A182440 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).

0, 14, 4, 0, 14, 7, 16, 1, 14, 8, 126, 40, 2, 14, 10, 770, 287, 60, 3, 14, 11, 4524, 1730, 420, 72, 4, 14, 12, 26404, 10141, 2522, 497, 88, 5, 14, 13, 153930, 59164, 14774, 2978, 602, 100, 6, 14, 14, 897206

Offset: 0

Author

Kenneth J Ramsey, Apr 28 2012

Keywords

• nonn
• tabl

Comments

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.

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.

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

Examples

			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.
		

Crossrefs

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

Programs

Mathematica

highTri = Compile[{{S1,_Integer}},Module[{xS0=0,xS1=S1},
While[xS1-xS0*(xS0+1)/2>xS0,xS0++];
xS0]];
overTri = Compile[{{S2,_Integer}},Module[{xS0=0,xS2=S2},
While[xS2-xS0*(xS0+1)/2>xS0,xS0++];
xS2 - (xS0*(1+xS0)/2)]];
K1 = 0;
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;
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];
  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[table[[b,a+1-b]]];b--];a++]][[2,1]];
list5

Formula

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)).

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

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).