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Revision History for A357831 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
a(n) = Sum_{k=0..floor(n/3)} 2^k * |Stirling1(n,3*k)|.
(history; published version)
#14 by Michael De Vlieger at Sat Oct 15 08:08:44 EDT 2022
STATUS

reviewed

approved

#13 by Joerg Arndt at Sat Oct 15 05:57:24 EDT 2022
STATUS

proposed

reviewed

#12 by Seiichi Manyama at Sat Oct 15 05:23:47 EDT 2022
STATUS

editing

proposed

#11 by Seiichi Manyama at Sat Oct 15 05:23:43 EDT 2022
CROSSREFS
STATUS

proposed

editing

#10 by Seiichi Manyama at Sat Oct 15 05:13:29 EDT 2022
STATUS

editing

proposed

#9 by Seiichi Manyama at Fri Oct 14 15:00:00 EDT 2022
FORMULA

a(n) = ( (2^(1/3))_n + (2^(1/3)*w)_n + (2^(1/3)*w^2)_n )/3, where (x)_n is the Pochhammer symbol.

#8 by Seiichi Manyama at Fri Oct 14 14:13:59 EDT 2022
FORMULA

Let w = exp(2*Pi*i/3) and set F(x) = (exp(x) + w^2*exp(w*x) + w*exp(w^2*x))/3 = x 1 + x^43/43! + x^76/76! + ... . Then the e.g.f. for the sequence is F(-2^(1/3) * log(1-x))/(2^(1/3)).

#7 by Seiichi Manyama at Fri Oct 14 13:50:37 EDT 2022
FORMULA

Let w = exp(2*Pi*i/3) and set F(x) = (exp(x) + w^2*exp(w*x) + w*exp(w^2*x))/3 = x + x^4/4! + x^7/7! + ... . Then the e.g.f. for the sequence is F(-2^(1/3) * log(1-x))/(2^(1/3)).

#6 by Seiichi Manyama at Fri Oct 14 13:03:23 EDT 2022
DATA

1, 0, 0, 2, 12, 70, 454, 3332, 27552, 254400, 2598852, 29125932, 355455468, 4693396656, 66671326176, 1013916648840, 16436063079552, 282920894841096, 5153797995148296, 99052313167391760, 2003040751641857856, 42513854724369719136, 944959706480298199824

#5 by Seiichi Manyama at Fri Oct 14 13:02:45 EDT 2022
PROG

(PARI) my(N=30, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N\3, 2^k*(-log(1-x))^(3*k)/(3*k)!)))