A027908 -id:A027908 - cp's OEIS frontend

A027907 Triangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2*n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1, 1, 7, 28, 77, 161, 266, 357, 393, 357, 266, 161, 77, 28, 7, 1, 1, 8, 36, 112, 266

Offset: 0

N. J. A. Sloane

When the rows are centered about their midpoints, each term is the sum of the three terms directly above it (assuming the undefined terms in the previous row are zeros). - N. J. A. Sloane, Dec 23 2021
T(n,k) = number of integer strings s(0),...,s(n) such that s(0)=0, s(n)=k, s(i) = s(i-1) + c, where c is 0, 1 or 2. Columns of T include A002426, A005717 and A014531.
Also number of ordered trees having n+1 leaves, all at level three and n+k+3 edges. Example: T(3,5)=3 because we have three ordered trees with 4 leaves, all at level three and 11 edges: the root r has three children; from one of these children two paths of length two are hanging (i.e., 3 possibilities) while from each of the other two children one path of length two is hanging. Diagonal sums are the tribonacci numbers; more precisely: Sum_{i=0..floor(2*n/3)} T(n-i,i) = A000073(n+2). - Emeric Deutsch, Jan 03 2004
T(n,k) = A111808(n,k) for 0 <= k <= n and T(n, 2*n-k) = A111808(n,k) for 0 <= k < n. - Reinhard Zumkeller, Aug 17 2005
The trinomial coefficients, T(n,i), are the absolute value of the coefficients of the chromatic polynomial of P_2 X P_n factored with x*(x-1)^i terms. Example: The chromatic polynomial of P_2 X P_2 is: x*(x-1) - 2*x*(x-1)^2 + x*(x-1)^3 and so T(1,0)=1, T(1,1)=2 and T(1,1) = 1. - Thomas J. Pfaff (tpfaff(AT)ithaca.edu), Oct 02 2006
T(n,k) is the number of distinct ways in which k unlabeled objects can be distributed in n labeled urns allowing at most 2 objects to fall into each urn. - N-E. Fahssi, Mar 16 2008
T(n,k) is the number of compositions of k into n parts p, each part 0 <= p <= 2. Adding 1 to each part, as a corollary, T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3. E.g., T(2,3)=2 since 5 = 3+2 = 2+3. - Steffen Eger, Jun 10 2011
Number of lattice paths from (0,0) to (n,k) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Number of lattice paths from (0,0) to (2*n-k,k) using steps (2,0), (1,1), (0,2). - Werner Schulte, Jan 25 2017
T(n,k) is number of distinct ways to sum the integers -1, 0 , and 1 n times to obtain n-k, where T(n,0) = T(n,2*n+1) = 1. - William Boyles, Apr 23 2017
T(n-1,k-1) is the number of 2-compositions of n with 0's having k parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
T(n,k) is the number of ways to obtain a sum of n+k when throwing a 3-sided die n times. Follows from the "T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3" comment above. - Feryal Alayont, Dec 30 2024

			The triangle T(n, k) begins:
  n\k 0   1   2   3   4   5   6   7   8   9 10 11 12
  0:  1
  1:  1   1   1
  2:  1   2   3   2   1
  3:  1   3   6   7   6   3   1
  4:  1   4  10  16  19  16  10   4   1
  5:  1   5  15  30  45  51  45  30  15   5  1
  6:  1   6  21  50  90 126 141 126  90  50 21  6  1
Concatenated rows:
G.f. = 1 + (x^2+x+1)*x + (x^2+x+1)^2*x^4 + (x^2+x+1)^3*x^9 + ...
     = 1 + (x + x^2 + x^3) + (x^4 + 2*x^5 + 3*x^6 + 2*x^7 + x^8) +
  (x^9 + 3*x^10 + 6*x^11 + 7*x^12 + 6*x^13 + 3*x^14 + x^15) + ... .
As a centered triangle, this begins:
           1
        1  1  1
     1  2  3  2  1
  1  3  6  7  6  3  1

