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%I #36 Sep 08 2022 08:45:09
%S 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,
%T 1,1,25,201,593,593,201,25,1,1,29,289,1181,1921,1181,289,29,1,1,33,
%U 393,2073,4881,4881,2073,393,33,1,1,37,513,3333,10497,15525,10497,3333,513,37,1
%N Pascal-(1,3,1) array.
%C 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.
%C 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
%H Vincenzo Librandi, <a href="/A081578/b081578.txt">Rows n = 0..100, flattened</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html">On Integer-Sequence-Based Constructions of Generalized Pascal Triangles</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL11/Barry/barry594.html">A note on Krawtchouk Polynomials and Riordan Arrays</a>, JIS 11 (2008) 08.2.2
%F 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).
%F Rows are the expansions of (1+3*x)^k/(1-x)^(k+1).
%F T(n,k) = Sum_{j=0..n} binomial(k,j-k)*binomial(n+k-j,k)*3^(j-k). - _Paul Barry_, Oct 23 2006
%F 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
%F From _G. C. Greubel_, May 26 2021: (Start)
%F T(n, k, m) = Hypergeometric2F1([-k, k-n], [1], m+1), for m = 3.
%F T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 3.
%F Sum_{k=0..n} T(n, k, 3) = A015518(n+1). (End)
%e Square array begins as:
%e 1, 1, 1, 1, 1, ... A000012;
%e 1, 5, 9, 13, 17, ... A016813;
%e 1, 9, 33, 73, 129, ... A081585;
%e 1, 13, 73, 245, 593, ... A081586;
%e 1, 17, 129, 593, 1921, ...
%e As a triangle this begins:
%e 1;
%e 1, 1;
%e 1, 5, 1;
%e 1, 9, 9, 1;
%e 1, 13, 33, 13, 1;
%e 1, 17, 73, 73, 17, 1;
%e 1, 21, 129, 245, 129, 21, 1;
%e 1, 25, 201, 593, 593, 201, 25, 1;
%e 1, 29, 289, 1181, 1921, 1181, 289, 29, 1;
%e 1, 33, 393, 2073, 4881, 4881, 2073, 393, 33, 1;
%e 1, 37, 513, 3333, 10497, 15525, 10497, 3333, 513, 37, 1; - _Philippe Deléham_, Mar 15 2014
%t Table[Hypergeometric2F1[-k, k-n, 1, 4], {n,0,10}, {k,0,n}]//Flatten (* _Jean-François Alcover_, May 24 2013 *)
%o (Haskell)
%o a081578 n k = a081578_tabl !! n !! k
%o a081578_row n = a081578_tabl !! n
%o a081578_tabl = map fst $ iterate
%o (\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
%o zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
%o -- _Reinhard Zumkeller_, Mar 16 2014
%o (Magma)
%o A081578:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
%o [A081578(n,k,3): k in [0..n], n in [0..12]]; // _G. C. Greubel_, May 26 2021
%o (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
%Y 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).
%K easy,nonn,tabl
%O 0,5
%A _Paul Barry_, Mar 23 2003
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