%I #15 Oct 19 2017 11:05:27
%S 1,1,1,1,3,1,1,7,6,3,1,15,25,30,12,1,31,90,195,180,60,1,63,301,1050,
%T 1680,1260,360,1,127,966,5103,12600,15960,10080,2520,1,255,3025,23310,
%U 83412,158760,166320,90720,20160,1,511,9330,102315,510300,1369620
%N Triangle read by rows in which row n gives number of ways to partition n labeled elements into k pie slices allowing the pie to be turned over (n >= 1, 1 <= k <= n).
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%F Take triangle of Stirling numbers of second kind (A008277) and multiply k-th column by A001710(k) (order of alternating group A_k).
%e Triangle begins:
%e 1,
%e 1, 1,
%e 1, 3, 1,
%e 1, 7, 6, 3,
%e 1, 15, 25, 30, 12,
%e 1, 31, 90, 195, 180, 60,
%e 1, 63, 301, 1050, 1680, 1260, 360.
%e ...
%e For row n = 4 we have the following "pies":
%e . 1
%e ./ \
%e 2 . 3 . 12 .. 12 . 123
%e .\ / .. / \ .(..)..(..)
%e . 4 .. 3--4 . 34 .. 4 .. (1234)
%e k=4 .. k=3 ..k=2 . k=2 . k=1
%e (3)....(6)...(3)..(4)... (1)
%p A001710 := proc(n) if n < 2 then 1; else n!/2 ; fi ; end: A008277 := proc(n,k) combinat[stirling2](n,k) ; end: A133800 := proc(n,k) A008277(n,k)*A001710(k-1) ; end: for n from 1 to 10 do for k from 1 to n do printf("%d, ",A133800(n,k)) ; od: od: # _R. J. Mathar_, Jan 18 2008
%t A001710[n_] := If[n<2, 1, n!/2]; A008277[n_, k_] := StirlingS2[n, k]; A133800[n_, k_] := A008277[n, k]*A001710[k-1]; Table[A133800[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 20 2014, after _R. J. Mathar_ *)
%t (* A (n >= 0, k >= 0)-based version: *)
%t A133800[n_, k_] := k! StirlingS2[n+1, k+1] / If[k>1, 2, 1];
%t Table[A133800[n,k], {n,0,9}, {k,0,n}] // Flatten (* _Peter Luschny_, Oct 19 2017 *)
%Y Row sums give A032262. Diagonals give A000225, A000392, A032263, A133799, A001710.
%K nonn,tabl,easy
%O 1,5
%A Barry Cipra and _N. J. A. Sloane_, Jan 17 2008
%E More terms from _R. J. Mathar_, Jan 18 2008