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A162509
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Row sums of the absolute values of a triangular array related to the Bernoulli numbers.
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6
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1, 1, 4, 20, 124, 932, 8284, 85220, 997084, 13082852, 190320604, 3040770020, 52937870044, 997533561572, 20228969244124, 439283696014820, 10170742982007004, 250110224694309092, 6510327792455418844, 178832105312143131620, 5169772417850111583964
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OFFSET
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0,3
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COMMENTS
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Let T(n,k) = sum_{v=0..k} (-1)^v*v*binomial(k,v)*(v+1)^(n-1) for n >= 1, k >= 1 and additionally T(0,0) = 1. Then a(n) = sum_{k=0..n} abs(T(n,k)).
a(n) appears to be the total number of subsets over all chains of the poset on the powerset of {1,2,...,n-1} ordered by set inclusion. That is, a(n) = Sum_{k=0..n} A038719(n,k)*(k+1). For example a(2)=4 because there are three chains: {}; {1}; {},{1}; and there are 4 total subsets. - Geoffrey Critzer, Nov 28 2014
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LINKS
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FORMULA
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G.f.: 1+x/(1-4x/(1-x/(1-6x/(1-2x/(1-8x/(1-3x/(1-10x/(1-4x/1-....)))))))) (continued fraction). - Philippe Deléham, Nov 22 2011
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(3*k+4) - 2*x^2*(k+1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 03 2013
a(n + 1) = sum {k >= 0} (k*(k + 1)^n)/2^(k + 1) for n >= 0. Comparison with the formula A000670(n) = sum {k >= 0} (k^n)/2^(k + 1) yields a(n + 1) = sum {k = 0..n} binomial(n,k)*A000670(k + 1). - Peter Bala, Jul 21 2014
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MAPLE
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A162508 := proc(n, k) local v; if n=0 and k=0 then 1 else
add((-1)^v*v*binomial(k, v)*(v+1)^(n-1), v=0..k) fi end:
a := proc(n) local k; add(abs(A162508(n, k)), k=0..n) end:
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MATHEMATICA
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t[0, 0] = 1; t[n_, k_] := Sum[(-1)^v*v*Binomial[k, v]*(v+1)^(n-1), {v, 0, k}]; a[n_] := Sum[Abs[t[n, k]], {k, 0, n}]; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 28 2013 *)
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PROG
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(Sage)
return add(abs(A162508(n, k)) for k in (0..n))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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