A182439 Table a(k,i), 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, 110, 14, 2, 8, 672, 95, 14, 3, 10, 3948, 568, 84, 14, 4, 11, 23042, 3325, 492, 81, 14, 5, 12, 134330, 19394, 2870, 472, 74, 14, 6, 13, 782964, 113051, 16730, 2751, 424, 71, 14, 7, 14, 4563480, 658924, 97512, 16034, 2464, 404, 68, 14, 8, 15

Offset: 0

Author

Kenneth J Ramsey, Apr 28 2012

Keywords

• nonn
• tabl

Comments

This is a square array 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) in the rest of the row. Each second-order recursive series term G(i) corresponds to 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. If the product is less than (r^2 + 3*r -2)/2 then assuming the row can take negative indices, the product can still be said to lie in the same row r. For instance, 0, 1, 3, and 6 are each a triangular number and appear as the first 4 terms of row 0 of square array A001477. Note that in the next row and to the left of the 1, 3, and 6 are 2, 4 and 7 so going down a row and to the left in the square array increases the value by 1. Going down to the next row and to the left again would be 3, 5, and 8 so 3 which is 2 more than 1 would be in row 2 if that row were made to take the indices (2,-1).

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 example, a(k,2+n) - a(k,2+n) = A001652(n) for n=0,1,2,3,... whereever a(k+1,0) - a(k,0) = 1.

Also, a(k+1,2+n) - a(k,2+n) is divisible by A143608(n) for n>0 for all k.

Examples

			     0,     0,    14,   110,   672,  3948, 23042,134330,782964,
     4,     1,    14,    95,   568,  3325, 19394,113051,658924,
     7,     2,    14,    84,   492,  2870, 16730, 97512,568344,
     8,     3,    14,    81,   472,  2751, 16034, 93453,544684,
    10,     4,    14,    74,   424,  2464, 14354, 83654,487564,
    11,     5,    14,    71,   404,  2345, 13658, 79595,463904,
    12,     6,    14,    68,   384,  2226, 12962, 75536,440244.
Note that 0*14, 14*110, 110*672, etc. are all triangular numbers and thus appear in row 0 of square array A001477; while, 1*14, 14*95, 95*568, 568*3325, etc. are all 4 more than a triangular number and appear in row 4 of square array A001477.
		

Crossrefs

Cf. A001108, A001652, A182431, A182440, A182441, A143608.

Programs

Maple

A182439 := proc(n,k)
        if k = 0 then
                A003056(14*n) ;
        elif k = 1 then
                n;
        elif k = 2 then
                14;
        else
                6*procname(n,k-1)-procname(n,k-2)+ 28+2*n-2-4*procname(n,0) ;
        end if;
end proc: # R. J. Mathar, Jul 09 2012

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
(* Second program: *)
A003056[n_] := Floor[(Sqrt[1 + 8n] - 1)/2];
T[n_, k_] := Switch[k, 0, A003056[14n], 1, n, 2, 14, _, 6T[n, k-1] - T[n, k-2] + 28 + 2n - 2 - 4T[n, 0]];
Table[T[n-k, k], {n, 0, 9}, {k, n, 0, -1}] (* Jean-François Alcover, May 09 2023, after R. J. Mathar *)

Formula

a(k,0) equals the largest m such that m*(m+1)/2 is equal to or less than 14*k, A003056(14*k).

a(k,1) = 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 = 28 + 2*k - 2 - 4*a(k,0).

a(k,i) = 7*a(k,i-1)-7*a(k,i-2)+a(k,i-3). - R. J. Mathar, Jul 09 2012