A182883 -id:A182883 - cp's OEIS frontend

A105868 Triangle read by rows, T(n,k) = C(n,k)*C(k,n-k).

1, 0, 1, 0, 2, 1, 0, 0, 6, 1, 0, 0, 6, 12, 1, 0, 0, 0, 30, 20, 1, 0, 0, 0, 20, 90, 30, 1, 0, 0, 0, 0, 140, 210, 42, 1, 0, 0, 0, 0, 70, 560, 420, 56, 1, 0, 0, 0, 0, 0, 630, 1680, 756, 72, 1, 0, 0, 0, 0, 0, 252, 3150, 4200, 1260, 90, 1, 0, 0, 0, 0, 0, 0, 2772, 11550, 9240, 1980, 110, 1, 0, 0

Offset: 0

Paul Barry, Apr 23 2005

Row sums are the central trinomial coefficients A002426.
Product of A007318 and this sequence is A008459.
Coefficient array for polynomials P(n,x) = x^n*F(1/2-n/2,-n/2;1;4/x). - Paul Barry, Oct 04 2008
Column sums give A001850. It appears that the sums along the antidiagonals of the triangle produce A182883. - Peter Bala, Mar 06 2013

			Triangle begins
  1;
  0,  1;
  0,  2,  1;
  0,  0,  6,  1;
  0,  0,  6, 12,  1;
  0,  0,  0, 30, 20, 1;

G. C. Greubel, Table of n, a(n) for the first 50 rows

Cf. A063007. A001850 (column sums), A182883.

[[Binomial(n,k)*Binomial(k,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jun 14 2015

gf := 1/((1 - x*y)^2 - 4*y^2*x)^(1/2):
yser := series(gf, y, 12): ycoeff := n -> coeff(yser, y, n):
row := n -> seq(coeff(expand(ycoeff(n)), x, k), k=0..n):
seq(row(n), n=0..7); # Peter Luschny, Oct 28 2020

Flatten[Table[Binomial[n,k]Binomial[k,n-k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Nov 12 2014 *)

G.f.: 1/(sqrt((1-x*y)^2-4*x^2*y)). - Vladimir Kruchinin, Oct 28 2020

A299500 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^(n-k)binomial(n,k)hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 7, 3, 1, 5, 16, 15, 4, 1, 8, 38, 46, 26, 5, 1, 13, 82, 141, 100, 40, 6, 1, 21, 173, 381, 375, 185, 57, 7, 1, 34, 352, 983, 1216, 820, 308, 77, 8, 1, 55, 701, 2400, 3704, 3101, 1575, 476, 100, 9, 1, 89, 1368, 5646, 10536, 10885, 6804, 2758, 696, 126, 10, 1

Offset: 0

Peter Luschny, Feb 11 2018

			The partial polynomials p_{n,k}(x) start:
[0] 1
[1] x,   1
[2] x^2, 2*x+1,        1
[3] x^3, 3*x^2+4*x,    3*x+2,             1
[4] x^4, 4*x^3+9*x^2,  6*x^2+12*x+1,      4*x+3,         1
[5] x^5, 5*x^4+16*x^3, 10*x^3+36*x^2+9*x, 10*x^2+24*x+3, 5*x+4, 1
.
The polynomials P_{n}(x) start:
[0]  1
[1]  1 +    x
[2]  2 +  2*x +    x^2
[3]  3 +  7*x +  3*x^2 +    x^3
[4]  5 + 16*x + 15*x^2 +  4*x^3 +   x^4
[5]  8 + 38*x + 46*x^2 + 26*x^3 + 5*x^4 + x^5
.
The triangle starts:
[0]  1
[1]  1,   1
[2]  2,   2,    1
[3]  3,   7,    3,    1
[4]  5,  16,   15,    4,    1
[5]  8,  38,   46,   26,    5,    1
[6] 13,  82,  141,  100,   40,    6,   1
[7] 21, 173,  381,  375,  185,   57,   7,   1
[8] 34, 352,  983, 1216,  820,  308,  77,   8, 1
[9] 55, 701, 2400, 3704, 3101, 1575, 476, 100, 9, 1'
.
The square array P_{n}(k) near k=0:
......  [k=-2] 1, -1,  2, -7,  17,  -44,  125,  -345,    958,   -2707, ...
A182883 [k=-1] 1,  0,  1, -2,   1,   -6,    7,   -12,     31,     -40, ...
A000045 [k=0]  1,  1,  2,  3,   5,    8,   13,    21,     34,      55, ...
A108626 [k=1]  1,  2,  5, 14,  41,  124,  383,  1200,   3799,   12122, ...
A299501 [k=2]  1,  3, 10, 37, 145,  588, 2437, 10251,  43582,  186785, ...
......  [k=3]  1,  4, 17, 78, 377, 1886, 9655, 50220, 264223, 1402108, ...

