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A105810 Inverse of a Fibonacci-Pascal matrix A105809.

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%I A105810 #11 Mar 21 2018 10:54:32
%S A105810 1,-1,1,0,-2,1,1,2,-3,1,-2,-1,5,-4,1,3,-1,-6,9,-5,1,-4,4,5,-15,14,-6,
%T A105810 1,5,-8,-1,20,-29,20,-7,1,-6,13,-7,-21,49,-49,27,-8,1,7,-19,20,14,-70,
%U A105810 98,-76,35,-9,1,-8,26,-39,6,84,-168,174,-111,44,-10,1,9,-34,65,-45,-78,252,-342,285,-155,54,-11,1
%N A105810 Inverse of a Fibonacci-Pascal matrix A105809.
%C A105810 First column is A105811, row sums are A105812, antidiagonal sums are (-1)^n.
%F A105810 Riordan array ((1+x-x^2)/(1+x)^2, x/(1+x)); Number triangle T(n, 0)=A105811(n), T(n, m)=-T(n-1, m-1)+T(n-1, m).
%F A105810 From _Wolfdieter Lang_, Oct 04 2014: (Start)
%F A105810 O.g.f. for row polynomials R(n,x) = sum(T(n,m)*x^m,m=0..n): (1 + z - z^2)/((1+z)*(1+(1-x)*z)) (Riordan property).
%F A105810 O.g.f. column m: x^m*(1 + x - x^2)/(1 + x)^(m+2), m >= 0.
%F A105810 The A-sequence of this Riordan triangle is [1, -1]. See the above given recurrence for T(n,m) for n>=1. The Z-sequence has o.g.f. -(1 - x^2)/(1 - x - x^2) and is -A132916(n+5) = -[1, 1, 1, 2, 3, 5, 8, 13, 21, 34,...]. See the W. Lang link under A006232 for Riordan A- and Z-sequences. (End)
%F A105810 T(n,k) = (-1)^(n+k)*(C(n, n-k) - Sum_{i = 2..n} C(n-i, n-k-i)), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0. - _Peter Bala_, Mar 21 2018
%e A105810 The triangle T(n,m) begins:
%e A105810 n\m   0   1   2    3   4    5    6    7     8    9   10  11  12 13 ...
%e A105810 0:    1
%e A105810 1:   -1   1
%e A105810 2:    0  -2   1
%e A105810 3:    1   2  -3    1
%e A105810 4:   -2  -1   5   -4   1
%e A105810 5:    3  -1  -6    9  -5    1
%e A105810 6:   -4   4   5  -15  14   -6    1
%e A105810 7:    5  -8  -1   20 -29   20   -7    1
%e A105810 8:   -6  13  -7  -21  49  -49   27   -8     1
%e A105810 9:    7 -19  20   14 -70   98  -76   35    -9    1
%e A105810 10:  -8  26 -39    6  84 -168  174 -111    44  -10    1
%e A105810 11:   9 -34  65  -45 -78  252 -342  285  -155   54  -11   1
%e A105810 12: -10  43 -99  110  33 -330  594 -627   440 -209   65 -12   1
%e A105810 13:  11 -53 142 -209  77  363 -924 1221 -1067  649 -274  77 -13  1
%e A105810 ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
%e A105810 -----------------------------------------------------------------------
%e A105810 Recurrence for T(n, 0) with row n-1 entries from Z-sequence (see a link given above): 3 = T(5, 0) = -(1*(-2) + 1*(-1) + 1*5 + 2*(-4) + 3*1) = 3.
%p A105810 C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if
%p A105810 end proc:
%p A105810 for n from 0 to 10 do
%p A105810     seq((-1)^(n+k)*(C(n, n-k) - add(C(n-i, n-k-i), i = 2..n)), k = 0..n);
%p A105810 end do; # _Peter Bala_, Mar 21 2018
%Y A105810 Cf. A105809, A105811, A105812, A248155 (alternating row sum). - _Wolfdieter Lang_, Oct 04 2014
%K A105810 easy,sign,tabl
%O A105810 0,5
%A A105810 _Paul Barry_, May 04 2005