A081577 -id:A081577 - cp's OEIS frontend

A098593 A triangle of Krawtchouk coefficients.

1, 1, 1, 1, 0, 1, 1, -1, -1, 1, 1, -2, -2, -2, 1, 1, -3, -2, -2, -3, 1, 1, -4, -1, 0, -1, -4, 1, 1, -5, 1, 3, 3, 1, -5, 1, 1, -6, 4, 6, 6, 6, 4, -6, 1, 1, -7, 8, 8, 6, 6, 8, 8, -7, 1, 1, -8, 13, 8, 2, 0, 2, 8, 13, -8, 1, 1, -9, 19, 5, -6, -10, -10, -6, 5, 19, -9, 1, 1, -10, 26, -2, -17, -20, -20, -20, -17, -2, 26, -10, 1, 1, -11, 34, -14, -29, -25

Offset: 0

Paul Barry, Sep 17 2004

Row sums are A009545(n+1), with e.g.f. exp(x)(cos(x)+sin(x)). Diagonal sums are A077948.
The rows are the diagonals of the Krawtchouk matrices. Coincides with the Riordan array (1/(1-x),(1-2x)/(1-x)). - Paul Barry, Sep 24 2004
Corresponds to Pascal-(1,-2,1) array, read by antidiagonals. The Pascal-(1,-2,1) array has n-th row generated by (1-2x)^n/(1-x)^(n+1). - Paul Barry, Sep 24 2004
A modified version (different signs) of this triangle is given by T(n,k) = Sum_{j=0..n} C(n-k,j)*C(k,j)*cos(Pi*(k-j)). - Paul Barry, Jun 14 2007

			Rows begin {1}, {1,1}, {1,0,1}, {1,-1,-1,1}, {1,-2,-2,-2,1}, ...
From _Paul Barry_, Oct 05 2010: (Start)
Triangle begins
  1,
  1,  1,
  1,  0,  1,
  1, -1, -1,  1,
  1, -2, -2, -2,  1,
  1, -3, -2, -2, -3,  1,
  1, -4, -1,  0, -1, -4,  1,
  1, -5,  1,  3,  3,  1, -5,  1,
  1, -6,  4,  6,  6,  6,  4, -6,  1,
  1, -7,  8,  8,  6,  6,  8,  8, -7,  1,
  1, -8, 13,  8,  2,  0,  2,  8, 13, -8,  1
Production matrix (related to large Schroeder numbers A006318) begins
  1,     1,
  0,    -1,     1,
  0,    -2,    -1,    1,
  0,    -6,    -2,   -1,   1,
  0,   -22,    -6,   -2,  -1,   1,
  0,   -90,   -22,   -6,  -2,  -1,  1,
  0,  -394,   -90,  -22,  -6,  -2, -1,  1,
  0, -1806,  -394,  -90, -22,  -6, -2, -1,  1,
  0, -8558, -1806, -394, -90, -22, -6, -2, -1, 1
Production matrix of inverse is
    -1,   1,
    -2,   1,  1,
    -4,   2,  1,  1,
    -8,   4,  2,  1,  1,
   -16,   8,  4,  2,  1, 1,
   -32,  16,  8,  4,  2, 1, 1,
   -64,  32, 16,  8,  4, 2, 1, 1,
  -128,  64, 32, 16,  8, 4, 2, 1, 1,
  -256, 128, 64, 32, 16, 8, 4, 2, 1, 1 (End)

P. Feinsilver and J. Kocik, Krawtchouk matrices from classical and quantum walks, Contemporary Mathematics, 287 2001, pp. 83-96.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Paul Barry, A note on Krawtchouk Polynomials and Riordan Arrays, JIS 11 (2008) 08.2.2
P. Feinsilver and J. Kocik, Krawtchouk polynomials and Krawtchouk matrices, arxiv:quant-ph/0702073, 2007.

