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A129334
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Triangle T(n,k) read by rows: inverse of the matrix PE = exp(P)/exp(1) given in A011971.
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0
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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, -9, -63, -42, 35, 35, 0, -7, 1, 50, -72, -252, -112, 70, 56, 0, -8, 1, 267, 450, -324, -756, -252, 126, 84, 0, -9, 1, 413, 2670, 2250, -1080, -1890, -504, 210, 120
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| 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.
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)).
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LINKS
| S. de Wannemacker, T. Laffey and R. Osburn, On a conjecture of Wilf
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FORMULA
| 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
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EXAMPLE
| Triangle starts:
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,
-9,-63,-42,35,35,0,-7,1,
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CROSSREFS
| 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.
Sequence in context: A136481 A100218 A098599 * A116399 A116405 A029352
Adjacent sequences: A129331 A129332 A129333 * A129335 A129336 A129337
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KEYWORD
| easy,tabl,sign
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AUTHOR
| Gottfried Helms (helms(AT)uni-kassel.de), Apr 08 2007
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EXTENSIONS
| Edited by Ralf Stephan, May 12 2007
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