A081579 Pascal-(1,4,1) array.
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
Examples
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
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.
Crossrefs
Programs
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Magma
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
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Mathematica
Table[Hypergeometric2F1[-k, k-n, 1, 5], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)
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Sage
flatten([[hypergeometric([-k, k-n], [1], 5).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021
Formula
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)
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