Cf. Pascal (1,m,1) array: A123562 (m = -3), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

T[n_, k_] := Sum[Binomial[n - k, k - j]*Binomial[k, j]*(-1)^(k - j), {j, 0, n}]; Table[T[n, k], {n, 0, 49}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 15 2017 *)

for(n=0,10, for(k=0,n, print1(sum(i=0,k, binomial(n-k, k-i) *binomial(k, i)*(-1)^(k-i)), ", "))) \\ G. C. Greubel, Oct 15 2017

T(n, k) = Sum_{i=0..k} binomial(n-k, k-i)*binomial(k, i)*(-1)^(k-i), k<=n.
T(n, k) = T(n-1, k) + T(n-1, k-1) - 2*T(n-2, k-1) (n>0). - Paul Barry, Sep 24 2004
T(n, k) = [k<=n]*Hypergeometric2F1(-k,k-n;1;-1). - Paul Barry, Jan 24 2011
E.g.f. for the n-th subdiagonal: exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} (-1)^k*binomial(n,k)* x^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 - 2*x + x^2/2) = 1 - x - 2*x^2/2! - 2*x^3/3! - x^4/4! + x^5/5! + .... - Peter Bala, Mar 05 2017

A081578 Pascal-(1,3,1) array.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 9, 9, 1, 1, 13, 33, 13, 1, 1, 17, 73, 73, 17, 1, 1, 21, 129, 245, 129, 21, 1, 1, 25, 201, 593, 593, 201, 25, 1, 1, 29, 289, 1181, 1921, 1181, 289, 29, 1, 1, 33, 393, 2073, 4881, 4881, 2073, 393, 33, 1, 1, 37, 513, 3333, 10497, 15525, 10497, 3333, 513, 37, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016813, A081585, A081586. Coefficients of the row polynomials in the Newton basis are given by A013611.

As a number triangle, this is the Riordan array (1/(1-x), x*(1+3*x)/(1-x)). It has row sums A015518(n+1) and diagonal sums A103143. - Paul Barry, Jan 24 2005

			Square array begins as:
  1,  1,   1,   1,    1, ... A000012;
  1,  5,   9,  13,   17, ... A016813;
  1,  9,  33,  73,  129, ... A081585;
  1, 13,  73, 245,  593, ... A081586;
  1, 17, 129, 593, 1921, ...
As a triangle this begins:
  1;
  1,  1;
  1,  5,   1;
  1,  9,   9,    1;
  1, 13,  33,   13,     1;
  1, 17,  73,   73,    17,     1;
  1, 21, 129,  245,   129,    21,     1;
  1, 25, 201,  593,   593,   201,    25,    1;
  1, 29, 289, 1181,  1921,  1181,   289,   29,   1;
  1, 33, 393, 2073,  4881,  4881,  2073,  393,  33,  1;
  1, 37, 513, 3333, 10497, 15525, 10497, 3333, 513, 37, 1; - _Philippe Deléham_, Mar 15 2014

Vincenzo Librandi, Rows n = 0..100, flattened
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
P. Barry, A note on Krawtchouk Polynomials and Riordan Arrays, JIS 11 (2008) 08.2.2

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

a081578 n k = a081578_tabl !! n !! k
a081578_row n = a081578_tabl !! n
a081578_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Mar 16 2014

A081578:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081578(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[Hypergeometric2F1[-k, k-n, 1, 4], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 4).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 3*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+3*x)^k/(1-x)^(k+1).
T(n,k) = Sum_{j=0..n} binomial(k,j-k)*binomial(n+k-j,k)*3^(j-k). - Paul Barry, Oct 23 2006
E.g.f. for the n-th subdiagonal of the triangle, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(4*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 8*x + 16*x^2/2) = 1 + 9*x + 33*x^2/2! + 73*x^3/3! + 129*x^4/4! + 201*x^5/5! + .... - Peter Bala, Mar 05 2017
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Hypergeometric2F1([-k, k-n], [1], m+1), for m = 3.
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 3.
Sum_{k=0..n} T(n, k, 3) = A015518(n+1). (End)

A081582 Pascal-(1,7,1) array.

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 17, 17, 1, 1, 25, 97, 25, 1, 1, 33, 241, 241, 33, 1, 1, 41, 449, 1161, 449, 41, 1, 1, 49, 721, 3297, 3297, 721, 49, 1, 1, 57, 1057, 7161, 14721, 7161, 1057, 57, 1, 1, 65, 1457, 13265, 44961, 44961, 13265, 1457, 65, 1, 1, 73, 1921, 22121, 108353, 192969, 108353, 22121, 1921, 73, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A017077, A081593, A081594. Coefficients of the row polynomials in the Newton basis are given by A013614.

