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%I #61 Dec 10 2016 22:36:38
%S 0,14,4,0,14,7,12,1,14,8,98,4,2,14,10,602,35,0,3,14,11,3540,218,0,4,4,
%T 14,12,20664,1285,2,21,4,5,14,13,120470,7504,14,122,14,8,6,14,14,
%U 702182,43751,84,711,74,35,12,7,14,15
%N Table, read by antidiagonals, in which the n-th row comprises A214206(n) 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).
%C This is a table related to the square array of nonnegative integers, A001477. Each row k contains the positive argument of the largest triangular number equal to or less than 14*k in column 0 and a corresponding 2nd-order recursive sequence G(k) beginning at position a(k,1). Each term G(i) is the same as a(k,i+1). If the product 14*k appears in row "r" of the square array A001477, then 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 table A001477. If the product is less than (r^2 +3*r -2)/2, then the product less r would be a triangular number, i.e., still lie in the same row assumed to contain all numbers n that equal a triangular number + r. For example, 3 is a triangular number and appears in row 0 of A001477, but if the rows could take negative indices, A001477(2,-1) would be a 3 so 3 can be said to also lie in row 2. See A182102 for a table of the arguments of the triangular numbers G(i)*G(i+1) - r.
%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, if a(k+1,0) - a(k,0) = 1 then a(k+1,i+1) - a(k,i+1) equals A182435(i) for all i. Also, for i>0, A143608(i) divides a(k+1,i+1)-a(k,i+1) for all k.
%F a(k,0) equals the largest m such that m*(m+1)/2 is less than or equal to 14*k.
%F a(k,1) = 14; a(k,2) = k.
%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*a(k,0).
%e The Table begins:
%e 0 14 0 12 98 602 3540 ...
%e 4 14 1 4 35 218 1285 ...
%e 7 14 2 0 0 2 14 ...
%e 8 14 3 4 21 122 711 ...
%e 10 14 4 4 14 74 424 ...
%e 11 14 5 8 35 194 1121 ...
%e 12 14 6 12 56 314 1818 ...
%e 13 14 7 16 77 434 2515 ...
%e 14 14 8 20 98 554 3212 ...
%e 15 14 9 24 119 674 3909 ...
%e 16 14 10 28 140 794 4606 ...
%e 17 14 11 32 161 914 5303 ...
%e 17 14 12 40 210 1202 6984 ...
%e ...
%e Note that 14*0,0*12,12*98, 98*602 etc are each 0 more than a triangular number and are in row 0 of square array A001477; while 14*1, 1*4, 4*35, 35*218 etc are each 4 more than a triangular number and thus can be said to 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 K1 - m + X); K3 = 6 K2 - K1 + X; 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[m,d]; Sow[K1,e]; 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, Sow[tab[[b,a+1-b]]]; b--]; a++]][[2,1]]; list5
%Y Cf. A182189, A182190, A182435.
%K nonn,tabl
%O 0,2
%A _Kenneth J Ramsey_, Apr 28 2012
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