A182119 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182439(k,i + 1) *A182439(k,i + 2) equals "r" + A000217(a(k,i)) for i<4, while a(k,i) = 0 for i>3.

0, 55, 4, 384, 51, 7, 2303, 328, 48, 8, 0, 1943, 287, 47, 10, 0, 0, 1680, 276, 45, 11, 0, 0, 0, 1611, 250, 44, 12, 0, 0, 0, 0, 1445, 239, 43, 13, 0, 0, 0, 0, 0, 1376, 228, 42, 14, 0, 0, 0, 0, 0, 1307, 213, 41, 15

Offset: 0

Author

Kenneth J Ramsey, Apr 12 2012

Keywords

• nonn
• tabl

Comments

The triangular product A000217(a(k,i)) for i < 4 + A002262(14*k) = the product of adjacent terms G(k,i+1)*G(k,i+2) where G is table A182439. The remainder of each row is padded with zeros. However, if for i > 3, a(k,i) were set to equal 7*a(k,i-1) - 7*a(k,i-2) + a(k,i-3) then the relation above would not be limited to i < 4.

Also, it is noted that the difference between adjacent rows of the respective elements, depends on the difference between the elements of column 0 in the respective rows. In the Mathematica program below, m is set to 14; however, regardless of it value of m, it is apparent that the series of differences corresponding to a difference of d in column 0, i.e. A(k+1,0) - A(k,0) = d, is defined as follows: D(0) = d, D(1) = - d, D(n) = 6*D(n-1) - D(n-2) -8*d + 4. The sequence of differences corresponding to a difference of 1 or 0 in column 0 form related series A182188 and A182190.

The Mathematica program below basically first computes only the nonnegative triangular arguments P. Then it changes at most two of the arguments P in each row k to the corresponding negative value, N = -P -1, in order to obtain the relation a(k,3) = a(k,0) - 7*a(k,1) + 7*a(k,2).

Crossrefs

Cf. A182439, A182188, A182190

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)]];
tt = SparseArray[{{12,1} -> 1,{1,12} -> 1}];
K1 = 0;
m = 14;While[K1<12,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;
o = overTri[m*K1]; tt[[1,K1+1]] =highTri[m*K1];
tt[[2,K1+1]] = highTri[m*K2-o];tt[[3,K1+1]] = highTri[K2*K3-o];tt[[4,K1+1]] = highTri[K3*K4-o];
K1++];k = 1;
While[k<13,z = 1; xx = 99; While[z<5 && xx == 99,
If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,Break[]];
If[z == 1,t = -tt[[z,k]]-1;tt[[z,k]] = t,s = -tt[[z-1,k]]-1;tt[[z-1,k]]=s;t =-tt[[z,k]]-1];tt[[z,k]] = t;
w = 1;While[w<5 && xx == 99,If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[]];If[w==z,w++];
t=-tt[[w,k]] - 1;tt[[w,k]]=t;If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[],
t = -tt[[w,k]] - 1];tt[[w,k]] = t;w++];z++];cc = tt[[1,k]] -6 tt[[2,k]] + tt[[3,k]];p = 5;While[p < 14-k,
tt[[p,k]] = 6 tt[[p-1,k]] - tt[[p-2,k]] + cc;p++]; k++];
a=1;list2 = Reap[While[a<12, b=a; While[b>4,Sow[0];b--];While[b>0, Sow[tt[[b, a+1-b]]]; b--]; a++]][[2, 1]];list2