			Rows begin
  1,  1,   1,    1,     1, ... A000012;
  1,  9,  17,   25,    33, ... A017077;
  1, 17,  97,  241,   449, ... A081593;
  1, 25, 241, 1161,  3297, ...
  1, 33, 449, 3297, 14721, ...
Triangle begins:
  1;
  1,  1;
  1,  9,    1;
  1, 17,   17,    1;
  1, 25,   97,   25,     1;
  1, 33,  241,  241,    33,    1;
  1, 41,  449, 1161,   449,   41,    1;
  1, 49,  721, 3297,  3297,  721,   49,  1;
  1, 57, 1057, 7161, 14721, 7161, 1057, 57, 1;

Vincenzo Librandi, Rows n = 0..100, flattened

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A143683 (m = 8).

Cf. A015519, A017077, A081593.

A081582:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081582(n,k,7): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[ Hypergeometric2F1[-k, k-n, 1, 8], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 8).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

T(n,k) = Sum_{j = 0..n-k} binomial(n-k,j)*binomial(k,j)*8^j.
Riordan array (1/(1 - x), x*(1 + 7*x)/(1 - x)).
Square array T(n, k) defined by T(n, 0) = T(0, k)=1, T(n, k) = T(n, k-1) + 7*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1 + 7*x)^k/(1 - x)^(k+1).
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 8). - Jean-François Alcover, May 24 2013
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(8*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 16*x + 64*x^2/2) = 1 + 17*x + 97*x^2/2! + 241*x^3/3! + 449*x^4/4! + 721*x^5/5! + .... - Peter Bala, Mar 05 2017
Sum_{k=0..n} T(n, k) = A015519(n+1). - G. C. Greubel, May 26 2021

A081579 Pascal-(1,4,1) array.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 11, 11, 1, 1, 16, 46, 16, 1, 1, 21, 106, 106, 21, 1, 1, 26, 191, 396, 191, 26, 1, 1, 31, 301, 1011, 1011, 301, 31, 1, 1, 36, 436, 2076, 3606, 2076, 436, 36, 1, 1, 41, 596, 3716, 9726, 9726, 3716, 596, 41, 1, 1, 46, 781, 6056, 21746, 33876, 21746, 6056, 781, 46, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016861, A081587, A081588. Coefficients of the row polynomials in the Newton basis are given by A013612.

			Square array begins as:
  1,  1,   1,    1,    1, ... A000012;
  1,  6,  11,   16,   21, ... A016861;
  1, 11,  46,  106,  191, ... A081587;
  1, 16, 106,  396, 1011, ... A081588;
  1, 21, 191, 1011, 3606, ...
As triangle this begins:
  1;
  1,  1;
  1,  6,   1;
  1, 11,  11,    1;
  1, 16,  46,   16,     1;
  1, 21, 106,  106,    21,     1;
  1, 26, 191,  396,   191,    26,     1;
  1, 31, 301, 1011,  1011,   301,    31,    1;
  1, 36, 436, 2076,  3606,  2076,   436,   36,   1;
  1, 41, 596, 3716,  9726,  9726,  3716,  596,  41,  1;
  1, 46, 781, 6056, 21746, 33876, 21746, 6056, 781, 46, 1; - _Philippe Deléham_, Mar 15 2014

Vincenzo Librandi, Rows n = 0..100, flattened
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

Cf. A016861, A063727, A081587, A081588.

A081579:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081579(n,k,4): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[Hypergeometric2F1[-k, k-n, 1, 5], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 5).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 4*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+4*x)^k/(1-x)^(k+1).
From Philippe Deléham, Mar 15 2014: (Start)
Riordan array (1/(1-x), x*(1+4*x)/(1-x)).
Sum_{k=0..n} T(n, k) = A063727(n). (End)
E.g.f. for the n-th subdiagonal of the triangle, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(5*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 10*x + 25*x^2/2) = 1 + 11*x + 46*x^2/2! + 106*x^3/3! + 191*x^4/4! + 301*x^5/5! + .... - Peter Bala, Mar 05 2017
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Hypergeometric2F1([-k, k-n], [1], m+1), for m = 4.
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 4. (End)

A081580 Pascal-(1,5,1) array.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 61, 19, 1, 1, 25, 145, 145, 25, 1, 1, 31, 265, 595, 265, 31, 1, 1, 37, 421, 1585, 1585, 421, 37, 1, 1, 43, 613, 3331, 6145, 3331, 613, 43, 1, 1, 49, 841, 6049, 17401, 17401, 6049, 841, 49, 1, 1, 55, 1105, 9955, 40105, 65527, 40105, 9955, 1105, 55, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016921, A081589, A081590. Coefficients of the row polynomials in the Newton basis are given by A013613.

