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A182355 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182441(k,i + 1) *A182441(k,i + 2) equals "r" + a(k,i)*(a(k,i) + 1)/2 for i<4, while a(k,i) = 0 for i>3. 1

%I #20 Aug 11 2015 17:42:43

%S -1,56,-5,399,60,-8,2400,463,63,-9,0,2816,512,64,-11,0,0,3135,531,66,

%T -12,0,0,0,3260,565,67,-13,0,0,0,0,3482,584,68,-14,0,0,0,0,0,3607,603,

%U 69,-15,0,0,0,0,0,0,3732,622

%N Table of triangular arguments such that if A002262(14*k) = "r" then the product A182441(k,i + 1) *A182441(k,i + 2) equals "r" + a(k,i)*(a(k,i) + 1)/2 for i<4, while a(k,i) = 0 for i>3.

%C The triangular product a(k,i)*(a(k,i)+1)/2 + A002262(14*k) for i<4 = the product of adjacent terms G(k,i+1)*G(k,i+2) where G is table A182441. 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.

%C 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 d of -1 is series A182193.

%C 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).

%t highTri = Compile[{{S1,_Integer}},Module[{xS0=0,xS1=S1},

%t While[xS1-xS0*(xS0+1)/2>xS0,xS0++];

%t xS0]];

%t overTri = Compile[{{S2,_Integer}},Module[{xS0=0,xS2=S2},

%t While[xS2-xS0*(xS0+1)/2>xS0,xS0++];

%t xS2 - (xS0*(1+xS0)/2)]];

%t tt = SparseArray[{{12,1} -> 0,{1,12} -> 0}];

%t K1 = 0;

%t m = 14;While[K1<12,J1=highTri[m*K1];X =2*(m+K1+(J1*2+1));

%t K2 = 6 m - K1 + X; K3 = 6 K2 - m + X;K4 = 6 K3 - K2 + X;

%t o = overTri[m*K1]; tt[[1,K1+1]] =highTri[m*K1];

%t tt[[2,K1+1]] = highTri[m*K2-o];tt[[3,K1+1]] = highTri[K2*K3-o];tt[[4,K1+1]] = highTri[K3*K4-o];

%t K1++];k = 1;

%t While[k<13,z = 1; xx = 99; While[z<5 && xx == 99,

%t If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,Break[]];

%t 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;

%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 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 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,

%t tt[[p,k]] = 6 tt[[p-1,k]] - tt[[p-2,k]] + cc;p++]; k++];

%t 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

%Y Cf. A182102, A182118, A182119, A182190, A182193, A182188, A182189.

%K sign,tabl

%O 0,2

%A _Kenneth J Ramsey_, Apr 25 2012

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