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 A129334 Triangle T(n,k) read by rows: inverse of the matrix PE = exp(P)/exp(1) given in A011971. 0

%I

%S 1,-1,1,0,-2,1,1,0,-3,1,1,4,0,-4,1,-2,5,10,0,-5,1,-9,-12,15,20,0,-6,1,

%T -9,-63,-42,35,35,0,-7,1,50,-72,-252,-112,70,56,0,-8,1,267,450,-324,

%U -756,-252,126,84,0,-9,1,413,2670,2250,-1080,-1890,-504,210,120

%N Triangle T(n,k) read by rows: inverse of the matrix PE = exp(P)/exp(1) given in A011971.

%C The structure of the triangle is A[r,c] = A000587(1+(r-c))*binomial(r-1,c-1) where row index r and column-index c start at 1.

%C Coefficients of polynomials defined recursively: P(0,x)=1, P(n+1,x)=x*P(n,x)-P(n,x+1). All generated polynomials appear to be irreducible. Polynomials evaluated at x=c give sequences with e.g.f. exp(1-cx-exp(-x)).

%H S. de Wannemacker, T. Laffey and R. Osburn, <a href="http://arXiv.org/abs/math.NT/0608085">On a conjecture of Wilf</a>

%F Let P be the lower-triangular Pascal-matrix, PE = exp(P-I) a matrix- exponential in exact integer arithmetic (or PE = lim exp(P)/exp(1) as limit of the exponential) then A= PE^-1 and a(n) = A[n, read sequentially]. - Gottfried Helms, Apr 08 2007

%e Triangle starts:

%e 1,

%e -1,1,

%e 0,-2,1,

%e 1,0,-3,1,

%e 1,4,0,-4,1,

%e -2,5,10,0,-5,1,

%e -9,-12,15,20,0,-6,1,

%e -9,-63,-42,35,35,0,-7,1,

%Y First column is A000587 (Uppuluri Carpenter numbers) which is also the negative of the row sums (=P(n, 1)). Polynomials evaluated at 2 are A074051, at -1 A109747.

%K easy,tabl,sign

%O 0,5

%A _Gottfried Helms_, Apr 08 2007

%E Edited by _Ralf Stephan_, May 12 2007

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Last modified August 21 23:04 EDT 2019. Contains 326169 sequences. (Running on oeis4.)