			Square array begins as:
  1,  1,   1,    1,    1, ... A000012;
  1,  7,  13,   19,   25, ... A016921;
  1, 13,  61,  145,  265, ... A081589;
  1, 19, 145,  595, 1585, ... A081590;
  1, 25, 265, 1585, 6145, ...
The triangle begins as:
  1;
  1,  1;
  1,  7,    1;
  1, 13,   13,    1;
  1, 19,   61,   19,     1;
  1, 25,  145,  145,    25,     1;
  1, 31,  265,  595,   265,    31,     1;
  1, 37,  421, 1585,  1585,   421,    37,    1;
  1, 43,  613, 3331,  6145,  3331,   613,   43,    1;
  1, 49,  841, 6049, 17401, 17401,  6049,  841,   49,  1;
  1, 55, 1105, 9955, 40105, 65527, 40105, 9955, 1105, 55, 1; - _Philippe Deléham_, Mar 15 2014

Vincenzo Librandi, Rows n = 0..100, flattened
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

Cf. A002532, A016921, A081589, A081590.

A081580:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081580(n,k,5): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[Hypergeometric2F1[-k, k-n, 1, 6], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 6).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 5*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+5*x)^k/(1-x)^(k+1).
From Paul Barry, Aug 28 2008: (Start)
Number triangle T(n,k) = Sum_{j=0..n-k} binomial(n-k,j)*binomial(k,j)*5^j.
Riordan array (1/(1-x), x*(1+5*x)/(1-x)). (End)
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 6). - Jean-François Alcover, May 24 2013
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(6*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 12*x + 36*x^2/2) = 1 + 13*x + 61*x^2/2! + 145*x^3/3! + 265*x^4/4! + 421*x^5/5! + .... - Peter Bala, Mar 05 2017
Sum_{k=0..n} T(n, k, 3) = A002532(n+1). - G. C. Greubel, May 26 2021

A081581 Pascal-(1,6,1) array.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 15, 15, 1, 1, 22, 78, 22, 1, 1, 29, 190, 190, 29, 1, 1, 36, 351, 848, 351, 36, 1, 1, 43, 561, 2339, 2339, 561, 43, 1, 1, 50, 820, 5006, 9766, 5006, 820, 50, 1, 1, 57, 1128, 9192, 28806, 28806, 9192, 1128, 57, 1, 1, 64, 1485, 15240, 68034, 116208, 68034, 15240, 1485, 64, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016993, A081591, A081592. Coefficients of the row polynomials in the Newton basis are given by A013614.

			Rows start as:
  1,  1,   1,    1,    1, ... A000012;
  1,  8,  15,   22,   29, ... A016993;
  1, 15,  78,  190,  351, ... A081591;
  1, 22, 190,  848, 2339, ...
  1, 29, 351, 2339, 9766, ...
The triangle starts as:
  1;
  1,  1;
  1,  8,   1;
  1, 15,  15,    1;
  1, 22,  78,   22,    1;
  1, 29, 190,  190,   29,   1;
  1, 36, 351,  848,  351,  36,  1;
  1, 43, 561, 2339, 2339, 561, 43, 1;

Vincenzo Librandi, Rows n = 0..100, flattened

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081582 (m = 7), A143683 (m = 8).

Cf. A016993, A081591, A083099.

A081581:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081581(n,k,6): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[Hypergeometric2F1[-k, k-n, 1, 7], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

t(n, k) = sum(j=0, n-k, binomial(n-k, j)*binomial(k, j)*7^j) \\ Michel Marcus, May 24 2013

flatten([[hypergeometric([-k, k-n], [1], 7).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 6*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+6*x)^k/(1-x)^(k+1).
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 7). - Jean-François Alcover, May 24 2013
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(7*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 14*x + 49*x^2/2) = 1 + 15*x + 78*x^2/2! + 190*x^3/3! + 351*x^4/4! + 561*x^5/5! + .... - Peter Bala, Mar 05 2017
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 6.
Sum_{k=0..n} T(n, k, 6) = A083099(n+1). (End)

A123562 Pascal-(1,-3,1) array, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, -1, 1, 1, -3, -3, 1, 1, -5, -3, -5, 1, 1, -7, 1, 1, -7, 1, 1, -9, 9, 11, 9, -9, 1, 1, -11, 21, 17, 17, 21, -11, 1, 1, -13, 37, 11, 1, 11, 37, -13, 1, 1, -15, 57, -15, -39, -39, -15, 57, -15, 1, 1, -17, 81, -69, -87, -81, -87, -69, 81, -17, 1

Offset: 0

Philippe Deléham, Nov 12 2006

Riordan array (1/(1-x), x*(1-3x)/(1-x)).

