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

0,5

Comments

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

Links

Table of n, a(n) for n=0..77.

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

Example

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.

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

Crossrefs

Cf. A105809, A105811, A105812, A248155 (alternating row sum). - Wolfdieter Lang, Oct 04 2014

Sequence in context: A317840 A048207 A359164 * A063726 A290267 A240750

Adjacent sequences: A105807 A105808 A105809 * A105811 A105812 A105813

Keyword

easy,sign,tabl

Author

Paul Barry, May 04 2005

Status

approved