cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A158815 Triangle T(n,k) read by rows, matrix product of A046899(row-reversed) * A130595.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 13, 5, 1, 1, 46, 16, 6, 1, 1, 166, 58, 19, 7, 1, 1, 610, 211, 71, 22, 8, 1, 1, 2269, 781, 261, 85, 25, 9, 1, 1, 8518, 2920, 976, 316, 100, 28, 10, 1, 1, 32206, 11006, 3676, 1196, 376, 116, 31, 11, 1, 1
Offset: 0

Views

Author

Gary W. Adamson and Roger L. Bagula, Mar 27 2009

Keywords

Comments

The left factor of the matrix product is the triangle which starts
1;
2, 1;
6, 3, 1;
20, 10, 4, 1;
a row-reversed version of A046899, equivalent to the triangular view of the array A092392. The right factor is the inverse of the matrix A007318, which is A130595.
Swapping the two factors, A007318^(-1) * A046899(row-reversed) would generate A158793.
Riordan array (f(x), g(x)) where f(x) is the g.f. of A026641 and where g(x) is the g.f. of A000957. - Philippe Deléham, Dec 05 2009
T(n,k) is the number of nonnegative paths consisting of upsteps U=(1,1) and downsteps D=(1,-1) of length 2n with k low peaks. (A low peak has its peak vertex at height 1.) Example: T(3,1)=5 counts UDUUUU, UDUUUD, UDUUDU, UDUUDD, UUDDUD. - David Callan, Nov 21 2011
Matrix product P^2 * Q * P^(-2), where P denotes Pascal's triangle A007318 and Q denotes A061554 (formed from P by sorting the rows into descending order). Cf. A158793 and A171243. - Peter Bala, Jul 13 2021

Examples

			The triangle starts
       1;
       1,     1;
       4,     1,     1;
      13,     5,     1,    1;
      46,    16,     6,    1,    1;
     166,    58,    19,    7,    1,   1;
     610,   211,    71,   22,    8,   1,   1;
    2269,   781,   261,   85,   25,   9,   1,  1;
    8518,  2620,   976,  316,  100,  28,  10,  1,  1;
   32206, 11006,  3676, 1196,  376, 116,  31, 11,  1, 1;
  122464, 41746, 13938, 4544, 1442, 441, 133, 34, 12, 1, 1;
  ...
		

Crossrefs

Programs

  • Maple
    A158815 := proc (n, k)
      add((-1)^(j+k)*binomial(2*n-j, n)*binomial(j, k), j = 0..n);
    end proc:
    seq(seq(A158815(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021
  • Mathematica
    T[n_,k_]:= T[n,k]= Sum[(-1)^(j+k)*Binomial[j,k]*Binomial[2*n-j,n], {j,0,n}];
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2021 *)
  • Sage
    def A158815(n,k): return sum( (-1)^(j+k)*binomial(2*n-j, n)*binomial(j, k) for j in (0..n) )
    flatten([[A158815(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Dec 22 2021

Formula

Sum_{k=0..n} T(n,k) = A046899(n).
T(n,0) = A026641(n).
Sum_{k=0..n} T(n,k)*x^k = A026641(n), A000984(n), A001700(n), A000302(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Dec 03 2009
T(n, k) = Sum_{j=0..n} binomial(j, k)*binomial(2*n-j, n). - Peter Bala, Jul 13 2021

A270447 Binomial transform(2) of Catalan numbers.

