A046899 -id:A046899 - cp's OEIS frontend

A239103 Triangular array read by rows, arising from enumeration of binary words containing n 0's and k 1's that avoid the pattern 1011101.

1, 2, 1, 6, 3, 1, 20, 10, 4, 1, 70, 35, 15, 5, 1, 248, 123, 54, 20, 6, 1, 894, 442, 198, 78, 26, 7, 1, 3264, 1611, 732, 300, 108, 33, 8, 1, 12036, 5936, 2727, 1150, 437, 146, 41, 9, 1, 44722, 22047, 10214, 4398, 1736, 617, 192, 50, 10, 1

Offset: 0

N. J. A. Sloane, Mar 25 2014

			Triangle begins:
     1
     2    1
     6    3   1
    20   10   4   1
    70   35  15   5   1
   248  123  54  20   6  1
   894  442 198  78  26  7 1
  3264 1611 732 300 108 33 8 1
  ...

Alois P. Heinz, Rows n = 0..200, flattened (first 16 rows from Chai Wah Wu)
D. Baccherini, D. Merlini, R. Sprugnoli, Binary words excluding a pattern and proper Riordan arrays, Discrete Math. 307 (2007), no. 9-10, 1021--1037. MR2292531 (2008a:05003).

See A046899 for a closely related triangle. Cf. A246971.

from itertools import combinations
A239103_list = []
for n in range(16):
    for k in range(n, -1, -1):
        c, d0 = 0, ['0']*(n+k)
        for x in combinations(range(n+k), n):
            d = list(d0)
            for i in x:
                d[i] = '1'
            if not '1011101' in ''.join(d):
                c += 1
        A239103_list.append(c) # Chai Wah Wu, Sep 12 2014

More terms from Chai Wah Wu, Sep 12 2014

A328901 Triangle T(n, k) read by rows: T(n, k) is the numerator of the rational Catalan number defined as binomial(n + k, n)/(n + k) for n > 0 and T(0, 0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 2, 10, 1, 1, 5, 5, 35, 1, 1, 3, 7, 14, 126, 1, 1, 7, 28, 21, 42, 77, 1, 1, 4, 12, 30, 66, 132, 1716, 1, 1, 9, 15, 165, 99, 429, 429, 6435, 1, 1, 5, 55, 55, 143, 1001, 715, 1430, 24310, 1, 1, 11, 22, 143, 1001, 1001, 1144, 2431, 4862, 46189, 1, 1, 6, 26, 91, 273, 728, 1768, 3978, 8398, 16796, 352716

Offset: 0

Stefano Spezia, Oct 30 2019

			n\k|   0   1   2   3   4   5   6
---+----------------------------
0  |   1
1  |   1   1
2  |   1   1   3
3  |   1   1   2  10
4  |   1   1   5   5  35
5  |   1   1   3   7  14 126
6  |   1   1   7  28  21  42  77
...

Stefano Spezia, First 141 rows of the triangle, flattened
D. Armstrong, N. A. Loehr, G. S. Warrington, Rational Parking Functions and Catalan Numbers, Annals of Combinatorics (2016), Volume 20, Issue 1, pp 21-58.
M. T. L. Bizley, Derivation of a new formula for the number of minimal lattice paths from (0, 0) to (km, kn) having just t contacts with the line my = nx and having no points above this line; and a proof of Grossman's formula for the number of paths which may touch but do not rise above this line, Journal of the Institute of Actuaries, Vol. 80, No. 1 (1954): 55-62.

Main diagonal gives A201058 (for n>0).

Cf. A000108, A046899, A051162, A328902 (denominator).

Flatten[Join[{1},Table[LCM[Binomial[n+k,n],n+k]/(n+k),{n,1,11},{k,0,n}]]]

T(n, k) = lcm(binomial(n + k, n), n + k)/(n + k) for n > 0.

