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Inverse of a Fibonacci-Pascal matrix A105809.
4

%I #11 Mar 21 2018 10:54:32

%S 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 1,5,-8,-1,20,-29,20,-7,1,-6,13,-7,-21,49,-49,27,-8,1,7,-19,20,14,-70,

%U 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 Inverse of a Fibonacci-Pascal matrix A105809.

%C First column is A105811, row sums are A105812, antidiagonal sums are (-1)^n.

%F 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 From _Wolfdieter Lang_, Oct 04 2014: (Start)

%F 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 O.g.f. column m: x^m*(1 + x - x^2)/(1 + x)^(m+2), m >= 0.

%F 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 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 The triangle T(n,m) begins:

%e n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...

%e 0: 1

%e 1: -1 1

%e 2: 0 -2 1

%e 3: 1 2 -3 1

%e 4: -2 -1 5 -4 1

%e 5: 3 -1 -6 9 -5 1

%e 6: -4 4 5 -15 14 -6 1

%e 7: 5 -8 -1 20 -29 20 -7 1

%e 8: -6 13 -7 -21 49 -49 27 -8 1

%e 9: 7 -19 20 14 -70 98 -76 35 -9 1

%e 10: -8 26 -39 6 84 -168 174 -111 44 -10 1

%e 11: 9 -34 65 -45 -78 252 -342 285 -155 54 -11 1

%e 12: -10 43 -99 110 33 -330 594 -627 440 -209 65 -12 1

%e 13: 11 -53 142 -209 77 363 -924 1221 -1067 649 -274 77 -13 1

%e ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014

%e -----------------------------------------------------------------------

%e 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 C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if

%p end proc:

%p for n from 0 to 10 do

%p seq((-1)^(n+k)*(C(n, n-k) - add(C(n-i, n-k-i), i = 2..n)), k = 0..n);

%p end do; # _Peter Bala_, Mar 21 2018

%Y Cf. A105809, A105811, A105812, A248155 (alternating row sum). - _Wolfdieter Lang_, Oct 04 2014

%K easy,sign,tabl

%O 0,5

%A _Paul Barry_, May 04 2005