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A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

%I A208904 #34 Aug 22 2015 01:57:18
%S A208904 1,3,1,5,6,1,7,19,9,1,9,44,42,12,1,11,85,138,74,15,1,13,146,363,316,
%T A208904 115,18,1,15,231,819,1059,605,165,21,1,17,344,1652,2984,2470,1032,224,
%U A208904 24,1,19,489,3060,7380,8378,4974,1624,292,27,1,21,670,5301,16488
%N A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.
%C A208904 For a discussion and guide to related arrays, see A208510.
%C A208904 Riordan array ((1+x)/(1-x)^2, x(1+x)/(1-x)^2) (follows from Kruchinin formula). - _Ralf Stephan_, Jan 02 2014
%C A208904 From _Peter Bala_, Jul 21 2014: (Start)
%C A208904 Let M denote the lower unit triangular array A099375 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
%C A208904 /I_k 0\
%C A208904 \ 0  M/
%C A208904 having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)
%F A208904 u(n,x)=u(n-1,x)+2x*v(n-1,x),
%F A208904 v(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
%F A208904 where u(1,x)=1, v(1,x)=1.
%F A208904 From _Vladimir Kruchinin_, Mar 11 2013: (Start)
%F A208904 T(n,k) = sum(i=0..n, binomial(i+k-1,2*k-1)*binomial(k,n-i))
%F A208904 ((x+x^2)/(1-x)^2)^k = sum(n>=k, T(n,k)*x^n).
%F A208904 T(n,2)=A005900(n).
%F A208904 T(2*n-1,n) / n = A003169(n).
%F A208904 T(2*n,n) = A156894(n), n>1.
%F A208904 sum(k=1..n, T(n,k)) = A003946(n).
%F A208904 sum(k=1..n, T(n,k)*(-1)^(n+k)) = A078050(n).
%F A208904 n*sum(k=1..n, T(n,k)/k) = A058481(n). (End)
%F A208904 Recurrence: T(n+1,k+1) = sum {i = 0..n-k} (2*i + 1)*T(n-i,k). - _Peter Bala_, Jul 21 2014
%e A208904 First five rows:
%e A208904 1
%e A208904 3...1
%e A208904 5...6....1
%e A208904 7...19...9....1
%e A208904 9...44...42...12...1
%e A208904 First five polynomials v(n,x):
%e A208904 1
%e A208904 3 + x
%e A208904 5 + 6x + x^2
%e A208904 7 + 19x + 9x^2 + x^3
%e A208904 9 + 44x + 42x^2 + 12x^3 + x^4
%e A208904 From _Peter Bala_, Jul 21 2014: (Start)
%e A208904 With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
%e A208904 /1        \/1        \/1        \      /1            \
%e A208904 |3 1      ||0 1      ||0 1      |      |3  1         |
%e A208904 |5 3 1    ||0 3 1    ||0 0 1    |... = |5  6  1      |
%e A208904 |7 5 3 1  ||0 5 3 1  ||0 0 3 1  |      |7 19  9  1   |
%e A208904 |9 7 5 3 1||0 7 5 3 1||0 0 5 3 1|      |9 44 42 12 1 |
%e A208904 |...      ||...      ||...      |      |...
%e A208904 (End)
%t A208904 u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t A208904 u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
%t A208904 v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
%t A208904 Table[Expand[u[n, x]], {n, 1, z/2}]
%t A208904 Table[Expand[v[n, x]], {n, 1, z/2}]
%t A208904 cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t A208904 TableForm[cu]
%t A208904 Flatten[%]    (* A208660 *)
%t A208904 Table[Expand[v[n, x]], {n, 1, z}]
%t A208904 cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t A208904 TableForm[cv]
%t A208904 Flatten[%]    (* A208904 *)
%Y A208904 Cf. A208660, A208510. A099375.
%K A208904 nonn,tabl
%O A208904 1,2
%A A208904 _Clark Kimberling_, Mar 03 2012