Original entry on oeis.org

1, 3, 11, 43, 174, 721, 3044, 13059, 56837, 250690, 1119612, 5059561, 23119628, 106753404, 497762380, 2342096579, 11113027686, 53138757319, 255892224332, 1240217043450, 6046131132030, 29631889507380, 145923474439800, 721733515299225, 3583733352377724
Offset: 0

Views

Author

Vladimir Kruchinin, Mar 17 2016

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[2*k,k]/(k+1) * Binomial[2*n-k,n], {k,0,n}], {n,0,25}] (* Vaclav Kotesovec, Mar 17 2016 *)
    a[n_] := ((2 n + 1) Binomial[2 n, n] (1 - Hypergeometric2F1[-1/2, -n - 1, -2 n - 1, 4]))/(2 (n + 1));
    Table[a[n], {n, 0, 24}] (* Peter Luschny, May 30 2022 *)
  • Maxima
    a(n):=sum((binomial(2*k,k)*binomial(2*n-k,n))/(k+1),k,0,n);
    
  • PARI
    a(n) = sum(i=0, n, (binomial(2*i, i)*binomial(2*n-i, n))/(i+1)); \\ Altug Alkan, Mar 17 2016

Formula

a(n) = Sum_{k=0..n} (T(n,k)*C(k)), where C(k) is Catalan numbers (A000108), T(n,k) - triangle of A092392.
a(n) = Sum_{k=0..n} ((binomial(2*k,k)/(k+1)*binomial(2*n-k,n))).
G.f.: C(C(x))*(1-C(x))^2/(((1-C(x))^2)-x)/x, where C(x)=(1-sqrt(1-4*x))/2.
Recurrence: 3*(n-1)*n*(n+1)*(2*n - 3)*a(n) = 16*(n-1)*n*(5*n^2 - 10*n + 3)*a(n-1) - 16*(n-1)*(2*n - 1)*(11*n^2 - 33*n + 24)*a(n-2) + 8*(2*n - 3)*(2*n - 1)*(4*n - 9)*(4*n - 7)*a(n-3). - Vaclav Kotesovec, Mar 17 2016
a(n) ~ 2^(4*n + 1/2) / (sqrt(Pi) * 3^(n - 1/2) * n^(3/2)). - Vaclav Kotesovec, Mar 17 2016
a(n) = [x^n] (1 - sqrt(1 - 4*x))/(2*x*(1 - x)^(n+1)). - Ilya Gutkovskiy, Nov 01 2017

A274404 Number T(n,k) of modified skew Dyck paths of semilength n with exactly k anti-down steps; triangle T(n,k), n>=0, 0<=k<=n-floor((1+sqrt(max(0,8n-7)))/2), read by rows.

Original entry on oeis.org

1, 1, 2, 5, 1, 14, 6, 42, 28, 3, 132, 120, 28, 1, 429, 495, 180, 20, 1430, 2002, 990, 195, 10, 4862, 8008, 5005, 1430, 165, 4, 16796, 31824, 24024, 9009, 1650, 117, 1, 58786, 125970, 111384, 51688, 13013, 1617, 70, 208012, 497420, 503880, 278460, 89180, 16016, 1386, 35
Offset: 0

Views

Author

Alois P. Heinz, Jun 20 2016

Keywords

Comments

A modified skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1) (up), D=(1,-1) (down) and A=(-1,1) (anti-down) so that A and D steps do not overlap.

Examples

			              /\
              \ \
T(3,1) = 1:   /  \
.
Triangle T(n,k) begins:
:     1;
:     1;
:     2;
:     5,     1;
:    14,     6;
:    42,    28,     3;
:   132,   120,    28,    1;
:   429,   495,   180,   20;
:  1430,  2002,   990,  195,   10;
:  4862,  8008,  5005, 1430,  165,   4;
: 16796, 31824, 24024, 9009, 1650, 117, 1;
		

Crossrefs

Columns k=0-3 give: A000108, A002694(n-1), A074922(n-2), A232224(n-3).
Row sums give A230823.
Last elements of rows give A092392(n-1) for n>0.

