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A093563 (6,1)-Pascal triangle.

%I A093563 #43 Nov 12 2024 15:59:40
%S A093563 1,6,1,6,7,1,6,13,8,1,6,19,21,9,1,6,25,40,30,10,1,6,31,65,70,40,11,1,
%T A093563 6,37,96,135,110,51,12,1,6,43,133,231,245,161,63,13,1,6,49,176,364,
%U A093563 476,406,224,76,14,1,6,55,225,540,840,882,630,300,90,15,1,6,61,280,765,1380
%N A093563 (6,1)-Pascal triangle.
%C A093563 The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
%C A093563 This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
%C A093563 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+5*z)/(1-(1+x)*z).
%C A093563 The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C A093563 For a closed-form formula for generalized Pascal's triangle see A228576. - _Boris Putievskiy_, Sep 09 2013
%D A093563 Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
%D A093563 Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.
%H A093563 Reinhard Zumkeller, <a href="/A093563/b093563.txt">Rows n = 0..125 of triangle, flattened</a>
%H A093563 Wolfdieter Lang, <a href="/A093563/a093563.pdf">First 10 rows and array of figurate numbers</a>
%F A093563 a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
%F A093563 Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
%F A093563 G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
%F A093563 T(n, k) = C(n, k) + 5*C(n-1, k). - _Philippe Deléham_, Aug 28 2005
%F A093563 exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*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 A093563 Triangle begins
%e A093563   1;
%e A093563   6,  1;
%e A093563   6,  7,  1;
%e A093563   6, 13,  8,  1;
%e A093563   6, 19, 21,  9,  1;
%e A093563   6, 25, 40, 30, 10,  1;
%e A093563   ...
%t A093563 lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* _Jean-François Alcover_, Sep 16 2011, after g.f. *)
%o A093563 (Haskell)
%o A093563 a093563 n k = a093563_tabl !! n !! k
%o A093563 a093563_row n = a093563_tabl !! n
%o A093563 a093563_tabl = [1] : iterate
%o A093563                (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
%o A093563 -- _Reinhard Zumkeller_, Aug 31 2014
%o A093563 (Python)
%o A093563 from math import comb, isqrt
%o A093563 def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # _Chai Wah Wu_, Nov 12 2024
%Y A093563 Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
%Y A093563 Cf. A007318, A093564 (d=7), A228196, A228576.
%Y A093563 The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.
%K A093563 nonn,easy,tabl
%O A093563 0,2
%A A093563 _Wolfdieter Lang_, Apr 22 2004