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Search: id:A129334
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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 1
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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; listen)
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OFFSET
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0,5
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COMMENT
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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
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S. de Wannemacker, T. Laffey and R. Osburn, On a conjecture of Wilf
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FORMULA
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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
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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
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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 this_sequence A116399 A116405 A029352
Adjacent sequences: A129331 A129332 A129333 this_sequence A129335 A129336 A129337
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KEYWORD
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easy,tabl,sign
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AUTHOR
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Gottfried Helms (helms(AT)uni-kassel.de), Apr 08 2007
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EXTENSIONS
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Edited by Ralf Stephan, May 12 2007
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