A328902 Triangle T(n, k) read by rows: T(n, k) is the denominator of the rational Catalan number defined as binomial(n + k, n)/(n + k) for 0 <= k <= n, n > 0; T(0, 0) = 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 1, 1, 3, 4, 1, 2, 1, 4, 5, 1, 1, 1, 1, 5, 6, 1, 2, 3, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 8, 1, 2, 1, 4, 1, 2, 1, 8, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 10, 1, 2, 1, 2, 5, 2, 1, 1, 1, 5, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 12, 1, 2, 3, 4, 1, 3, 1, 2, 3, 1, 1, 6

Offset: 0

Stefano Spezia, Oct 30 2019

			n\k| 0 1 2 3 4 5 6
---+--------------
0  | 1
1  | 1 1
2  | 2 1 2
3  | 3 1 1 3
4  | 4 1 2 1 4
5  | 5 1 1 1 1 5
6  | 6 1 2 3 1 1 1
...

Stefano Spezia, First 141 rows of the triangle, flattened
D. Armstrong, N. A. Loehr, G. S. Warrington, Rational Parking Functions and Catalan Numbers, Annals of Combinatorics (2016), Volume 20, Issue 1, pp 21-58.
M. T. L. Bizley, Derivation of a new formula for the number of minimal lattice paths from (0, 0) to (km, kn) having just t contacts with the line my = nx and having no points above this line; and a proof of Grossman's formula for the number of paths which may touch but do not rise above this line, Journal of the Institute of Actuaries, Vol. 80, No. 1 (1954): 55-62.

Cf. A000108, A028310 (1st column), A046899, A051162, A328901 (numerator).

Flatten[Join[{1},Table[(n+k)/GCD[n+k,Binomial[n+k,n]],{n,1,12},{k,0,n}]]]

A328902(n,k)=if(n,(n+k)/gcd(binomial(n+k,n),n+k),1) \\ M. F. Hasler, Nov 04 2019

T(n, k) = (n + k)/gcd(binomial(n + k, n), n + k) for n > 0.

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

Peter Luschny, Mar 21 2024

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.

			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]

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.
Cf. A046899, A092392, A371395, A244038, A371399.
Cf. A008459, A371401.

T := (n, k) -> binomial(k + n, k) * binomial(2*n - k, n):
seq(print(seq(T(n, k), k = 0..n)), n = 0..8);

T[n_, k_] := Hypergeometric2F1[-n, -k, 1, 1] Hypergeometric2F1[-n, -n +k, 1, 1];
Table[T[n, k], {n, 0, 7}, {k, 0, n}]

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).

A223540 Matrix T(m,n) = nim-product(2^m,2^n) read by rows of lower triangle.

Original entry on oeis.org

1, 2, 3, 4, 8, 6, 8, 12, 11, 13, 16, 32, 64, 128, 24, 32, 48, 128, 192, 44, 52, 64, 128, 96, 176, 75, 141, 103, 128, 192, 176, 208, 141, 198, 185, 222, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 384, 512, 768, 2048, 3072, 8192, 12288

Offset: 0

Tilman Piesk, Mar 21 2013

More compact representation of the symmetric matrix A223541.

This sequence is related to A223541, as A046899 is to A007318.

Tilman Piesk, First 128 rows of the matrix, flattened
Tilman Piesk, Walsh permutation; nimber multiplication (Wikiversity)

Cf. A223541.

A295612 a(n) = Sum_{k=0..n} binomial(n+k,k)^k.

Original entry on oeis.org

1, 3, 40, 8105, 24053106, 1016507243472, 622366942086680904, 5608321882919220905812521, 752711651805019773658037206391596, 1518219710649896586598445898967340890577318, 46343146356260529633020448755386347142785083052620084

Offset: 0

Ilya Gutkovskiy, Nov 24 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..41
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A001700, A046899, A112028, A112029, A167008, A219562, A219563, A219564.

