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%I #42 Aug 28 2019 16:06:17
%S 1,4,1,4,5,1,4,9,6,1,4,13,15,7,1,4,17,28,22,8,1,4,21,45,50,30,9,1,4,
%T 25,66,95,80,39,10,1,4,29,91,161,175,119,49,11,1,4,33,120,252,336,294,
%U 168,60,12,1,4,37,153,372,588,630,462,228,72,13,1,4,41,190,525,960,1218
%N (4,1) Pascal triangle.
%C The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
%C This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
%C This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x) = (1+3*z)/(1-(1+x)*z).
%C The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C For a closed-form formula for generalized Pascal's triangle see A228576. - _Boris Putievskiy_, Sep 09 2013
%C The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - _Peter Bala_, Mar 02 2018
%D Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
%D Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.
%H Reinhard Zumkeller, <a href="/A093561/b093561.txt">>Rows n = 0..125 of triangle, flattened</a>
%H P. Bala, <a href="/A081577/a081577.pdf">A note on the diagonals of a proper Riordan Array</a>
%H W. Lang, <a href="/A093561/a093561.txt">First 10 rows and array of figurate numbers </a>.
%F a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*n+m)*binomial(n+m-1, m-1)/m if m>=1.
%F Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
%F G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
%F T(n, k) = C(n, k) + 3*C(n-1, k). - _Philippe Deléham_, Aug 28 2005
%F exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - _Peter Bala_, Dec 22 2014
%e Triangle begins
%e [1];
%e [4, 1];
%e [4, 5, 1];
%e [4, 9, 6, 1];
%e ...
%o (Haskell)
%o a093561 n k = a093561_tabl !! n !! k
%o a093561_row n = a093561_tabl !! n
%o a093561_tabl = [1] : iterate
%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
%o -- _Reinhard Zumkeller_, Aug 31 2014
%Y Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
%Y Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.
%Y Cf. A007318, A093562 (d=5), A228196, A228576.
%K nonn,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Apr 22 2004
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