array. Then s_k(n)=sum{i=0,...,k}a_(k,i)* binomial(n,k) is the n-th element of the k-th column of A213742. For example, s_1(n)=binomial(n,1)=n is the first column of A213742 for n>1, s_2(n)=binomial(n,1)+binomial(n,2)is the second column of A213742 for n>1, etc. In particular (see comment in A213742) in cases k=4,5,6,7,8, s_k(n) is A005718(n+2), A005719(n), A005720(n), A001919(n), A064055(n+3), respectively. - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012

			Rows start
1, 0,  0,  0,  0,  0, ...;
1, 1,  1,  0,  0,  0,  0, ...;
1, 2,  3,  2,  1,  0,  0, ...;
1, 3,  6,  7,  6,  3,  1, 0, ...;
1, 4, 10, 16, 19, 16, 10, 4, 1, ...; etc.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3

Visible version of A027907. Row sums are 3^n, i.e. A000244. Central diagonal is A002426. Cf. A071676 for a slight variation.

T[n_, k_] := Sum[Binomial[n - k - j, j]*Binomial[k, n - k - j], {j, 0,
Floor[(n - k)/2]}]; Table[T[n, k], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-1, k-2) starting with T(0, 0)=1. See A027907 for more.

As a number triangle, T(n, k) = Sum_{i=0..floor((n-k)/2)} C(n-k-i, i) * C(k, n-k-i). - Paul Barry, Apr 26 2005

A089627 T(n,k) = binomial(n,2k)binomial(2*k,k) for 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 6, 0, 0, 1, 12, 6, 0, 0, 1, 20, 30, 0, 0, 0, 1, 30, 90, 20, 0, 0, 0, 1, 42, 210, 140, 0, 0, 0, 0, 1, 56, 420, 560, 70, 0, 0, 0, 0, 1, 72, 756, 1680, 630, 0, 0, 0, 0, 0, 1, 90, 1260, 4200, 3150, 252, 0, 0, 0, 0, 0, 1, 110, 1980, 9240, 11550, 2772, 0, 0, 0, 0, 0, 0, 1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0, 1, 156, 4290, 34320, 90090, 72072, 12012, 0, 0, 0, 0, 0, 0, 0, 1, 182, 6006, 60060, 210210, 252252, 84084, 3432, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 31 2003

The rows of this triangle are the gamma vectors of the n-dimensional type B associahedra (Postnikov et al., p.38 ). Cf. A055151 and A101280. - Peter Bala, Oct 28 2008
T(n,k) is the number of Grand Motzkin paths of length n having exactly k upsteps (1,1). Cf. A109189, A055151. - Geoffrey Critzer, Feb 05 2014
The result Sum_{k = 0..floor(n/2)} C(n,2*k)*C(2*k,k)*x^k = (sqrt(1 - 4*x))^n* P(n,1/sqrt(1 - 4*x)) expressing the row polynomials of this triangle in terms of the Legendre polynomials P(n,x) is due to Catalan. See Laden, equation 7.10, p. 56. - Peter Bala, Mar 18 2018

			Triangle begins:
  1
  1,   0
  1,   2,    0
  1,   6,    0,     0
  1,  12,    6,     0,     0
  1,  20,   30,     0,     0,     0
  1,  30,   90,    20,     0,     0,   0
  1,  42,  210,   140,     0,     0,   0, 0
  1,  56,  420,   560,    70,     0,   0, 0, 0
  1,  72,  756,  1680,   630,     0,   0, 0, 0, 0
  1,  90, 1260,  4200,  3150,   252,   0, 0, 0, 0, 0
  1, 110, 1980,  9240, 11550,  2772,   0, 0, 0, 0, 0, 0
  1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0
Relocating the zeros to be evenly distributed and interpreting the triangle as the coefficients of polynomials
                     1
                     1
                 1 + 2 q^2
                 1 + 6 q^2
            1 + 12 q^2 +  6 q^4
            1 + 20 q^2 + 30 q^4
       1 + 30 q^2 +  90 q^4 +  20 q^6
       1 + 42 q^2 + 210 q^4 + 140 q^6
  1 + 56 q^2 + 420 q^4 + 560 q^6 + 70 q^8
then the substitution q^k -> 1/(floor(k/2)+1) gives the Motzkin numbers A001006.
- _Peter Luschny_, Aug 29 2011

Alois P. Heinz, Rows n = 0..140, flattened
Rui Duarte and António Guedes de Oliveira, A Famous Identity of Hajós in Terms of Sets, Journal of Integer Sequences, Vol. 17 (2014), #14.9.1.
Hyman N. Laden, An historical, and critical development of the theory of Legendre polynomials before 1900, Master of Arts Thesis, University of Maryland 1938.
Shi-Mei Ma, On gamma-vectors and the derivatives of the tangent and secant functions, arXiv:1304.6654 [math.CO], 2013.
A. Postnikov, V. Reiner, and L. Williams, Faces of generalized permutohedra, arXiv:math/0609184 [math.CO], 2006-2007.