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Columns of T include A002426, A005717, A014531, A005581, A005712, etc. See also A035000, A008287.
First differences are in A025177. Pairwise sums are in A025564.
Cf. A007318 (Pascal's triangle); A000073, A001006, A123531, A055217, A027908, A027913, A027914, A027023.

a027907 n k = a027907_tabf !! n !! k
a027907_row n = a027907_tabf !! n
a027907_tabf = [1] : iterate f [1, 1, 1] where
   f row = zipWith3 (((+) .) . (+))
                    (row ++ [0, 0]) ([0] ++ row ++ [0]) ([0, 0] ++ row)
a027907_list = concat a027907_tabf
-- Reinhard Zumkeller, Jul 06 2014, Jan 22 2013, Apr 02 2011

A027907 := proc(n,k) expand((1+x+x^2)^n) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A027907(n,k),k=0..2*n),n=0..5) ; # R. J. Mathar, Jun 13 2011
T := (n,k) -> simplify(GegenbauerC(`if`(kPeter Luschny, May 08 2016

Table[CoefficientList[Series[(Sum[x^i, {i, 0, 2}])^n, {x, 0, 2 n}], x], {n, 0, 10}] // Grid (* Geoffrey Critzer, Mar 31 2010 *)
Table[Sum[Binomial[n, i]Binomial[n - i, k - 2i], {i, 0, n}], {n, 0, 10}, {k, 0, 2n}] (* Adi Dani, May 07 2011 *)
T[ n_, k_] := If[ n < 0, 0, Coefficient[ (1 + x + x^2)^n, x, k]]; (* Michael Somos, Nov 08 2016 *)
Flatten[DeleteCases[#,0]&/@CellularAutomaton[{Total[#] &, {}, 1}, {{1}, 0}, 8] ] (* Giorgos Kalogeropoulos, Nov 09 2021 *)

trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
create_list(trinomial(n,k),n,0,8,k,0,2*n); /* Emanuele Munarini, Mar 15 2011 */

create_list(ultraspherical(k,-n,-1/2),n,0,6,k,0,2*n); /* Emanuele Munarini, Oct 18 2016 */

{T(n, k) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, k))}; /* Michael Somos, Jun 27 2003 */

