A182431 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).
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, 14, 12, 20664, 1285, 2, 21, 4, 5, 14, 13, 120470, 7504, 14, 122, 14, 8, 6, 14, 14, 702182, 43751, 84, 711, 74, 35, 12, 7, 14, 15
Offset: 0
Examples
The Table begins: 0 14 0 12 98 602 3540 ... 4 14 1 4 35 218 1285 ... 7 14 2 0 0 2 14 ... 8 14 3 4 21 122 711 ... 10 14 4 4 14 74 424 ... 11 14 5 8 35 194 1121 ... 12 14 6 12 56 314 1818 ... 13 14 7 16 77 434 2515 ... 14 14 8 20 98 554 3212 ... 15 14 9 24 119 674 3909 ... 16 14 10 28 140 794 4606 ... 17 14 11 32 161 914 5303 ... 17 14 12 40 210 1202 6984 ... ... 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.
Programs
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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 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[tab[[b,a+1-b]]]; b--]; a++]][[2,1]]; list5
Formula
a(k,0) equals the largest m such that m*(m+1)/2 is less than or equal to 14*k.
a(k,1) = 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*a(k,0).
Comments