Row sums A002426. Antidiagonal sums A098479.

Cf. A055151, A063007, A109187, A101280.

for i from 0 to 12 do seq(binomial(i, j)*binomial(i-j, j), j=0..i) od; # Zerinvary Lajos, Jun 07 2006
# Alternatively:
R := (n, x) -> simplify(hypergeom([1/2 - n/2, -n/2], [1], 4*x)):
Trow := n -> seq(coeff(R(n,x), x, j), j=0..n):
seq(print(Trow(n)), n=0..9); # Peter Luschny, Mar 18 2018

nn=15;mxy=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y);Map[Select[#,#>0&]&, CoefficientList[Series[1/(1-x-2y x^2mxy),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)

T(n,k) = binomial(n,2*k)*binomial(2*k,k);
concat(vector(15, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 01 2018

T(n,k) = n!/((n-2*k)!*k!*k!).
E.g.f.: exp(x)*BesselI(0, 2*x*sqrt(y)). - Vladeta Jovovic, Apr 07 2005
O.g.f.: ( 1 - x - sqrt(1 - 2*x + x^2 - 4*x^2*y))/(2*x^2*y). - Geoffrey Critzer, Feb 05 2014
R(n, x) = hypergeom([1/2 - n/2, -n/2], [1], 4*x) are the row polynomials. - Peter Luschny, Mar 18 2018
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^i*binomial(n, i)*binomial(n-i, k-i)^2. Cf. A063007(n,k) = Sum_{i = 0..k} binomial(n, i)^2*binomial(n-i, k-i).
T(n,k) = A063007(n-k,k); that is, the diagonals of this table are the rows of A063007. (End)

A084615 a(n) = sum of absolute values of coefficients of (1+x-3x^2)^n.

Original entry on oeis.org

1, 5, 23, 99, 401, 1525, 6345, 27331, 122083, 520805, 2117293, 8301441, 34517395, 147850771, 628707981, 2675100397, 10920387779, 43701876543, 180872758207, 769658883325, 3243501133481, 13617178197183, 56148348498199

Offset: 0

Paul D. Hanna, Jun 01 2003

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A002426, A084600 - A084614.

A084614:= func< n,k | (&+[Binomial(n, k-j)*Binomial(k-j, j)*(-3)^j: j in [0..k]]) >;
[(&+[Abs(A084614(n,k)): k in [0..2*n]]): n in [0..50]]; // G. C. Greubel, Mar 25 2023

Table[Total[Abs[CoefficientList[Expand[(1+x-3x^2)^n],x]]],{n,0,30}] (* Harvey P. Dale, Mar 26 2013 *)

for(n=0,40,S=0; for(k=0,2*n,t=polcoeff((1+x-3*x^2)^n,k,x); S=S+abs(t)); print1(S","))

@CachedFunction
def A084614(n,k): return sum(binomial(n,k-j)*binomial(k-j,j)*(-3)^j for j in range(k+1))
def A084615(n): return sum(abs(A084614(n,k)) for k in range(2*n+1))
[A084615(n) for n in range(50)] # G. C. Greubel, Mar 25 2023

a(n) = Sum_{k=0..2*n} abs( Sum_{j=0..k} binomial(n,k-j)*binomial(k-j,j)*(-3)^j ). - G. C. Greubel, Mar 25 2023

A027914 T(n,0) + T(n,1) + ... + T(n,n), T given by A027907.

