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%I #15 Sep 08 2022 08:45:28
%S 1,2,3,3,9,6,4,18,24,10,5,30,60,50,15,6,45,120,150,90,21,7,63,210,350,
%T 315,147,28,8,84,336,700,840,588,224,36,9,108,504,1260,1890,1764,1008,
%U 324,45,10,135,720,2100,3780,4410,3360,1620,450,55
%N Triangle read by rows: T(n,k) = k*(k+1)*binomial(n,k)/2 (1 <= k <= n).
%C Row sums = A001793: (1, 5, 18, 56, 160, 432, ...).
%C Triangle is P*M, where P is the Pascal triangle as an infinite lower triangular matrix and M is an infinite bidiagonal matrix with (1,3,6,10,...) in the main diagonal and in the subdiagonal.
%C This number triangle can be used as a control sequence when listing combinations of subsets as in Pascals triangle by assigning a number to each element that corresponds to the n:th subset that the element belongs to. One then gets number blocks whose sums are the terms in this number triangle. - _Mats Granvik_, Jan 14 2009
%H G. C. Greubel, <a href="/A124932/b124932.txt">Rows n = 1..100 of triangle, flattened</a>
%F T(n,k) = binomial(k+1,2)*binomial(n,k). - _G. C. Greubel_, Nov 19 2019
%e First few rows of the triangle:
%e 1;
%e 2, 3;
%e 3, 9, 6;
%e 4, 18, 24, 10;
%e 5, 30, 60, 50, 15;
%e 6, 45, 120, 150, 90, 21;
%e 7, 63, 210, 350, 315, 147, 28;
%e ...
%e From _Mats Granvik_, Dec 18 2009: (Start)
%e The numbers in this triangle are sums of the following recursive number blocks:
%e 1................................
%e .................................
%e 11.....12........................
%e .................................
%e 111....112....123................
%e .......122.......................
%e .................................
%e 1111...1112...1123...1234........
%e .......1122...1223...............
%e .......1222...1233...............
%e .................................
%e 11111..11112..11123..11234..12345
%e .......11122..11223..12234.......
%e .......11222..12223..12334.......
%e .......12222..11233..12344.......
%e ..............12233..............
%e ..............12333..............
%e .................................
%e (End)
%p T:=(n,k)->k*(k+1)*binomial(n,k)/2: for n from 1 to 12 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
%t Table[Binomial[k + 1, 2]*Binomial[n, k], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, Nov 19 2019 *)
%o (PARI) T(n,k) = binomial(k+1,2)*binomial(n,k); \\ _G. C. Greubel_, Nov 19 2019
%o (Magma) B:=Binomial; [B(k+1,2)*B(n,k): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Nov 19 2019
%o (Sage) b=binomial; [[b(k+1,2)*b(n,k) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Nov 19 2019
%o (GAP) B:=Binomial;; Flat(List([1..12], n-> List([1..n], k-> B(k+1,2)* B(n,k) ))); # _G. C. Greubel_, Nov 19 2019
%Y Cf. A001793.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, Nov 12 2006
%E Edited by _N. J. A. Sloane_, Nov 24 2006
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