Cf. A000045, A108625, A108626, A182883, A299499, A299501, A299502.

CoeffList := p -> op(PolynomialTools:-CoefficientList(p,x)):
PrintPoly := p -> print(sort(expand(p),x,ascending)):
T := (n,k) -> x^(n-k)*binomial(n, k)*hypergeom([-k,k-n,k-n], [1,-n], 1/x):
P := [seq(add(simplify(T(n,k)),k=0..n), n=0..10)]:
seq(CoeffList(p), p in P); # seq(PrintPoly(p), p in P);
R := proc(n,k) option remember; # Recurrence
if n < 4 then return [1,k+1,(k+1)^2+1,(k+1)^3+4*k+2][n+1] fi; ((2-n)*R(n-4,k)-
(3-2*n)*(k-1)*R(n-3,k)+(k^2+2*k-1)*(1-n)*R(n-2,k)+(2*n-1)*(k+1)*R(n-1,k))/n end:
for k from -2 to 3 do lprint(seq(R(n,k), n=0..9)) od;

Let P_{n}(x) = Sum_{k=0..n} p_{n,k}(x) then
2^n*P_{n}(1/2) = A299502(n).
P_{n}(-1) = A182883(n). P_{n}(0) = A000045(n+1).
P_{n}(1) = A108626(n).  P_{n}(2) = A299501(n).
The general case: for fixed k the sequence P_{n}(k) with n >= 0 has the generating function ogf(k, x) = (1-2*(k+1)*x + (k^2+2*k-1)*x^2 - 2*(k-1)*x^3 + x^4)^(-1/2).  The example section shows the start of this square array of sequences.
These sequences can be computed by the recurrence P(n,k) = ((2-n)*P(n-4,k)-(3-2*n)*(k-1)*P(n-3,k)+(k^2+2*k-1)*(1-n)*P(n-2,k)+(2*n-1)*(k+1)*P(n-1,k))/n with initial values 1, k+1, (k+1)^2+1 and (k+1)^3+4*k+2.
The partial polynomials p_{n,k}(x) reduce for x = 1 to A108625 (seen as a triangle).

A383571 Expansion of 1/sqrt((1-x^3)^2 - 4*x^4).

Original entry on oeis.org

1, 0, 0, 1, 2, 0, 1, 6, 6, 1, 12, 30, 21, 20, 90, 141, 100, 210, 561, 672, 672, 1681, 3206, 3528, 5125, 11622, 17892, 21253, 38172, 74052, 102565, 141680, 268092, 454741, 622182, 979836, 1790361, 2784366, 3993132, 6741593, 11587758, 17380116, 26551097, 45489082, 74098518

Offset: 0

Seiichi Manyama, Apr 30 2025

nonn

Number of lattice paths from (0,0) to (n,n) using steps (4,0),(0,4),(3,3).

Diagonal of the rational function 1 / (1 - x^4 - y^4 - x^3*y^3).

Cf. A053442, A376791.

Cf. A002426, A182883, A383572.

a(n) = sum(k=0, n\3, binomial(n-2*k, k)*binomial(k, n-3*k));

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

A383572 Expansion of 1/sqrt((1-x^4)^2 - 4*x^5).

