|
%I #18 Sep 08 2022 08:45:23
%S 1,1,1,1,2,1,1,4,4,1,1,7,10,7,1,1,11,19,19,11,1,1,16,31,37,31,16,1,1,
%T 22,46,61,61,46,22,1,1,29,64,91,101,91,64,29,1,1,37,85,127,151,151,
%U 127,85,37,1,1,46,109,169,211,226,211,169,109,46,1
%N Triangle T(n,m) read by rows: T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1.
%C From _Paul Barry_, Jan 07 2009: (Start)
%C This triangle follows a general construction method as follows: Let a(n) be an integer sequence with a(0)=1, a(1)=1. Then T(n,k,r) := [k<=n](1+r*a(k)*a(n-k)) defines a symmetrical triangle.
%C Row sums are n + 1 + r*Sum_{k=0..n} a(k)*a(n-k) and central coefficients are 1+r*a(n)^2.
%C Here a(n) = C(n+1,2) and r=1.
%C Row sums are A154322 and central coefficients are A154323. (End)
%H G. C. Greubel, <a href="/A113582/b113582.txt">Rows n=0..100 of triangle, flattened</a>
%F T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1.
%e {1},
%e {1, 1},
%e {1, 2, 1},
%e {1, 4, 4, 1},
%e {1, 7, 10, 7, 1},
%e {1, 11, 19, 19, 11, 1},
%e {1, 16, 31, 37, 31, 16, 1},
%e {1, 22, 46, 61, 61, 46, 22, 1},
%e {1, 29, 64, 91, 101, 91, 64, 29, 1},
%e {1, 37, 85, 127, 151, 151, 127, 85, 37, 1},
%e {1, 46, 109, 169, 211, 226, 211, 169, 109, 46, 1}
%t t[n_, m_] = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]//Flatten
%o (Magma) /* As triangle: */ [[(n-m)*(n-m+1)*m*(m+1)/4+1: m in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Sep 12 2016
%o (PARI) for(n=0,15, for(k=0,n, print1((n-k)*(n-k+1)*k*(k+1)/4 + 1, ", "))) \\ _G. C. Greubel_, Aug 31 2018
%K nonn,tabl,easy
%O 1,5
%A _Roger L. Bagula_, Aug 25 2008
|