Original entry on oeis.org

1, 2, 6, 17, 50, 147, 435, 1290, 3834, 11411, 34001, 101400, 302615, 903632, 2699598, 8068257, 24121674, 72137547, 215786649, 645629160, 1932081885, 5782851966, 17311097568, 51828203475, 155188936431, 464732722872

Offset: 0

Clark Kimberling

Let b(n)=a(n) mod 2; then b(n)=1/2+(-1)^n*(1/2-A010060(floor(n/2))). - Benoit Cloitre, Mar 23 2004
Binomial transform of A027306. Inverse binomial transform of = A032443. Hankel transform is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 20 2005
Sums of rows of the triangle in A111808. - Reinhard Zumkeller, Aug 17 2005
Number of 3-ary words of length n in which the number of 1's does not exceed the number of 0's. - David Scambler, Aug 14 2012
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Moa Apagodu and Ji-Cai Liu, Congruence properties for the trinomial coefficients, arXiv:1907.13547 [math.NT], 2019.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 24.

Cf. A025191, A027915, A081673, A092255, A055217, A005773.

a027914 n = sum $ take (n + 1) $ a027907_row n
-- Reinhard Zumkeller, Jan 22 2013

a := n -> simplify((3^n + GegenbauerC(n,-n,-1/2))/2):
seq(a(n), n=0..25); # Peter Luschny, May 12 2016

CoefficientList[ Series[ (1 + x + Sqrt[1 - 2x - 3x^2])/(2 - 4x - 6x^2), {x, 0, 26}], x] (* Robert G. Wilson v, Jul 21 2015 *)
Table[(3^n + Hypergeometric2F1[1/2 - n/2, -n/2, 1, 4])/2, {n, 0, 20}] (* Vladimir Reshetnikov, May 07 2016 *)
f[n_] := Plus @@ Take[ CoefficientList[ Sum[x^k, {k, 0, 2}]^n, x], n +1]; Array[f, 26, 0] (* Robert G. Wilson v, Jan 30 2017 *)

a(n)=sum(i=0,n,polcoeff((1+x+x^2)^n,i,x))

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,if(i+j+k-n,0,(n!/i!/j!/k!)))))

x='x+O('x^99); Vec((1+x+(1-2*x-3*x^2)^(1/2))/(2*(1-2*x-3*x^2))) \\ Altug Alkan, May 12 2016

a(n) = ( 3^n + A002426(n) )/2; lim n -> infinity a(n+1)/a(n) = 3; 3^n < 2*a(n) < 3^(n+1). - Benoit Cloitre, Sep 28 2002
From Benoit Cloitre, Jan 26 2003: (Start)
a(n) = (1/2) *( Sum{k = 0..n} binomial(n,k)*binomial(n-k,k) + 3^n );
a(n) = Sum_{k = 0..n} Sum_{i = 0..k} binomial(n,i)*binomial(n-i,k);
a(n) = 3^n/2*(1+c/sqrt(n)+O(n^-1/2)) where c=0.5... (End)
c = sqrt(3/Pi)/2 = 0.4886025119... - Vaclav Kotesovec, May 07 2016
a(n) = n!*Sum(i+j+k=n, 1/(i!*j!*k!)) 0<=i<=n, 0<=k<=j<=n. - Benoit Cloitre, Mar 23 2004
G.f.: (1+x+sqrt(1-2x-3x^2))/(2(1-2x-3x^2)); a(n) = Sum_{k = 0..n} floor((k+2)/2)*Sum_{i = 0..floor((n-k)/2)} C(n,i)*C(n-i,i+k)* ((k+1)/(i+k+1)). - Paul Barry, Sep 23 2005; corrected Jan 20 2008
D-finite with recurrence: n*a(n) +(-5*n+4)*a(n-1) +3*(n-2)*a(n-2) +9*(n-2)*a(n-3)=0. - R. J. Mathar, Dec 02 2012
G.f.: (1+x+1/G(0))/(2*(1-2*x-3*x^2)), where G(k)= 1 + x*(2+3*x)*(4*k+1)/(4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 30 2013
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 3*x - 1/(1 - x) )^n.
1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + ... is the o.g.f. for A005773. (End)
a(n) = (3^n + GegenbauerC(n,-n,-1/2))/2. - Peter Luschny, May 12 2016

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A038622 Triangular array that counts rooted polyominoes.

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A081671 Expansion of e.g.f. exp(4x) * I_0(2x).

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A111808 Left half of trinomial triangle (A027907), triangle read by rows.

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A094061 Number of n-moves paths of a king starting and ending at the origin of an infinite chessboard.

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A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

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A089627 T(n,k) = binomial(n,2k)binomial(2*k,k) for 0 <= k <= n, triangle read by rows.