Table[Sum[Binomial[n + k, k]^k, {k, 0, n}], {n, 0, 10}]
Table[Sum[((n + k)!/(n! k!))^k, {k, 0, n}], {n, 0, 10}]

a(n) = sum(k=0, n, binomial(n+k,k)^k); \\ Michel Marcus, Nov 25 2017

a(n) = Sum_{k=0..n} A046899(n,k)^k.

a(n) ~ 2^(2*n^2) / (exp(1/8) * Pi^(n/2) * n^(n/2)). - Vaclav Kotesovec, Nov 25 2017

A130746 Triangle read by rows: T(n,m) = binomial(n+m,1+n), 1<=m<=n.

Original entry on oeis.org

1, 1, 4, 1, 5, 15, 1, 6, 21, 56, 1, 7, 28, 84, 210, 1, 8, 36, 120, 330, 792, 1, 9, 45, 165, 495, 1287, 3003, 1, 10, 55, 220, 715, 2002, 5005, 11440, 1, 11, 66, 286, 1001, 3003, 8008, 19448, 43758, 1, 12, 78, 364, 1365, 4368, 12376, 31824, 75582, 167960

Offset: 1

Roger L. Bagula, Jul 12 2007

Row sums are A002054.

			1;
1, 4;
1, 5, 15;
1, 6, 21, 56;
1, 7, 28, 84, 210;
1, 8, 36, 120, 330, 792;
1, 9, 45, 165, 495, 1287, 3003;
1, 10, 55, 220, 715, 2002, 5005, 11440;

Cf. A178300, A046899, A059481, A002054, A000292.

Table[Table[Binomial[n + i, i + 1], {n, 1, i}], {i, 1, 10}] Flatten[%]

A171824 Triangle T(n,k)= binomial(n + k,n) + binomial(2*n-k,n) read by rows.

Original entry on oeis.org

2, 3, 3, 7, 6, 7, 21, 14, 14, 21, 71, 40, 30, 40, 71, 253, 132, 77, 77, 132, 253, 925, 469, 238, 168, 238, 469, 925, 3433, 1724, 828, 450, 450, 828, 1724, 3433, 12871, 6444, 3048, 1452, 990, 1452, 3048, 6444, 12871, 48621, 24320, 11495, 5225, 2717, 2717, 5225, 11495, 24320, 48621

Offset: 0

Roger L. Bagula, Dec 19 2009

			Triangle begins as:
       2;
       3,     3;
       7,     6,     7;
      21,    14,    14,    21;
      71,    40,    30,    40,   71;
     253,   132,    77,    77,  132,  253;
     925,   469,   238,   168,  238,  469, 925;
    3433,  1724,   828,   450,  450,  828, 1724,  3433;
   12871,  6444,  3048,  1452,  990, 1452, 3048,  6444, 12871;
   48621, 24320, 11495,  5225, 2717, 2717, 5225, 11495, 24320, 48621;
  184757, 92389, 43824, 19734, 9009, 6006, 9009, 19734, 43824, 92389, 184757;

G. C. Greubel, Table of n, a(n) for n = 0..1325

Row sums are A000984(n+1).

Cf. A001700, A007318, A054142, A085478.

T:= func< n,k | Binomial(n+k,n) + Binomial(2*n-k,n) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 29 2021

T[n_, k_] = Binomial[n+k, k] + Binomial[2*n-k, n-k];
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten

def T(n, k): return binomial(n+k,n) + binomial(2*n-k,n)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 29 2021

T(n,k) = A046899(n,k) + A092392(n,k).

Sum_{k=0..n} T(n,k) = binomial(2*n+2, n+1) = 2*A001700(n) = A000984(n+1). - G. C. Greubel, Apr 29 2021

Formula and row sums reference added by the Assoc. Editors of the OEIS, Feb 24 2010

A143131 Binomial transform of [1, 4, 10, 20, 0, 0, 0, ...].

