|
%I #37 Aug 28 2019 16:11:38
%S 1,7,1,7,8,1,7,15,9,1,7,22,24,10,1,7,29,46,34,11,1,7,36,75,80,45,12,1,
%T 7,43,111,155,125,57,13,1,7,50,154,266,280,182,70,14,1,7,57,204,420,
%U 546,462,252,84,15,1,7,64,261,624,966,1008,714,336,99,16,1,7,71,325,885
%N (7,1) Pascal triangle.
%C The array F(7;n,m) gives in the columns m>=1 the figurate numbers based on A016993, including the 9-gonal numbers A001106, (see the W. Lang link).
%C This is the seventh member, d=7, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-3, for d=1..6.
%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+6*z)/(1-(1+x)*z).
%C The SW-NE diagonals give A022097(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 6. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%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="/A093564/b093564.txt">Rows n = 0..125 of triangle, flattened</a>
%H W. Lang, <a href="/A093564/a093564.txt">First 10 rows and array of figurate numbers </a>.
%F a(n, m)=F(7;n-m, m) for 0<= m <= n, otherwise 0, with F(7;0, 0)=1, F(7;n, 0)=7 if n>=1 and F(7;n, m):=(7*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)=7 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
%F G.f. column m (without leading zeros): (1+6*x)/(1-x)^(m+1), m>=0.
%F T(n, k) = C(n, k) + 6*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)*(7 + 15*x + 9*x^2/2! + x^3/3!) = 7 + 22*x + 46*x^2/2! + 80*x^3/3! + 125*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 [7, 1];
%e [7, 8, 1];
%e [7, 15, 9, 1];
%e ...
%p N:= 20: # to get the first N rows
%p T:=Matrix(N,N):
%p T[1,1]:= 1:
%p for m from 2 to N do
%p T[m,1]:= 7:
%p T[m,2..m]:= T[m-1,1..m-1] + T[m-1,2..m];
%p od:
%p for m from 1 to N do
%p convert(T[m,1..m],list)
%p od; # _Robert Israel_, Dec 28 2014
%o (Haskell)
%o a093564 n k = a093564_tabl !! n !! k
%o a093564_row n = a093564_tabl !! n
%o a093564_tabl = [1] : iterate
%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [7, 1]
%o -- _Reinhard Zumkeller_, Sep 01 2014
%Y Row sums: A000079(n+2), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 6 for n=2 and 0 otherwise.
%Y The column sequences give for m=1..9: A016993, A001106 (9-gonal), A007584, A051740, A051877, A050403, A027818, A034266, A055994.
%Y Cf. A093565 (d=8).
%K nonn,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Apr 22 2004
|
