%I #25 Sep 15 2024 03:35:30
%S 1,2,6,21,85,410,2366,16065,125665,1112074,10976174,119481285,
%T 1421542629,18348340114,255323504918,3809950976993,60683990530209,
%U 1027542662934898,18430998766219318,349096664728623317,6962409983976703317,145841989688186383338,3201192743180799343822
%N Sum sequence A000522 then subtract 0,1,2,3,4,5,...
%C 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
%F a(n) = A006231(n) + 1 = A002104(n) - (n-1). - _Franklin T. Adams-Watters_, Aug 29 2006
%F E.g.f.: exp(x)*(log(1/(1-x)) - x + 1). - _Geoffrey Critzer_, Nov 07 2015
%e A000522 begins 1 2 5 16 65 326 ...
%e with sums 1 3 8 24 89 415 ...
%e so sequence begins 1 2 6 21 85 410 ...
%e .
%e From _Peter Luschny_, Nov 19 2020: (Start):
%e The combinatorial interpretation is illustrated by this computation of a(5):
%e 5! / aut([5]) = 120 / A339033(5, 1) = 120/5 = 24
%e 5! / aut([4, 1]) = 120 / A339033(5, 2) = 120/4 = 30
%e 5! / aut([3, 1, 1]) = 120 / A339033(5, 3) = 120/6 = 20
%e 5! / aut([2, 1, 1, 1]) = 120 / A339033(5, 4) = 120/12 = 10
%e 5! / aut([1, 1, 1, 1, 1]) = 120 / A339033(5, 5) = 120/120 = 1
%e --------------------------------------------------------------
%e Sum: a(5) = 85
%e (End)
%t 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 *)
%o (PARI) A000522(n)={ return( sum(k=0,n,n!/k!)) ; } A121726(n)={ return(sum(k=0,n-1,A000522(k))-n+1) ; } { for(n=1,25, print1(A121726(n),",") ; ) ; } \\ _R. J. Mathar_, Sep 02 2006
%o (SageMath)
%o def A121726(n):
%o def h(n, k):
%o if n == k: return 1
%o return factorial(n)//((n + 1 - k)*factorial(k - 1))
%o return sum(h(n, k) for k in (1..n))
%o print([A121726(n) for n in (1..23)])
%o # Demonstrates the combinatorial view:
%o def A121726(n):
%o if n == 0: return 1
%o f = factorial(n); S = 0
%o for k in (0..n):
%o for p in Partitions(n, max_part=k, inner=[k], length=n+1-k):
%o S += (f // p.aut())
%o return S
%o print([A121726(n) for n in (1..23)]) # _Peter Luschny_, Nov 20 2020
%Y Also the row sums of A092271.
%Y Cf. A000522, A006231, A002104, A339016, A339033.
%K easy,nonn
%O 1,2
%A _Alford Arnold_, Aug 17 2006
%E More terms from _Franklin T. Adams-Watters_, Aug 29 2006
%E More terms from _R. J. Mathar_, Sep 02 2006