G.f.: 1/(1-z*(1+w+w^2)).
T(n,k) = Sum_{r=0..floor(k/3)} (-1)^r*binomial(n, r)*binomial(k-3*r+n-1, n-1).
Recurrence: T(0,0) = 1; T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k-0), with T(n,k) = 0 if k < 0 or k > 2*n:
T(i,0) = T(i, 2*i) = 1 for i >= 0, T(i, 1) = T(i, 2*i-1) = i for i >= 1 and for i >= 2 and 2 <= j <= i-2, T(i, j) = T(i-1, j-2) + T(i-1, j-1) + T(i-1, j).
The row sums are powers of 3 (A000244). - Gerald McGarvey, Aug 14 2004
T(n,k) = Sum_{i=0..floor(k/2)} binomial(n, 2*i+n-k) * binomial(2*i+n-k, i). - Ralf Stephan, Jan 26 2005
T(n,k) = Sum_{j=0..n} binomial(n, j) * binomial(j, k-j). - Paul Barry, May 21 2005
T(n,k) = Sum_{j=0..n} binomial(k-j, j) * binomial(n, k-j). - Paul Barry, Nov 04 2005
From Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006: (Start)
T(n,k) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(2*n-2*j, k-j); (G. E. Andrews (1990)) obtained by expanding ((1+x)^2 - x)^n.
T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(n-j, k-2*j); obtained by expanding ((1+x) + x^2)^n.
T(n,k) = (-1)^k*Sum_{j=0..n} (-3)^j * binomial(n,j) * binomial(2*n-2*j, k-j); obtained by expanding ((1-x)^2 + 3*x)^n.
T(n,k) = (1/2)^k * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k-2*j); obtained by expanding ((1+x/2)^2 + (3/4)*x^2)^n.
T(n,k) = (2^k/4^n) * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k); obtained by expanding ((1/2+x)^2 + 3/4)^n using T(n,k) = T(2*n-k). (End)
From Paul D. Hanna, Apr 18 2012: (Start)
Let A(x) be the g.f. of the flattened sequence, then:
G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x+x^2)^n.
G.f.: A(x) = Sum_{n>=0} x^n*(1+x+x^2)^n * Product_{k=1..n} (1 - (1+x+x^2) * x^(4*k-3)) / (1 - (1+x+x^2)*x^(4*k-1)).
G.f.: A(x) = 1/(1 - x*(1+x+x^2)/(1 + x*(1-x^2)*(1+x+x^2)/(1 - x^5*(1+x+x^2)/(1 + x^3*(1-x^4)*(1+x+x^2)/(1 - x^9*(1+x+x^2)/(1 + x^5*(1-x^6)*(1+x+x^2)/(1 - x^13* (1+x+x^2)/(1 + x^7*(1-x^8)*(1+x+x^2)/(1 - ...))))))))), a continued fraction.
(End)
Triangle: G.f. = Sum_{n>=0} (1+x+x^2)^n * x^(n^2) * y^n. - Daniel Forgues, Mar 16 2015
From Peter Luschny, May 08 2016: (Start)
T(n+1,n)/(n+1) = A001006(n) (Motzkin) for n>=0.
T(n,k) = H(n, k) if k < n else H(n, 2*n-k) where H(n,k) = binomial(n,k)*hypergeom([(1-k)/2, -k/2], [n-k+1], 4).
T(n,k) = GegenbauerC(m, -n, -1/2) where m=k if k < n else 2*n-k. (End)
T(n,k) = (-1)^k * C(2*n,k) * hypergeom([-k, -(2*n-k)], [-n+1/2], 3/4), for all k with 0 <= k <= 2n. - Robert S. Maier, Jun 13 2023
T(n,n) = Sum_{k=0..2*n} (-1)^k*(T(n,k))^2 and T(2*n,2*n) = Sum_{k=0..2*n} (T(n,k))^2 for n >= 0. - Werner Schulte, Nov 08 2016
T(n,n) = A002426(n), central trinomial coefficients. - M. F. Hasler, Nov 02 2019
Sum_{k=0..n-1} T(n, 2*k) = (3^n-1)/2. - Tony Foster III, Oct 06 2020

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

Original entry on oeis.org

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

A143927 G.f. satisfies: A(x) = (1 + xA(x) + x^2A(x)^2)^2.

Original entry on oeis.org

1, 2, 7, 28, 123, 572, 2769, 13806, 70414, 365636, 1926505, 10273870, 55349155, 300783420, 1646828655, 9075674700, 50304255210, 280248358964, 1568399676946, 8813424968192, 49709017472751, 281306750922072, 1596802663432503

Offset: 0

Paul D. Hanna, Sep 08 2008

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.

Cf. A006605, A027908, A143926.

Table[GegenbauerC[n,-2n-2,-1/2]/(n+1),{n,0,12}] (* Emanuele Munarini, Oct 20 2016 *)
n = 20;
A = Sum[a[k] x^k, {k, 0, n}] + x O[x]^n;
Table[a[k], {k, 0, n}] /. Reverse[Solve[LogicalExpand[(1 + x A + x^2 A^2)^2 == A]]] (* Emanuele Munarini, Oct 20 2016 *)

makelist(ultraspherical(n,-2*n-2,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */

{a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=(1+x*A+x^2*A^2)^2);polcoeff(A,n)}

Self-convolution of A006605.
Bisection of A143926.
a(n) = ((24*n+12)*A006605(n) + (3*n+5)*A006605(n+1))/(13*n+17). - Mark van Hoeij, Jul 14 2010
a(n) = (1/(n+1))*Sum_{j=0..2*n+2} (binomial(j,2*j-3*n-4)*binomial(2*n+2 ,j)). - Vladimir Kruchinin, Dec 24 2010
a(n) = GegenbauerPoly(n,-2n-2,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = T(2*n+2, n)/(n+1), where T(n,k) = A027907(n,k). - Emanuele Munarini, Oct 20 2016
The g.f. A(x) satisfies 1 + x*A'(x)/A(x) = 1 + 2*x + 10*x^2 + 50*x^3 + 266*x^3 + ..., the g.f. of A027908. - Peter Bala, Aug 03 2023

A370159 Coefficient of x^n in the expansion of ( (1+x) * (1+x+x^2)^2 )^n.

