

A168423


Triangle read by rows: expansion of (1  x)/(exp(t)*(1  x*exp(t*(1  x))))


0



1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 21, 21, 1, 1, 1, 1, 51, 161, 51, 1, 1, 1, 1, 113, 813, 813, 113, 1, 1, 1, 1, 239, 3361, 7631, 3361, 239, 1, 1, 1, 1, 493, 12421, 53833, 53833, 12421, 493, 1, 1, 1, 1, 1003, 42865, 320107, 607009, 320107, 42865
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OFFSET

0,14


COMMENTS

This sequence was derived from the Eulerian number umbral calculus expansion and A046802 by taking the exp(t) term and inverting it.
What is interesting here is the '1,1' terms that appear.
I had thought I would get "1,5,1" not "1,7,1" from this function.
An OEIS search came up with A046739 which has the same internal symmetric number structure.
Inverse binomial transform of Eulerian numbers A123125. [Paul Barry, May 10 2011]


LINKS

Table of n, a(n) for n=0..63.


FORMULA

E.g.f. sum(T(n,k) t^n/n! x^k) = p(x,t) = (1  x)/(exp(t)*(1  x*exp(t*(1  x))))
T(n,k)=sum{j=0..n, (1)^(nj)*C(n,j)*A123125(j,k)}. [Paul Barry, May 10 2011]


EXAMPLE

{1},
{1, 1},
{1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 7, 1},
{1, 1, 1, 21, 21, 1},
{1, 1, 1, 51, 161, 51, 1},
{1, 1, 1, 113, 813, 813, 113, 1},
{1, 1, 1, 239, 3361, 7631, 3361, 239, 1},
{1, 1, 1, 493, 12421, 53833, 53833, 12421, 493, 1},
{1, 1, 1, 1003, 42865, 320107, 607009, 320107, 42865, 1003, 1}


MATHEMATICA

p[t_] = (1  x)/(Exp[t]*(1  x*Exp[t*(1  x)]))
a = Table[ CoefficientList[FullSimplify[ExpandAll[n!*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]]], x], {n, 0, 10}];
Flatten[a]


CROSSREFS

Cf. A046802, A046739, A000166 (row sums), A123125.
Sequence in context: A284097 A091258 A174544 * A336844 A072101 A088840
Adjacent sequences: A168420 A168421 A168422 * A168424 A168425 A168426


KEYWORD

sign,tabl


AUTHOR

Roger L. Bagula, Nov 25 2009


STATUS

approved



