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A121726
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Sum sequence A000522 then subtract 0,1,2,3,4,5,...
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3
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1, 2, 6, 21, 85, 410, 2366, 16065, 125665, 1112074, 10976174, 119481285, 1421542629, 18348340114, 255323504918, 3809950976993, 60683990530209, 1027542662934898, 18430998766219318, 349096664728623317, 6962409983976703317, 145841989688186383338, 3201192743180799343822
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OFFSET
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1,2
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COMMENTS
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Let aut(p) denote the size of the centralizer of the partition p (see A339016 for the definition). Then a(n) = Sum_{p in P} n!/aut(p), where P are the partitions of n with largest part k and length n + 1 - k. - Peter Luschny, Nov 19 2020
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LINKS
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FORMULA
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EXAMPLE
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with sums 1 3 8 24 89 415 ...
so sequence begins 1 2 6 21 85 410 ...
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The combinatorial interpretation is illustrated by this computation of a(5):
5! / aut([5]) = 120 / A339033(5, 1) = 120/5 = 24
5! / aut([4, 1]) = 120 / A339033(5, 2) = 120/4 = 30
5! / aut([3, 1, 1]) = 120 / A339033(5, 3) = 120/6 = 20
5! / aut([2, 1, 1, 1]) = 120 / A339033(5, 4) = 120/12 = 10
5! / aut([1, 1, 1, 1, 1]) = 120 / A339033(5, 5) = 120/120 = 1
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Sum: a(5) = 85
(End)
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MATHEMATICA
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f[list_] :=Total[list]!/Apply[Times, list]/Apply[Times, Map[Length, Split[list]]!]; Table[Total[Map[f, Select[Partitions[n], Count[#, Except[1]] == 1 &]]] + 1, {n, 1, 20}] (* Geoffrey Critzer, Nov 07 2015 *)
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PROG
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(SageMath)
def h(n, k):
if n == k: return 1
return factorial(n)//((n + 1 - k)*factorial(k - 1))
return sum(h(n, k) for k in (1..n))
print([A121726(n) for n in (1..23)])
# Demonstrates the combinatorial view:
if n == 0: return 1
f = factorial(n); S = 0
for k in (0..n):
for p in Partitions(n, max_part=k, inner=[k], length=n+1-k):
S += (f // p.aut())
return S
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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