Original entry on oeis.org

1, 3, 19, 132, 963, 7228, 55264, 428067, 3347843, 26378079, 209065644, 1664967747, 13312423056, 106798422942, 859244421187, 6930167382832, 56015610380931, 453628706358333, 3679805451367471, 29895358350622638, 243204082036270588, 1980931117038586824

Offset: 0

Seiichi Manyama, Feb 11 2024

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A027908, A370160.

Cf. A369477, A370194.

a[n_]:=SeriesCoefficient[((1+x)*(1+x+x^2)^2)^n,{x,0,n}]; Array[a,22,0] (* Stefano Spezia, Apr 30 2024 *)

a(n, s=2, t=2, u=1) = sum(k=0, n\s, binomial(t*n, k)*binomial((t+u)*n-k, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k) * binomial(3*n-k,n-2*k).

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x / ((1+x) * (1+x+x^2)^2) ). See A369477.

A370160 Coefficient of x^n in the expansion of ( (1+x)^2 * (1+x+x^2)^2 )^n.

Original entry on oeis.org

1, 4, 32, 286, 2688, 26004, 256322, 2559960, 25816576, 262307824, 2681024032, 27534988936, 283926200706, 2937573629800, 30480431060160, 317053438632786, 3305105501423616, 34519689280675808, 361146528603877520, 3784045825018539968, 39702608870540290688

Offset: 0

Seiichi Manyama, Feb 11 2024

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..970

Cf. A027908, A370159.

Cf. A369478, A370195.

a[n_]:=SeriesCoefficient[((1+x)^2*(1+x+x^2)^2)^n,{x,0,n}]; Array[a,21,0] (* Stefano Spezia, Apr 30 2024 *)

a(n, s=2, t=2, u=2) = sum(k=0, n\s, binomial(t*n, k)*binomial((t+u)*n-k, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k) * binomial(4*n-k,n-2*k).

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x / ((1+x)^2 * (1+x+x^2)^2) ). See A369478.

A027913 T(n,[ n/2 ]), T given by A027907.

Original entry on oeis.org

1, 1, 2, 3, 10, 15, 50, 77, 266, 414, 1452, 2277, 8074, 12727, 45474, 71955, 258570, 410346, 1481108, 2355962, 8533660, 13599915, 49402850, 78855339, 287134346, 458917850, 1674425300, 2679183405, 9792273690, 15683407785

Offset: 0

Clark Kimberling

nonn

The median coefficient in the expansion of (1 + x + x^2)^n. - Vladimir Reshetnikov, Nov 21 2020

Robert Israel, Table of n, a(n) for n = 0..2550

Cf. A027907, A027908.

seq(simplify(GegenbauerC(floor(n/2),-n,-1/2)), n=0..100); # Robert Israel, Oct 20 2016

Table[GegenbauerC[Floor[n/2], -n, -1/2] + KroneckerDelta[n, 0], {n, 0,
100}] (* Emanuele Munarini, Oct 20 2016 *)

makelist(ultraspherical(floor(n/2),-n,-1/2),n,0,12); /* Emanuele Munarini, Oct 18 2016 */

a(n) = GegenbauerC(floor(n/2), -n, -1/2). - Emanuele Munarini, Oct 18 2016

G.f.: g(t) = (1+(t+t^2)*A(t^2)+t^4*A(t^2)^2)/(1-t^2*A(t^2)-3*t^4*A(t^2)^2), where A(t) is the g.f. of A143927 and satisfies A(t) = [1 + x*A(t) + t^2*A(t)^2]^2. - Emanuele Munarini, Oct 20 2016

A344396 a(n) = binomial(2n + 1, n)hypergeom([-(n + 1)/2, -n/2], [n + 2], 4).