Programs

  • Maple
    b:= proc(x, y, t, n) option remember; expand(`if`(y>n, 0,
          `if`(n=y, `if`(t=2, 0, 1), b(x+1, y+1, 0, n-1)+
          `if`(t<>1 and x>0, b(x-1, y+1, 2, n-1)*z, 0)+
          `if`(t<>2 and y>0, b(x+1, y-1, 1, n-1), 0))))
        end:
    T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(0$3, 2*n)):
    seq(T(n), n=0..14);
  • Mathematica
    b[x_, y_, t_, n_] := b[x, y, t, n] = Expand[If[y > n, 0,
         If[n == y, If[t == 2, 0, 1], b[x + 1, y + 1, 0, n - 1] +
         If[t != 1 && x > 0, b[x - 1, y + 1, 2, n - 1] z, 0] +
         If[t != 2 && y > 0, b[x + 1, y - 1, 1, n - 1], 0]]]];
    T[n_] := CoefficientList[b[0, 0, 0, 2n], z];
    T /@ Range[0, 14] // Flatten (* Jean-François Alcover, Mar 27 2021, after Alois P. Heinz *)

Formula

Sum_{k>0} k * T(n,k) = A274405(n).

A119304 Triangle read by rows: T(n,k) = binomial(4n-k,n-k), 0 <= k <= n.

Original entry on oeis.org

1, 4, 1, 28, 7, 1, 220, 55, 10, 1, 1820, 455, 91, 13, 1, 15504, 3876, 816, 136, 16, 1, 134596, 33649, 7315, 1330, 190, 19, 1, 1184040, 296010, 65780, 12650, 2024, 253, 22, 1, 10518300, 2629575, 593775, 118755, 20475, 2925, 325, 25, 1, 94143280, 23535820
Offset: 0

Views

Author

Paul Barry, May 13 2006

Keywords

Examples

			Triangle begins
       1;
       4,     1;
      28,     7,    1;
     220,    55,   10,    1;
    1820,   455,   91,   13,   1;
   15504,  3876,  816,  136,  16,  1;
  134596, 33649, 7315, 1330, 190, 19, 1;
		

Crossrefs

Rows sums are A052203. First column is A005810. Inverse of A119305.

Programs

  • Mathematica
    Flatten[Table[Binomial[4n-k,n-k],{n,0,9},{k,0,n}]] (* Indranil Ghosh, Feb 26 2017 *)
  • PARI
    tabl(nn) = {for (n=0,nn,for (k=0,n,print1(binomial(4*n-k,n-k),", ");); print(););} \\ Indranil Ghosh, Feb 26 2017
    
  • Python
    from sympy import binomial
    i=0
    for n in range(12):
        for k in range(n+1):
            print(str(i)+" "+str(binomial(4*n-k,n-k)))
            i+=1 # Indranil Ghosh, Feb 26 2017

Formula

Riordan array (1/(1-4f(x)),f(x)) where f(x)(1-f(x))^3 = x.
From Peter Bala, Jun 04 2024: (Start)
'Horizontal' recurrence equation: T(n, 0) = binomial(4*n,n) and for k >= 1, T(n, k) = Sum_{i = 1..n+1-k} i*(i+1)/2 * T(n-1, k-2+i).
T(n, k) = Sum_{j = 0..n} binomial(n+j-1, j)*binomial(3*n-k-j, 2*n). (End)

A159965 Riordan array (1/sqrt(1-4x), (1-2x-(1-3x)c(x))/(x*sqrt(1-4x))), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 20, 21, 8, 1, 70, 84, 45, 11, 1, 252, 330, 220, 78, 14, 1, 924, 1287, 1001, 455, 120, 17, 1, 3432, 5005, 4368, 2380, 816, 171, 20, 1, 12870, 19448, 18564, 11628, 4845, 1330, 231, 23, 1, 48620, 75582, 77520, 54264, 26334, 8855, 2024, 300, 26, 1
Offset: 0

Views

Author

Paul Barry, Apr 28 2009

Keywords

Comments

Product of A007318 and A114422. Product of A007318^2 and A116382. Row sums are A108080.
Diagonal sums are A108081.
Riordan array (1/sqrt(1 - 4*x), x*c(x)^3) obtained from A092392 by taking every third column starting from column 0; x*c(x)^3 is the o.g.f. for A000245. - Peter Bala, Nov 24 2015

Examples

			Triangle begins
1,
2, 1,
6, 5, 1,
20, 21, 8, 1,
70, 84, 45, 11, 1,
252, 330, 220, 78, 14, 1,
924, 1287, 1001, 455, 120, 17, 1,
3432, 5005, 4368, 2380, 816, 171, 20, 1
		

Crossrefs

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n+k, n+2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 27 2015

Formula

Number triangle T(n,k) = Sum_{j = 0..n} binomial(n+k,j-k)*binomialC(n,j).
T(n,k) = binomial(2*n + k, n + 2*k). - Peter Bala, Nov 24 2015

A264773 Triangle T(n,k) = binomial(4*n - 3*k, 3*n - 2*k), 0 <= k <= n.

