This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
The triangle begins:
1: 1
2: 1 1
3: 2 3 1
4: 6 8 6 1
5: 24 30 20 10 1
6: 120 144 90 40 15 1
...
From _Peter Luschny_, Nov 19 2020: (Start):
The combinatorial interpretation is illustrated by this computation of row 6:
6! / aut([6]) = 720 / A339033(6, 1) = 720/6 = 120 = T(6, 1)
6! / aut([5, 1]) = 720 / A339033(6, 2) = 720/5 = 144 = T(6, 2)
6! / aut([4, 1, 1]) = 720 / A339033(6, 3) = 720/8 = 90 = T(6, 3)
6! / aut([3, 1, 1, 1]) = 720 / A339033(6, 4) = 720/18 = 40 = T(6, 4)
6! / aut([2, 1, 1, 1, 1]) = 720 / A339033(6, 5) = 720/48 = 15 = T(6, 5)
6! / aut([1, 1, 1, 1, 1, 1]) = 720 / A339033(6, 6) = 720/720 = 1 = T(6, 6)
-------------------------------------------------------------------------------
Sum: 410 = A121726(6)
(End)
f[list_] :=Total[list]!/Apply[Times, list]/Apply[Times, Map[Length, Split[list]]!];Table[Append[Map[f, Select[Partitions[n], Count[#, Except[1]] == 1 &]], 1], {n,1, 10}] // Grid (* Geoffrey Critzer, Nov 07 2015 *)
def A092271(n, k): if n == k: return 1 return factorial(n) // ((n + 1 - k)*factorial(k - 1)) for n in (1..9): print(n, [A092271(n, k) for k in (1..n)]) def A092271Row(n): if n == 0: return [1] f = factorial(n); S = [] for k in range(n,0,-1): for p in Partitions(n, max_part=k, inner=[k], length=n+1-k): S.append(f // p.aut()) return S for n in (1..9): print(A092271Row(n)) # Peter Luschny, Nov 20 2020
A000522 begins 1 2 5 16 65 326 ...
with sums 1 3 8 24 89 415 ...
so sequence begins 1 2 6 21 85 410 ...
.
From _Peter Luschny_, Nov 19 2020: (Start):
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
--------------------------------------------------------------
Sum: a(5) = 85
(End)
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 *)
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
def A121726(n): 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: def A121726(n): 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 print([A121726(n) for n in (1..23)]) # Peter Luschny, Nov 20 2020
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