Original entry on oeis.org

1, 5, 25, 133, 726, 4037, 22737, 129285, 740554, 4266830, 24701425, 143567173, 837212650, 4896136845, 28703894775, 168640510725, 992671051482, 5853000551090, 34562387229046, 204368928058958, 1209916827501876, 7170955214476509, 42543879586512435, 252638095187722437

Offset: 0

Peter Luschny, May 19 2021

nonn

Related to the Motzkin triangle A064189 counting certain lattice paths.

Cf. A027908, A064189, A026300, A344394, A327871.

alias(C=binomial):
a := n -> add(C(2*n + 1, j)*(C(2*n + 1 - j, j + n) - C(2*n + 1 - j, j + n + 2)), j = 0..2*n+1): seq(a(n), n=0..23);

a[n_] := Binomial[2 n + 1, n] Hypergeometric2F1[-(n + 1)/2, -n/2, n + 2, 4];
Table[a[n], {n, 0, 23}]

a(n) = Sum_{j=0..2*n+1} C(2*n + 1, j)*(C(2*n + 1 - j, j + n) - C(2*n + 1 - j, j + n + 2)).
a(n) = A064189(2*n+1, n).
a(n) = A026300(2*n+1, n+1).
a(n) ~ sqrt((5242 + 18674/sqrt(13))/2187) * ((70 + 26*sqrt(13))/27)^n / sqrt(Pi*n). - Vaclav Kotesovec, May 19 2021
From Peter Bala, Aug 03 2023: (Start)
P-recursive: 3*(13*n - 4)*(3*n + 2)*(3*n + 1)*(n + 1)*a(n) = 2*(2*n + 1)*(455*n^3 + 315*n^2 - 44*n - 24)*a(n-1) + 36*(13*n + 9)*(2*n + 1)*(2*n - 1)*n*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = (1/2)*A027908(n+1). (End)

A372370 Coefficient of x^n in the expansion of ( (1+x+x^2)^2 / (1+x) )^n.

Original entry on oeis.org

1, 1, 5, 13, 53, 176, 677, 2451, 9333, 34978, 133580, 508806, 1953701, 7509178, 28981643, 112046213, 434289525, 1686080622, 6557830310, 25542229740, 99622788428, 389023326600, 1520817551742, 5951305115982, 23310374278437, 91380414955176, 358506409488102

Offset: 0

Seiichi Manyama, Apr 28 2024

nonn

Cf. A027908, A370159, A370160.
Cf. A002426, A351858, A372382.
Cf. A166135.

a[n_]:=SeriesCoefficient[((1+x+x^2)^2/(1+x))^n,{x,0,n}]; Array[a,27,0] (* Stefano Spezia, Apr 30 2024 *)

a(n, s=2, t=2, u=-1) = sum(k=0, n\s, binomial(t*n, k)*binomial((t+u)*n-k, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k) * binomial(n-k,n-2*k).

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1+x) / (1+x+x^2)^2 ).

Showing 1-8 of 8 results.

cp's OEIS Frontend

A027907 Triangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2*n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

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

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A143927 G.f. satisfies: A(x) = (1 + x*A(x) + x^2*A(x)^2)^2.

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A370159 Coefficient of x^n in the expansion of ( (1+x) * (1+x+x^2)^2 )^n.

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A370160 Coefficient of x^n in the expansion of ( (1+x)^2 * (1+x+x^2)^2 )^n.

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A027913 T(n,[ n/2 ]), T given by A027907.

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A344396 a(n) = binomial(2*n + 1, n)*hypergeom([-(n + 1)/2, -n/2], [n + 2], 4).

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A143927 G.f. satisfies: A(x) = (1 + xA(x) + x^2A(x)^2)^2.

A344396 a(n) = binomial(2n + 1, n)hypergeom([-(n + 1)/2, -n/2], [n + 2], 4).