Original entry on oeis.org

1, 4, 1, 28, 5, 1, 220, 36, 6, 1, 1820, 286, 45, 7, 1, 15504, 2380, 364, 55, 8, 1, 134596, 20349, 3060, 455, 66, 9, 1, 1184040, 177100, 26334, 3876, 560, 78, 10, 1, 10518300, 1560780, 230230, 33649, 4845, 680, 91, 11, 1, 94143280, 13884156, 2035800, 296010, 42504, 5985, 816, 105, 12, 1
Offset: 0

Views

Author

Peter Bala, Nov 30 2015

Keywords

Comments

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. for A002293 and f(x) = g(x)/(4 - 3*g(x)) = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + ... is the o.g.f. for A005810.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 3 and b = 2. See A092392, A264772, A264774 and A113139 for further examples.

Examples

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+-----------------------------------------------
   0  |       1
   1  |       4      1
   2  |      28      5     1
   3  |     220     36     6    1
   4  |    1820    286    45    7   1
   5  |   15504   2380   364   55   8   1
   6  |  134596  20349  3060  455  66   9   1
   7  | 1184040 177100 26334 3876 560  78  10   1
...
		

Crossrefs

A005810 (column 0), A052203 (column 1), A257633 (column 2), A224274 (column 3), A004331 (column 4). Cf. A002293, A007318, A092392 (C(2n-k,n)), A119301 (C(3n-k,n-k)), A264772, A264774.

Programs

  • Magma
    /* As triangle: */ [[Binomial(4*n-3*k, 3*n-2*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
  • Maple
    A264773:= proc(n,k) binomial(4*n - 3*k, 3*n - 2*k); end proc:
    seq(seq(A264773(n,k), k = 0..n), n = 0..10);
  • Mathematica
    A264773[n_,k_] := Binomial[4*n - 3*k, n - k];
    Table[A264773[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Feb 06 2024 *)

Formula

T(n,k) = binomial(4*n - 3*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(4*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(3*n + 1)*binomial(4*n,n)*x^n.

A264774 Triangle T(n,k) = binomial(5*n - 4*k, 4*n - 3*k), 0 <= k <= n.

Original entry on oeis.org

1, 5, 1, 45, 6, 1, 455, 55, 7, 1, 4845, 560, 66, 8, 1, 53130, 5985, 680, 78, 9, 1, 593775, 65780, 7315, 816, 91, 10, 1, 6724520, 736281, 80730, 8855, 969, 105, 11, 1, 76904685, 8347680, 906192, 98280, 10626, 1140, 120, 12, 1, 886163135, 95548245, 10295472, 1107568, 118755, 12650, 1330, 136, 13, 1
Offset: 0

Views

Author

Peter Bala, Nov 30 2015

Keywords

Comments

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + ... is the o.g.f. for A002294 and f(x) = g(x)/(5 - 4*g(x)) = 1 + 5*x + 45*x^2 + 455*x^3 + 4845*x^4 + ... is the o.g.f. for A001449.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 4 and b = 3. See A092392, A264772, A264773 and A113139 for further examples.

Examples

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+---------------------------------------------
   0  |       1
   1  |       5      1
   2  |      45      6     1
   3  |     455     55     7    1
   4  |    4845    560    66    8   1
   5  |   53130   5985   680   78   9   1
   6  |  593775  65780  7315  816  91  10   1
   7  | 6724520 736281 80730 8855 969 105  11  1
...
		

