%I #47 Dec 10 2016 16:57:53

%S 0,0,4,14,1,7,114,14,2,8,700,131,14,3,10,4116,820,144,14,4,11,24026,

%T 4837,912,149,14,5,12,140070,28250,5390,948,158,14,6,13,816424,164711,

%U 31490,5607,1012,163,14,7,14,4758504

%N Table, read by antidiagonals, in which the n-th row comprises A214206(n) in position 0 followed by a second order recursive series G in which each product G(i)*G(i+1) lies in the same row of A001477 (interpreted as a square array - see below).

%C This is a table related to the square array of the nonnegative integers (A001477). Each row k contains A003056(14*k) in column 0 and a corresponding 2nd order recursive sequence G(k) beginning at position a(k,1). That is each term G(i) is a(k,i+1). If A002262(14*n) is "r", the product of adjacent terms G(i)*G(i+1) if greater than (r^2 + 3*r - 2)/2, is always in row "r" of the square array A001477. If the product is less than (r^2 + 3*r -2)/2, then the product could still be said to lie in the same row r since the product is equal to the sum of a triangular number + r, which is a property of all numbers in row r of the square array A002262.

%C A property of this table is that a(k+1,i)-a(k,i) directly depends on the value of a(k+1,0)-a(k,0) in the same manner regardless of the value of k. For instance, wherever a(k+1,0)-a(k,0) = 0, a(k+1,i+1)-a(k,i+1) = A212329. Also, a(k+1,n+2)-a(k,n+2) is divisible by A143608(n).

%F a(k,0) equals the largest m such that m*(m+1)/2 <= 14*k (A003056(14*k)).

%F a(k,1) equals k; a(k,2) = 14.

%F For i > 2, a(k,i) = 6*a(k,i-1) -a (k,i-2) + G_k where G_k is a constant equal to 28 + 2*k + 2 + 4*A003056(14*k).

%e For i>0 a(0,i) * a(0,i+1) = 0*14,14*114,114*700,700*4116,etc. which are all triangular numbers and lie in row 0 of square array A001477, while a(1,i)*a(1.i+1) = 1*14, 14*131, 131*820, 820*4837 etc. which are all 4 more than a triangular number and lie in row 4 of square array A001477.

%t highTri = Compile[{{S1,_Integer}}, Module[{xS0=0, xS1=S1}, While[xS1-xS0*(xS0+1)/2 > xS0, xS0++]; xS0]];

%t overTri = Compile[{{S2,_Integer}}, Module[{xS0=0, xS2=S2}, While[xS2-xS0*(xS0+1)/2 > xS0, xS0++]; xS2 - (xS0*(1+xS0)/2)]];

%t K1 = 0; m = 14; tab=Reap[While[K1<16,J1=highTri[m*K1]; X = 2*(m+K1+(J1*2+1)); K2 = (6 m - K1 + X); K3 = 6 K2 - m + X;

%t K4 = 6 K3 - K2 + X; K5 = 6 K4 -K3 + X; K6 = 6*K5 - K4 + X; K7 = 6*K6-K5+X; K8 = 6*K7-K6+X; Sow[J1,c]; Sow[K1,d]; Sow[m,e];

%t Sow[K2,f]; Sow[K3,g]; Sow[K4,h];

%t Sow[K5,i]; Sow[K6,j]; Sow[K7,k]; Sow[K8,l]; K1++]][[2]]; a=1; list5 = Reap[While[a<11, b=a; While[b>0,

%t Sow[tab[[b,a+1-b]]]; b--]; a++]][[2,1]]; list5

%Y Cf. A212329, A182431, A182439, A182440, A143608.

%K nonn,tabl

%O 0,3

%A _Kenneth J Ramsey_, Apr 28 2012