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

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

0,5

Comments

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

Links

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

Formula

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)

Example

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

Mathematica

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

Prog

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

Crossrefs

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).

Sequence in context: A296128 A131061 A157169 * A184883 A279003 A210651

Adjacent sequences: A081575 A081576 A081577 * A081579 A081580 A081581

Keyword

easy,nonn,tabl

Author

Paul Barry, Mar 23 2003

Status

approved