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%I #68 Sep 08 2022 08:46:17
%S 1,2,14,3,28,50,4,42,100,110,5,56,150,220,194,6,70,200,330,388,302,7,
%T 84,250,440,582,604,434,8,98,300,550,776,906,868,590,9,112,350,660,
%U 970,1208,1302,1180,770,10,126,400,770,1164,1510,1736,1770,1540,974,11,140,450,880,1358,1812,2170,2360,2310,1948,1202
%N Triangle read by rows: T(n,k) = 2*(6*k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.
%C The row sums of the triangle provide the positive terms of A000583.
%C Similar triangles can be generated by the formula P(n,k,m) = (Q(k+1,m)-Q(k,m))*(n+1-k), where Q(i,r) = i^r-(i-1)^r, 0 < k <= n, and P(n,0,m) = n+1. T(n,k) is the case m=4, that is T(n,k) = P(n,k,4).
%H Michael De Vlieger, <a href="/A276189/b276189.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150, flattened)
%F T(n,n-h) = (h+1)*A005914(n-h) for 0 <= h <= n. Therefore, the main diagonal of the triangle is A005914.
%F Sum_{k=0..n} T(n,k) = T(n,0)^4 = A000583(n+1).
%e Triangle starts:
%e ----------------------------------------------
%e n \ k | 0 1 2 3 4 5 6 7
%e ----------------------------------------------
%e 0 | 1;
%e 1 | 2, 14;
%e 2 | 3, 28, 50;
%e 3 | 4, 42, 100, 110;
%e 4 | 5, 56, 150, 220, 194;
%e 5 | 6, 70, 200, 330, 388, 302;
%e 6 | 7, 84, 250, 440, 582, 604, 434;
%e 7 | 8, 98, 300, 550, 776, 906, 868, 590;
%e ...
%t Table[If[k == 0, n + 1, 2 (6 k^2 + 1) (n + 1 - k)], {n, 0, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Aug 29 2016 *)
%o (Magma) [IsZero(k) select n+1 else 2*(6*k^2+1)*(n+1-k): k in [0..n], n in [0..10]];
%o (Magma) /* As triangle (see the second comment): */ m:=4; Q:=func<i,r | i^r-(i-1)^r>; P:=func<n,k,m | IsZero(k) select n+1 else (Q(k+1,m)-Q(k,m))*(n+1-k)>; [[P(n,k,m): k in [0..n]]: n in [0..10]];
%Y Cf. A000583, A005914, A276158.
%K nonn,tabl
%O 0,2
%A _Stefano Maruelli_, Aug 24 2016
%E Corrected, rewritten and extended by _Bruno Berselli_, Aug 31 2016
%E a(40) ff. corrected by _Georg Fischer_, Nov 08 2021
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