Crossrefs

Cf. A001449 (column 0), A079589(column 1). Cf. A002294, A007318, A092392 (C(2n-k,n)), A113139, A119301 (C(3n-k,n-k)), A264772, A264773.

Programs

  • Magma
    /* As triangle */ [[Binomial(5*n-4*k, 4*n-3*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
  • Maple
    A264774:= proc(n,k) binomial(5*n - 4*k, 4*n - 3*k); end proc:
    seq(seq(A264774(n,k), k = 0..n), n = 0..10);
  • Mathematica
    Table[Binomial[5 n - 4 k, 4 n - 3 k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2015 *)

Formula

T(n,k) = binomial(5*n - 4*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(5*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(4*n + 1)*binomial(5*n,n)*x^n.

A293468 a(n) = Sum_{k=0..n} k!*binomial(2*n-k, n).

Original entry on oeis.org

1, 3, 11, 44, 189, 880, 4542, 26712, 182793, 1461368, 13477650, 140564536, 1627370146, 20621925504, 283161372284, 4182215376240, 66065933347425, 1111053154779720, 19814069772086730, 373435157945506680, 7415765258637418950, 154751460071567005920, 3385387828167428482020
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 09 2017

Keywords

Crossrefs

Programs

  • Maple
    seq(simplify( GAMMA(n+1/2)*4^n*hypergeom([1,1,-n],[-2*n],1)/(sqrt(Pi)*n!)),n=0..30); # Robert Israel, Oct 09 2017
  • Mathematica
    Table[Sum[k! Binomial[2 n - k, n], {k, 0, n}], {n, 0, 22}]
    Table[Sum[Gamma[k + 1] Gamma[2 n - k  + 1]/(Gamma[n + 1] Gamma[n - k + 1]), {k, 0, n}], {n, 0, 22}]
    Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 + ContinuedFractionK[-Floor[(k + 1)/2] x, 1, {k, 1, n}]), {x, 0, n}], {n, 0, 22}]
    Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) Sum[k! x^k, {k, 0, n}], {x, 0, n}], {n, 0, 22}]
    A293468[n_] := DifferenceRoot[Function[{a,k}, {(k+1)(k-n)a[k] + (k(n-2)-k^2+3n)
    a[k+1] + (k-2n) a[k+2] == 0, a[0] == 0, a[1] == Binomial[2n, n]}]][1+n];
    Table[A293468[n], {n, 0, 22}] (* Peter Luschny, Oct 09 2017 *)

Formula

a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - x/(1 - x/(1 - 2*x/(1 - 2*x/(1 - 3*x/(1 - 3*x/(1 - ...)))))))), a continued fraction.
a(n) = Gamma(n+1/2)*4^n*hypergeom([1,1,-n],[-2n],1)/(sqrt(Pi)*n!). - Robert Israel, Oct 09 2017
a(n) ~ exp(1) * n!. - Vaclav Kotesovec, Oct 18 2017

A371400 Triangle read by rows: T(n, k) = binomial(k + n, k)*binomial(2*n - k, n).

Original entry on oeis.org

1, 2, 2, 6, 9, 6, 20, 40, 40, 20, 70, 175, 225, 175, 70, 252, 756, 1176, 1176, 756, 252, 924, 3234, 5880, 7056, 5880, 3234, 924, 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432, 12870, 57915, 135135, 212355, 245025, 212355, 135135, 57915, 12870
Offset: 0

Views

Author

Peter Luschny, Mar 21 2024

Keywords

Comments

The main diagonal and column 0 of the triangle are the central binomial coefficients, which are the sums of the squares of Pascal's triangle entries. This sum representation can be generalized, and all terms can be seen as sums of coefficients of some polynomials. (See the Example section.)
To see this, consider T(n, k) as the value of the polynomials P(n, k)(x) at x = 1, where P(n, k)(x) = H([-n, -k], [1], x)*H([-n, -n + k], [1], x) and H denotes the hypergeometric sum 2F1. For instance column 0 is given by the row sums of A008459, and column 1 by the row sums of A371401.