			Triangle begins:
  1;
  1,   1;
  1,  -1,   1;
  1,  -3,  -3,   1;
  1,  -5,  -3,  -5,   1;
  1,  -7,   1,   1,  -7,   1;
  1,  -9,   9,  11,   9,  -9,  1;
  1, -11,  21,  17,  17,  21, -11,   1;
  1, -13,  37,  11,   1,  11,  37, -13,   1;

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

Cf. Pascal (1,m,1) array: A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

T[n_, k_] := Sum[Binomial[n - j, k]*Binomial[k, j]*(-3)^j, {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 15 2017 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, binomial(n-j,k)*  binomial(k,j)*(-3)^j), ", "))) \\ G. C. Greubel, Oct 15 2017

Sum_{k=0..n} T(n,k) = A088137(n+1).
T(n,k) = T(n-1,k-1) + T(n-1,k) - 3*T(n-2,k-1), n>0.
From Paul Barry, Jan 24 2011: (Start)
T(n,k) = Sum_{j=0..n} binomial(n-j,k)*binomial(k,j)*(-3)^j.
T(n,k) = [k<=n]*Hypergeometric2F1(-k,k-n,1,-2). (End)
E.g.f. for the n-th subdiagonal: exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(-2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 - 4*x + 4*x^2/2) = 1 - 3*x - 3*x^2/2! + x^3/3! + 9*x^4/4! + 21*x^5/5! + .... - Peter Bala, Mar 05 2017

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Abdullahi Umar, Aug 25 2008

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A007051, A118801, A135233, columns: A000225, A000337, A055580, A027608.
Cf. also A038207, A250118, A250119.
Cf. A007318, A167374, A145661, A029635.

A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011

Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A143683 Pascal-(1,8,1) array.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 19, 19, 1, 1, 28, 118, 28, 1, 1, 37, 298, 298, 37, 1, 1, 46, 559, 1540, 559, 46, 1, 1, 55, 901, 4483, 4483, 901, 55, 1, 1, 64, 1324, 9856, 21286, 9856, 1324, 64, 1, 1, 73, 1828, 18388, 67006, 67006, 18388, 1828, 73, 1, 1, 82, 2413, 30808, 164242, 304300, 164242, 30808, 2413, 82, 1

Offset: 0

Paul Barry, Aug 28 2008

			Square array begins as:
  1,  1,    1,     1,      1,       1,        1, ... A000012;
  1, 10,   19,    28,     37,      46,       55, ... A017173;
  1, 19,  118,   298,    559,     901,     1324, ...
  1, 28,  298,  1540,   4483,    9856,    18388, ...
  1, 37,  559,  4483,  21286,   67006,   164242, ...
  1, 46,  901,  9856,  67006,  304300,  1004590, ...
  1, 55, 1324, 18388, 164242, 1004590,  4443580, ...
Antidiagonal triangle begins as:
  1;
  1,  1;
  1, 10,   1;
  1, 19,  19,    1;
  1, 28, 118,   28,    1;
  1, 37, 298,  298,   37,   1;
  1, 46, 559, 1540,  559,  46,  1;
  1, 55, 901, 4483, 4483, 901, 55, 1;

Reinhard Zumkeller, Rows n = 0..125 of table, flattened

Cf.Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7).

Cf. A003683, A017173, A143680.

a143683 n k = a143683_tabl !! n !! k
a143683_row n = a143683_tabl !! n
a143683_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) (map (* 8) ([0] ++ us ++ [0])) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Mar 16 2014

A143683:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A143683(n,k,8): k in [0..n], n in [0..12]]; // G. C. Greubel, May 27 2021

Table[Hypergeometric2F1[-k, k-n, 1, 9], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 9).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 27 2021

Square array: T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 8*T(n-1, k-1) + T(n-1, k).
Number triangle: T(n,k) = Sum_{j=0..n-k} binomial(n-k,j)*binomial(k,j)*9^j.
Rows are the expansions of (1+8*x)^k/(1-x)^(k+1).
Riordan array (1/(1-x), x*(1+8*x)/(1-x)).
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 9). - Jean-François Alcover, May 24 2013
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(9*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 18*x + 81*x^2/2) = 1 + 19*x + 118*x^2/2! + 298*x^3/3! + 559*x^4/4! + 901*x^5/5! + .... - Peter Bala, Mar 05 2017
Sum_{k=0..n} T(n,k) = A003683(n+1). - G. C. Greubel, May 27 2021

