A182440 - OEIS

%I #32 Dec 10 2016 17:04:17

%S 0,14,4,0,14,7,16,1,14,8,126,40,2,14,10,770,287,60,3,14,11,4524,1730,

%T 420,72,4,14,12,26404,10141,2522,497,88,5,14,13,153930,59164,14774,

%U 2978,602,100,6,14,14,897206

%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).

%C This is a table related to A001477 interpreted as a square array of the onnegative 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) such that G(i) = 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 square array A001477.

%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, a(k+1,i+1)-a(k,i+1 = A210695(i) if a(k + 1,0) - a(k,0) = 1; while a(k+1,i+1)-a(k,i+1 = A001108(i) if a(k+1,0) - a(k,0) = 0.

%C A related property is that a(k+1,1+n) - a(k,1+n) is divisible by A143608(n) for all k.

%F a(k,0) equals the positive argument of the largest triangular number equal to or less than 14*k (= A214206(k) which = A003056(14*k)).

%F a(k,1) equals 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*A214206(k).

%e For i = 1,2,3,4 ..., a(1,i)*a(1,i+1) = 14*1,1*40,40*287,287*1730, ...; and, each product is 4 more than a triangular number and thus lies in row 4 of square array A001477.

%t highTri = Compile[{{S1,_Integer}},Module[{xS0=0,xS1=S1},

%t While[xS1-xS0*(xS0+1)/2>xS0,xS0++];

%t xS0]];

%t overTri = Compile[{{S2,_Integer}},Module[{xS0=0,xS2=S2},

%t While[xS2-xS0*(xS0+1)/2>xS0,xS0++];

%t xS2 - (xS0*(1+xS0)/2)]];

%t K1 = 0;

%t m = 14;table=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;

%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[m,d];

%t 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];

%t K1++]][[2]];

%t a=1;

%t list5 = Reap[While[a<11,b=a;

%t While[b>0,Sow[table[[b,a+1-b]]];b--];a++]][[2,1]];

%t list5

%Y Cf. A001108, A210695, A143608, A182431, A182439, A182441

%K nonn,tabl

%O 0,2

%A _Kenneth J Ramsey_, Apr 28 2012