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).
Original entry on oeis.org
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
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.
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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
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).
Original entry on oeis.org
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
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.
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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
-
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 *)
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).
Original entry on oeis.org
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
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.
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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
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).
Original entry on oeis.org
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
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.
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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
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