Original entry on oeis.org

1, 5, 19, 63, 157, 321, 575, 939, 1433, 2077, 2891, 3895, 5109, 6553, 8247, 10211, 12465, 15029, 17923, 21167, 24781, 28785, 33199, 38043, 43337, 49101, 55355, 62119, 69413, 77257, 85671, 94675, 104289, 114533, 125427, 136991, 149245

Offset: 1

Gary W. Adamson, Jul 27 2008

			a(4) = 63 = (1, 3, 3, 1) dot (1, 4, 10, 20) = (1 + 12 + 30 + 20).

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A046899.

LinearRecurrence[{4,-6,4,-1},{1,5,19,63},37] (* or *) Rest[CoefficientList[Series[x*(1+x+5*x^2+13*x^3)/(1-x)^4,{x,0,37}],x]] (* or *) a[n_]:=(-39 + 77*n - 45*n^2 + 10*n^3)/3;Array[a,37] (* James C. McMahon, Aug 17 2025 *)

Binomial transform of [1, 4, 10, 20, 0, 0, 0, ...], where (1, 4, 10, 20) = row 3 of triangle A046899
a(n) = (-39 + 77*n - 45*n^2 + 10*n^3)/3. - T. D. Noe, Aug 22 2008
G.f.: x*(1+x+5*x^2+13*x^3)/(1-x)^4. - Colin Barker, Mar 23 2012

More terms from T. D. Noe, Aug 22 2008

A143132 Binomial transform of [1, 5, 15, 35, 70, 0, 0, 0, ...].

Original entry on oeis.org

1, 6, 26, 96, 321, 876, 2006, 4026, 7321, 12346, 19626, 29756, 43401, 61296, 84246, 113126, 148881, 192526, 245146, 307896, 382001, 468756, 569526, 685746, 818921, 970626, 1142506, 1336276, 1553721, 1796696, 2067126, 2367006, 2698401, 3063446

Offset: 1

Gary W. Adamson, Jul 27 2008

nonn

Conjecture: rightmost digit of terms is cyclic: (1, 6, 6, 6, ... repeat).

			a(4) = 96 = (1, 3, 3, 1) dot (1, 5, 15, 35) = (1 + 15 + 45 + 35).

Cf. A046899.

Binomial transform of [1, 5, 15, 35, 70, 0, 0, 0, ...] where (1, 5, 15, 35, 70) = row 4 of triangle A046899.
From R. J. Mathar, Jul 31 2008: (Start)
O.g.f.: (1 + x + 6x^2 + 16x^3 + 46x^4)/(1-x)^5.
a(n) = 46 - 200*n + 330*A000217(n) - 245*A000292(n) + 70*A000332(n+3). (End)
a(n) = (552 - 1190*n + 895*n^2 - 280*n^3 + 35*n^4)/12. - T. D. Noe, Aug 22 2008

More terms from T. D. Noe, Aug 22 2008

cp's OEIS Frontend

A239103 Triangular array read by rows, arising from enumeration of binary words containing n 0's and k 1's that avoid the pattern 1011101.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Python

Extensions

A328901 Triangle T(n, k) read by rows: T(n, k) is the numerator of the rational Catalan number defined as binomial(n + k, n)/(n + k) for n > 0 and T(0, 0) = 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A328902 Triangle T(n, k) read by rows: T(n, k) is the denominator of the rational Catalan number defined as binomial(n + k, n)/(n + k) for 0 <= k <= n, n > 0; T(0, 0) = 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A223540 Matrix T(m,n) = nim-product(2^m,2^n) read by rows of lower triangle.

Views

Author

Keywords

Comments

Links

Crossrefs

A295612 a(n) = Sum_{k=0..n} binomial(n+k,k)^k.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A130746 Triangle read by rows: T(n,m) = binomial(n+m,1+n), 1<=m<=n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A171824 Triangle T(n,k)= binomial(n + k,n) + binomial(2*n-k,n) read by rows.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

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