Original entry on oeis.org

1, 0, 0, 0, 1, 2, 0, 0, 1, 6, 6, 0, 1, 12, 30, 20, 1, 20, 90, 140, 71, 30, 210, 560, 631, 294, 420, 1680, 3151, 2828, 1680, 4200, 11551, 16704, 13272, 12672, 34651, 72162, 86064, 69960, 102961, 252362, 423390, 446160, 429001, 805508, 1685970, 2393820, 2419561

Offset: 0

Seiichi Manyama, Apr 30 2025

nonn

Number of lattice paths from (0,0) to (n,n) using steps (5,0),(0,5),(4,4).

Diagonal of the rational function 1 / (1 - x^5 - y^5 - x^4*y^4).

Cf. A002426, A182883, A383571.

Cf. A376792.

a(n) = sum(k=0, n\4, binomial(n-3*k, k)*binomial(k, n-4*k));

a(n) = Sum_{k=0..floor(n/4)} binomial(n-3*k,k) * binomial(k,n-4*k).

A182882 Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,0)-steps of weight 1. L_n is the set of lattice paths of weight n that start at (0,0) , end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step with weight 1; an (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 1, 6, 3, 0, 1, 6, 3, 12, 4, 0, 1, 7, 24, 6, 20, 5, 0, 1, 12, 34, 60, 10, 30, 6, 0, 1, 31, 60, 100, 120, 15, 42, 7, 0, 1, 40, 185, 180, 230, 210, 21, 56, 8, 0, 1, 91, 260, 645, 420, 455, 336, 28, 72, 9, 0, 1, 170, 636, 980, 1715, 840, 812, 504, 36, 90, 10, 0, 1

Offset: 0

Emeric Deutsch, Dec 11 2010

Sum of entries in row n is A051286(n).
T(n,0)=A182883(n).
Sum(k*T(n,k), k=0..n)=A182884(n).

			T(3,1)=2. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them have exactly one h step.
Triangle starts:
1;
0,1;
1,0,1;
2,2,0,1;
1,6,3,0,1;
6,3,12,4,0,1

M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.

Cf. A051286, A182883, A182884.

G:=1/sqrt(1-2*t*z-2*z^2+t^2*z^2+2*t*z^3+z^4-4*z^3): Gser:=simplify(series(G,z=0,15)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 11 do seq(coeff(P[n],t,k),k=0..n) od; # yields sequence in triangular form

G.f.: G(t,z) =1/sqrt(1-2tz-2z^2+t^2*z^2+2t*z^3+z^4-4z^3).

A299499 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^kbinomial(n, k)hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 11, 16, 9, 4, 1, 26, 44, 34, 14, 5, 1, 63, 122, 111, 60, 20, 6, 1, 153, 341, 351, 225, 95, 27, 7, 1, 376, 940, 1103, 796, 400, 140, 35, 8, 1, 931, 2581, 3384, 2764, 1561, 651, 196, 44, 9, 1, 2317, 7064, 10224, 9304, 5915, 2772, 994, 264, 54, 10, 1

Offset: 0

Peter Luschny, Feb 11 2018

			The partial polynomials p_{n,k}(x) start:
[0] 1
[1] 1, x
[2] 1, 2*x+ 1,    x^2
[3] 1, 3*x+ 4,  3*x^2+ 2*x,      x^3
[4] 1, 4*x+ 9,  6*x^2+12*x+1,  4*x^3+ 3*x^2,       x^4
[5] 1, 5*x+16, 10*x^2+36*x+9, 10*x^3+24*x^2+3*x, 5*x^4+4*x^3, x^5
.
The polynomials P_{n}(x) start:
[0]   1
[1]   1 +    x
[2]   2 +  2*x +    x^2
[3]   5 +  5*x +  3*x^2 +    x^3
[4]  11 + 16*x +  9*x^2 +  4*x^3 +   x^4
[5]  26 + 44*x + 34*x^2 + 14*x^3 + 5*x^4 + x^5
.
The triangle starts:
[0]   1
[1]   1,    1
[2]   2,    2,    1
[3]   5,    5,    3,    1
[4]  11,   16,    9,    4,    1
[5]  26,   44,   34,   14,    5,   1
[6]  63,  122,  111,   60,   20,   6,   1
[7] 153,  341,  351,  225,   95,  27,   7,  1
[8] 376,  940, 1103,  796,  400, 140,  35,  8, 1
[9] 931, 2581, 3384, 2764, 1561, 651, 196, 44, 9, 1
.
The square array P_{n}(k) near k=0:
......  [k=-2] 1, -1,  2, -1,  -1,   10,  -25,    51,    -68,     41, ...
A182883 [k=-1] 1,  0,  1,  2,   1,    6,    7,    12,     31,     40, ...
A051286 [k=0]  1,  1,  2,  5,  11,   26,   63,   153,    376,    931, ...
A108626 [k=1]  1,  2,  5, 14,  41,  124,  383,  1200,   3799,  12122, ...
A299443 [k=2]  1,  3, 10, 35, 127,  474, 1807,  6999,  27436, 108541, ...
......  [k=3]  1,  4, 17, 74, 329, 1490, 6855, 31956, 150607, 716236, ...

