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Search: id:A081733
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| A081733 |
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T[n,k]=Triangle of terms in Euler-polynomials 2^n EulerE[n,x=1/2]summing to n-th Euler number EulerE[n]. |
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+0
2
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| 0, -1, 1, 0, -2, 1, 2, 0, -3, 1, 0, 8, 0, -4, 1, -16, 0, 20, 0, -5, 1, 0, -96, 0, 40, 0, -6, 1, 272, 0, -336, 0, 70, 0, -7, 1, 0, 2176, 0, -896, 0, 112, 0, -8, 1, -7936, 0, 9792, 0, -2016, 0, 168, 0, -9, 1, 0, -79360, 0, 32640, 0, -4032, 0, 240, 0, -10, 1, 353792, 0, -436480, 0, 89760, 0, -7392, 0, 330, 0, -11, 1, 0, 4245504, 0
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Sum of row n equals EulerE[n].
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FORMULA
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T(n,k) = C(n,k)2^(n-k)E_{n-k}(0) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. [From Peter Luschny (peter(AT)luschny.de), Jan 25 2009]
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EXAMPLE
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The coefficient lists of the first 4 Euler polynomials are {-1/2, 1}, {0, -1, 1}, {1/4, 0, -3/2, 1}, {0, 1, 0, -2, 1}. Multiply by 2^n and by (1/2)^{0,1,2,..n} to get {-1, 1}, {0, -2, 1}, {2, 0, -3, 1}, {0, 8, 0, -4, 1}
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MAPLE
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Contribution from Peter Luschny (peter(AT)luschny.de), Jan 25 2009: (Start)
T := (n, k) -> 2^(n-k)*coeff(euler(n, x), x, k):
T := (n, k) -> 2^(n-k)*binomial(n, k)*euler(n-k, 0): (End)
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MATHEMATICA
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Table[2^n (1/2)^(Range[0, n]) CoefficientList[EulerE[n, x], x], {n, 16}]
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CROSSREFS
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Cf. A002425, A058940, A006519.
Sequence in context: A127471 A071485 A127969 this_sequence A102587 A159834 A147786
Adjacent sequences: A081730 A081731 A081732 this_sequence A081734 A081735 A081736
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KEYWORD
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sign,tabl
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AUTHOR
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Wouter Meeussen (wouter.meeussen(AT)pandora.be), Apr 06 2003
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