%I #48 Feb 06 2022 06:31:07
%S 1,0,1,1,1,1,1,3,2,1,3,6,6,3,1,6,15,15,10,4,1,15,36,40,29,15,5,1,36,
%T 91,105,84,49,21,6,1,91,232,280,238,154,76,28,7,1,232,603,750,672,468,
%U 258,111,36,8,1,603,1585,2025,1890,1398,837,405,155,45,9,1,1585,4213,5500
%N Inverse binomial matrix applied to A039599.
%C Reverse of A071947 - related to lattice paths. First column is A005043.
%C Triangle T(n,k), 0 <= k <= n, defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - _Philippe Deléham_, Feb 27 2007
%C This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise from choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - _Philippe Deléham_, Sep 25 2007
%C Riordan array (f(x),x*g(x)), where f(x)is the o.g.f. of A005043 and g(x)is the o.g.f. of A001006. - _Philippe Deléham_, Nov 22 2009
%C Riordan array ((1+x-sqrt(1-2x-3x^2))/(2x(1+x)), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse of Riordan array ((1+x)/(1+x+x^2),x/(1+x+x^2)). E.g.f. of column k is exp(x)*(Bessel_I(k,2x)-Bessel_I(k+1,2x)).
%C Diagonal sums are A187306.
%C Simultaneous equations using the first n rows solve for diagonal lengths of odd N = (2n+1) regular polygons, with constants c^0, c^1, c^2, ...; where c = 1 + 2*cos( 2*Pi/N) = sin(3*Pi/N)/sin(Pi/N) = the third longest diagonal of N>5. By way of example, take the first 4 rows relating to the 9-gon (nonagon), N=(2*4 + 1), with c = 1 + 2*cos(2*Pi/9) = 2.5320888.... The simultaneous equations are (1,0,0,0) = 1; (0,1,0,0) = c; (1,1,1,0) = c^2, (1,3,2,1) = c^3. The answers are 1, 2.532..., 2.879..., and 1.879...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. - _Gary W. Adamson_, Sep 07 2011
%H G. C. Greubel, <a href="/A089942/b089942.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H P. Barry and A. Hennessy, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Barry1/barry202.html">Four-term Recurrences, Orthogonal Polynomials and Riordan Arrays</a>, Journal of Integer Sequences, 2012, article 12.4.2. - From _N. J. A. Sloane_, Sep 21 2012
%H E. Deutsch, L. Ferrari and S. Rinaldi, <a href="http://dx.doi.org/10.1016/j.aam.2004.05.002">Production Matrices</a>, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
%H D. Merlini, D. G. Rogers, R. Sprugnoli and M. C. Verri, <a href="http://dx.doi.org/10.4153/CJM-1997-015-x">On some alternative characterizations of Riordan arrays</a>, Canad J. Math., 49 (1997), 301-320.
%H Sun, Yidong; Ma, Luping <a href="https://doi.org/10.1016/j.ejc.2014.01.004">Minors of a class of Riordan arrays related to weighted partial Motzkin paths</a>. Eur. J. Comb. 39, 157-169 (2014) Table 2.2
%F G.f.: (1+z-q)/[(1+z)(2z-t+tz+tq)], where q = sqrt(1-2z-3z^2).
%F Sum_{k>=0} T(m,k)*T(n,k) = T(m+n,0) = A005043(m+n). - _Philippe Deléham_, Mar 22 2007
%F Sum_{k=0..n} T(n,k)*(2k+1) = 3^n. - _Philippe Deléham_, Mar 22 2007
%F Sum_{k=0..n} T(n,k)*2^k = A112657(n). - _Philippe Deléham_, Apr 01 2007
%F T(n,2k) + T(n,2k+1) = A109195(n,k). - _Philippe Deléham_, Nov 11 2008
%F T(n,k) = GegenbauerC(n-k,-n+1,-1/2) - GegenbauerC(n-k-1,-n+1,-1/2) for 1 <= k <= n. - _Peter Luschny_, May 12 2016
%e Triangle begins
%e 1,
%e 0, 1,
%e 1, 1, 1,
%e 1, 3, 2, 1,
%e 3, 6, 6, 3, 1,
%e 6, 15, 15, 10, 4, 1,
%e 15, 36, 40, 29, 15, 5, 1,
%e 36, 91, 105, 84, 49, 21, 6, 1,
%e 91, 232, 280, 238, 154, 76, 28, 7, 1
%e Production matrix is
%e 0, 1,
%e 1, 1, 1,
%e 0, 1, 1, 1,
%e 0, 0, 1, 1, 1,
%e 0, 0, 0, 1, 1, 1,
%e 0, 0, 0, 0, 1, 1, 1,
%e 0, 0, 0, 0, 0, 1, 1, 1,
%e 0, 0, 0, 0, 0, 0, 1, 1, 1,
%e 0, 0, 0, 0, 0, 0, 0, 1, 1, 1
%p T:= (n,k) -> simplify(GegenbauerC(n-k,-n+1,-1/2)-GegenbauerC(n-k-1,-n+1,-1/2)): for n from 1 to 9 do seq(T(n,k), k=1..n) od; # _Peter Luschny_, May 12 2016
%p # Or by recurrence:
%p T := proc(n, k) option remember;
%p if n = k then 1 elif k < 0 or n < 0 or k > n then 0
%p elif k = 0 then T(n-1, 1) else T(n-1, k-1) + T(n-1, k) + T(n-1, k+1) fi end:
%p for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # _Peter Luschny_, May 25 2021
%t T[n_, k_] := GegenbauerC[n - k, -n + 1, -1/2] - GegenbauerC[n - k - 1, -n + 1, -1/2]; Table[T[n, k], {n,1,10}, {k,1,n}] // Flatten (* _G. C. Greubel_, Feb 28 2017 *)
%Y Row sums give A002426 (central trinomial coefficients).
%K nonn,tabl
%O 0,8
%A _Paul Barry_, Nov 16 2003
%E Edited by _Emeric Deutsch_, Mar 04 2004