A121726 - OEIS

%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