|
Search: id:A094587
|
|
|
| A094587 |
|
Triangle of permutation coefficients arranged with 1's on the diagonal. |
|
+0
17
|
|
| 1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
(list; table; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
COMMENT
|
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers sum{k=0..n, (-1)^(n+k)n!/k!}. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry (pbarry(AT)wit.ie), Mar 27 2007
Comments from Tom Copeland (tcjpn(AT)msn.com), Nov 01 2007: (Start) T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = sum(j=0,...,n) Binom(n,j) * j! * x^(n-j) = sum(j=0,...,n) (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0] , umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = sum(j=0,...,n) (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x) , formally
5) B(x) = sum(j=0,1,...) (xDx)^j A(x)
6) B(x) = sum(j=0,1,...) x^j * D^j * x^j A(x)
7) B(x) = sum(j=0,1,...) j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x, and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
|
|
REFERENCES
|
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 207.
|
|
LINKS
|
P. Luschny, Variants of Variations.
|
|
FORMULA
|
T(n, k)=n!/k! if n >= k >= 0 else 0.
T(n, k) = Sum[i=k..n, |S1(n+1, i+1)S2(i, k)| * (-1)^i ], with S1, S2 the Stirling numbers.
1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24 - Paul Barry (pbarry(AT)wit.ie), Mar 27 2007
|
|
EXAMPLE
|
Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}....
|
|
CROSSREFS
|
Adjacent sequences: A094584 A094585 A094586 this_sequence A094588 A094589 A094590
Sequence in context: A134558 A137381 A109316 this_sequence A135878 A121284 A108076
|
|
KEYWORD
|
easy,nonn,tabl
|
|
AUTHOR
|
Paul Barry (pbarry(AT)wit.ie), May 13 2004
|
|
|
Search completed in 0.002 seconds
|
