A182441 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).

0, 0, 4, 14, 1, 7, 114, 14, 2, 8, 700, 131, 14, 3, 10, 4116, 820, 144, 14, 4, 11, 24026, 4837, 912, 149, 14, 5, 12, 140070, 28250, 5390, 948, 158, 14, 6, 13, 816424, 164711, 31490, 5607, 1012, 163, 14, 7, 14, 4758504

Offset: 0

Author

Kenneth J Ramsey, Apr 28 2012

Keywords

• nonn
• tabl

Comments

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.

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).

Examples

			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.
		

Crossrefs

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

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

Formula

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

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

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).