A054120 Triangular array T(n,k): start with T(n,0)=T(n,n)=1 for n >= 0; recursively, draw vertical lines through T(n-1,k-1) if present and T(n-1,k) if present; then T(n,k) is the sum of T(i,j) that lie on or between the lines and not below T(n,k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 9, 18, 9, 1, 1, 12, 39, 39, 12, 1, 1, 15, 69, 114, 69, 15, 1, 1, 18, 108, 261, 261, 108, 18, 1, 1, 21, 156, 507, 750, 507, 156, 21, 1, 1, 24, 213, 879, 1779, 1779, 879, 213, 24, 1, 1, 27, 279, 1404, 3672, 5058

Offset: 0

Clark Kimberling

Conjecture: T(n,k) = T(n-1,k-1) + 2*T(n-2,k-1) + T(n-1,k) (except for T(0,0) = 1 and T(2,1) = 3 and assuming that T(n,k) = 0 for elements outside the triangular array). - Gerald McGarvey, Sep 20 2007
Conjecture: T(n,k) = A081577(n,k) - A081577(n-2,k-1). (A081577 is Pascal-(1,2,1) array). - Gerald McGarvey, Sep 20 2007
From Russell Jay Hendel, Jun 02 2015: (Start)
We prove the 1st McGarvey conjecture, (A) T(n,k)=T(n-1,k-1)+2*T(n-2,k-1)+T(n-1,k), with McGarvey's added modifications that n >= 3 and that T(n,k)=0 for n<0, k<0, and k>n. We make explicit that the rows and columns of the triangle start at 0 and furthermore, that both diagonals are at 45 degrees to the perpendicular (in both directions). Note that we use letters (A), (B), (C) to indicate equations.
We indicate the set of numbers corresponding to the half-line starting at T(n,k) with(B) L(n,k)={T(n,k), T(n-2,k-1), T(n-4,k-2),T(n-6,k-3),...}. There is no loss of generality in extending this line upwards indefinitely since most of the numbers lying on a half-ray are 0 anyway. If S_1, S_2, S_3 are sets, then the notation SUM(S_1,S_2, S_3) will indicate the sum of all elements in S_1, S_2 and S_3.
Using two applications of (B) we have the identity (C) L(n,k)={T(n,k)} UNION L(n-2,k-1). The definition of T(n,k) is given by (D) T(n,k)=SUM(L(n-1,k-1),L(n-2,k-1),L(n-1,k)). Combining (D) with (C), we immediately have (E), T(n,k)=T(n-1,k-1)+T(n-2,k-1)+T(n-1,k)+SUM(L(n-3,k-2), L(n-4,k-2), L(n-3,k-1)). By another application of (D), we have (F), T(n-2,k-1)=SUM(L(n-3,k-2),L(n-4,k-2),L(n-3,k-1)). Combining (E) and (F), we obtain (A) as required.
We illustrate the proof with n=5 and k=3. We first calculate T(3,2). By (D), we have (G) T(3,2)=6=SUM(L(2,1),L(1,1),L(2,2)) with L(2,1)={3,1,0,0,0,...}, L(1,1)={1,0,0,....} and L(2,2)={1,0,0,...}. By another application of (D), we have T(5,3)=39=(SUM(L(4,2),L(3,2),L(4,3)) where L(4,2)={18,3,1,...}={18} UNION L(2,1), L(3,2)={6,1,0,...}={6} UNION L(1,1), and  L(4,3) = {1,0,0,...}={1} UNION L(2,2). Combining this last equation with (G), we have T(5,3)=39=18+6+1+SUM(L(2,1),L(1,1),L(2,2))=18+6+1+6=18+2*6+1 as required.
Note: There are two parts to the definition of the triangle:(i) T(n,0)=1, and ii) equation (D). Since (D) does not apply to  T(0,0), (A) does not hold there. Since (D) applied to T(2,1) refers back to T(0,0), (A) does not hold there either. This explains the McGarvey requirement that n>=3. (End)
Consider the array with g.f. (1-u*v)/(1-u-v-2*u*v). The triangle appears to be that symmetric array read by antidiagonals. - R. J. Mathar, Jan 26 2022

			Rows:
1;
1,1;
1,3,1;
1,6,6,1;
1,9,18,9,1;
1,12,39,39,12,1;

Row sums: A052945. A054122 (diagonal), A052392 (subdiagonal).

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