Cf. A051286, A108625, A108626, A182883, A298611, A299443, A299500.

CoeffList := p -> op(PolynomialTools:-CoefficientList(p,x)):
PrintPoly := p -> print(sort(expand(p),x,ascending)):
T := (n,k) -> x^k*binomial(n, k)*hypergeom([-k,k-n,k-n], [1,-n], 1/x):
P := [seq(add(simplify(T(n,k)),k=0..n), n=0..10)]:
seq(CoeffList(p), p in P); seq(PrintPoly(p), p in P);
R := proc(n,k) option remember; # Recurrence
if n < 4 then return [1,k+1,(k+1)^2+1,(k+1)^3+2*k+4][n+1] fi; ((2-n)*R(n-4,k)+
(3-2*n)*(k-1)*R(n-3,k)+(k^2+2*k-1)*(1-n)*R(n-2,k)+(2*n-1)*(k+1)*R(n-1,k))/n end:
for k from -2 to 3 do lprint(seq(R(n,k), n=0..9)) od;

nmax = 10;
p[n_, k_, x_] := x^k*Binomial[n, k]*HypergeometricPFQ[{-k, k-n, k-n}, {1, -n}, 1/x];
p[n_, x_] := Sum[p[n, k, x], {k, 0, n}];
Table[CoefficientList[p[n, x], x], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Feb 26 2018 *)

Let P_{n}(x) = Sum_{k=0..n} p_{n,k}(x) then
2^n*P_{n}(1/2) = A298611(n).
P_{n}(-1) = A182883(n), P_{n}(0) = A051286(n).
P_{n}( 1) = A108626(n), P_{n}(2) = A299443(n).
The general case: for fixed k the sequence P_{n}(k) with n >= 0 has the generating function ogf(k, x) = (1-2*(k+1)*x + (k^2+2*k-1)*x^2 + 2*(k-1)*x^3 + x^4)^(-1/2).  The example section shows the start of this square array of sequences.
These sequences can be computed by the recurrence P(n,k) = ((2-n)*P(n-4,k)+(3-2*n)*(k-1)*P(n-3,k)+(k^2+2*k-1)*(1-n)*P(n-2,k)+(2*n-1)*(k+1)*P(n-1,k))/n with initial values 1, k+1, (k+1)^2+1 and (k+1)^3+2*k+4.
The partial polynomials p_{n,k}(x) reduce for x = 1 to A108625 (seen as a triangle).

cp's OEIS Frontend

A105868 Triangle read by rows, T(n,k) = C(n,k)*C(k,n-k).

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A299500 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^(n-k)*binomial(n,k)*hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.

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A383571 Expansion of 1/sqrt((1-x^3)^2 - 4*x^4).

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A383572 Expansion of 1/sqrt((1-x^4)^2 - 4*x^5).

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A299499 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^k*binomial(n, k)*hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.

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A299500 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^(n-k)binomial(n,k)hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.

A299499 Triangle read by rows, T(n,k) = [x^k] Sum_{k=0..n} p_{n,k}(x) where p_{n,k}(x) = x^kbinomial(n, k)hypergeom([-k, k-n, k-n], [1, -n], 1/x), for 0 <= k <= n.