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#232 by Alois P. Heinz at Sat Jun 01 11:45:09 EDT 2024
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#231 by Robert C. Lyons at Sat Jun 01 11:42:10 EDT 2024
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#230 by Robert C. Lyons at Sat Jun 01 11:42:08 EDT 2024
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| PROG
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(SageSageMath)
a(n):= if n=0 then 1 else (sum((n+k-1)!*sum(((-1)^(j)/(k-j)!*sum((1/i! *stirling1(n-i+j-1, j-i))/(n-i+j-1)!, i, 0, j))*2^(n-j-1), j, 0, k), k, 0, n-1)); \\ _)); /* _Vladimir Kruchinin_, Nov 11 2012 */
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approved
editing
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#229 by Joerg Arndt at Fri Apr 30 01:35:01 EDT 2021
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#228 by Michel Marcus at Fri Apr 30 00:44:27 EDT 2021
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#227 by Michael De Vlieger at Thu Apr 29 16:25:15 EDT 2021
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#226 by Michael De Vlieger at Thu Apr 29 16:25:13 EDT 2021
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| LINKS
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Paul Barry, <a href="https://arxiv.org/abs/2101.06713">On the inversion of Riordan arrays</a>, arXiv:2101.06713 [math.CO], 2021.
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approved
editing
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#225 by Peter Luschny at Sat Feb 13 09:14:36 EST 2021
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#224 by Peter Luschny at Sat Feb 13 09:14:30 EST 2021
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| FORMULA
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Similarly, c(0)=1, c(n)=-a(n+1) are the alternating row sums of the second-order Eulerian numbers A340556, or c(n) = E2poly(n,-1). - Peter Luschny, Feb 13 2021
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reviewed
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#223 by Joerg Arndt at Sat Feb 13 08:55:47 EST 2021
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