A105810 Inverse of a Fibonacci-Pascal matrix A105809.
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, 1, 5, -8, -1, 20, -29, 20, -7, 1, -6, 13, -7, -21, 49, -49, 27, -8, 1, 7, -19, 20, 14, -70, 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
Offset: 0
Examples
The triangle T(n,m) begins: n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ... 0: 1 1: -1 1 2: 0 -2 1 3: 1 2 -3 1 4: -2 -1 5 -4 1 5: 3 -1 -6 9 -5 1 6: -4 4 5 -15 14 -6 1 7: 5 -8 -1 20 -29 20 -7 1 8: -6 13 -7 -21 49 -49 27 -8 1 9: 7 -19 20 14 -70 98 -76 35 -9 1 10: -8 26 -39 6 84 -168 174 -111 44 -10 1 11: 9 -34 65 -45 -78 252 -342 285 -155 54 -11 1 12: -10 43 -99 110 33 -330 594 -627 440 -209 65 -12 1 13: 11 -53 142 -209 77 363 -924 1221 -1067 649 -274 77 -13 1 ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014 ----------------------------------------------------------------------- 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.
Crossrefs
Programs
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Maple
C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if end proc: for n from 0 to 10 do seq((-1)^(n+k)*(C(n, n-k) - add(C(n-i, n-k-i), i = 2..n)), k = 0..n); end do; # Peter Bala, Mar 21 2018
Formula
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).
From Wolfdieter Lang, Oct 04 2014: (Start)
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).
O.g.f. column m: x^m*(1 + x - x^2)/(1 + x)^(m+2), m >= 0.
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)
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
Comments