|
%I #14 Aug 01 2023 07:46:33
%S 1,1,2,1,5,4,1,9,18,8,1,14,51,54,16,1,20,115,215,145,32,1,27,225,650,
%T 750,363,64,1,35,399,1645,2870,2310,868,128,1,44,658,3668,8995,10724,
%U 6538,2012,256,1,54,1026,7434,24381,40257,35658,17442,4563,512,1,65,1530,13980,59115,129150,156135,109020,44595,10185,1024
%N The following triangle is based on Pascal's triangle. The r-th term of the n-th row is sum of C(n,r) successive integers so that the sum of all the terms of the row is (2^n)*(2^n+1)/2, the 2^n -th triangular number. Sequence contains the triangle read by rows.
%C The leading diagonal contains 2^n.
%F T(n,0) = 1, T(n,k) = C(A008949(n,k)+1, 2) - C(A008949(n,k-1)+1, 2) = C(n,k)*(A008949(n+1,k)+1)/2 for k>0. - _Franklin T. Adams-Watters_, Sep 27 2006
%e Row for n = 3 is 1, (2+3+4), (5+6+7), 8.
%e Triangle begins:
%e 1
%e 1 2
%e 1 5 4
%e 1 9 18 8
%e 1 14 51 54 16
%e ...
%t A008949[n_, k_] := Sum[Binomial[n, j], {j, 0, k}];
%t A112358[n_, k_] := If[k == 0, 1, Binomial[A008949[n, k] + 1, 2] - Binomial[A008949[n, k - 1] + 1, 2]];
%t Table[A112358[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 01 2023 *)
%Y Cf. A112356, A112357, A112359.
%Y Cf. A008949.
%K easy,nonn,tabl
%O 0,3
%A _Amarnath Murthy_, Sep 05 2005
%E More terms from Amber Reardon (alr5041(AT)psu.edu) and Vincent M. DelPrince (vmd5003(AT)psu.edu), Oct 04 2005
|