Examples

			Triangle starts:
[0]    1;
[1]    2,     2;
[2]    6,     9,     6;
[3]   20,    40,    40,    20;
[4]   70,   175,   225,   175,    70;
[5]  252,   756,  1176,  1176,   756,   252;
[6]  924,  3234,  5880,  7056,  5880,  3234,   924;
[7] 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432;
.
Because of the symmetry, only the sum representation of terms with k <= n/2 are shown.
0:                 [1]
1:               [1+1]
2:             [1+4+1],               [1+4+4]
3:           [1+9+9+1],            [1+9+21+9]
4:      [1+16+36+16+1],       [1+16+66+76+16],        [1+16+76+96+36]
5: [1+25+100+100+25+1], [1+25+160+340+205+25], [1+25+190+460+400+100]
		

Crossrefs

Column 0 and main diagonal are A000984.
Column 1 and subdiagonal are A097070.
Row sums are A045721.
The even bisection of the alternating row sums is A005809.
The central terms are A188662.

Programs

  • Maple
    T := (n, k) -> binomial(k + n, k) * binomial(2*n - k, n):
    seq(print(seq(T(n, k), k = 0..n)), n = 0..8);
  • Mathematica
    T[n_, k_] := Hypergeometric2F1[-n, -k, 1, 1] Hypergeometric2F1[-n, -n +k, 1, 1];
    Table[T[n, k], {n, 0, 7}, {k, 0, n}]

Formula

T(n, k) = A046899(n, k) * A092392(n, k).
T(n, k) = A046899(n, k) * A046899(n, n - k).
T(n, k) = A092392(n, k) * A092392(n, n - k).
T(n, k) = A371395(n, k) * (n + 1).
T(n, k) = hypergeom([-n, -k], [1], 1) * hypergeom([-n, -n + k], [1], 1).
2^n*Sum_{k=0..n} T(n, k)*(1/2)^k = A244038(n).
2^n*Sum_{k=0..n} T(n, k)*(-1/2)^k = A371399(n).

A176992 Triangle T(n,m) = binomial(2n-k+1, n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 10, 4, 1, 35, 15, 5, 1, 126, 56, 21, 6, 1, 462, 210, 84, 28, 7, 1, 1716, 792, 330, 120, 36, 8, 1, 6435, 3003, 1287, 495, 165, 45, 9, 1, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1, 352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1
Offset: 0

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Author

Roger L. Bagula, Dec 08 2010

Keywords

Comments

Row sums are A001791.
Obtained from A059481 by removal of the last two terms in each row, followed by row reversal.
Riordan array (c(x)/sqrt(1 - 4*x), x*c(x)) where c(x) is the g.f. of A000108. - Philippe Deléham, Jul 12 2015

Examples

			Triangle begins:
       1;
       3,      1;
      10,      4,     1;
      35,     15,     5,     1;
     126,     56,    21,     6,    1;
     462,    210,    84,    28,    7,     1;
    1716,    792,   330,   120,   36,     8,    1;
    6435,   3003,  1287,   495,   165,   45,    9,   1;
   24310,  11440,  5005,  2002,   715,  220,   55,  10,  1;
   92378,  43758, 19448,  8008,  3003, 1001,  286,  66, 11,  1;
  352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1;
		

Crossrefs

Cf. Similar triangle: A033184, A054445.
Cf. A178300 (reversal).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n-k+1,n+1): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 12 2015
  • Maple
    A176992 := proc(n,k) binomial(1+2*n-k,n+1) ; end proc: # R. J. Mathar, Dec 09 2010
  • Mathematica
    p[t_, j_] = ((-1)^(j + 1)/2)*Sum[Binomial[k - j - 1, j + 1]*t^k, {k, 0, Infinity}];
    Flatten[Table[CoefficientList[ExpandAll[p[t, j]], t], {j, 0, 10}]]

Formula

n-th row of the triangle = top row of M^n, where M is the following infinite square production matrix:
3, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
... - Philippe Deléham, Jul 12 2015
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