0,1

Row sums are:

{2, 4, 8, 24, 96, 576, 5184, 72576, 1596672, 55883520, 3129477120,...}.

If you take:

with H(i) as quantum Magnetic fields:

Product[1+H(i)*x,{i,0,n}]

The sequence that results is Stirling number like.

Making that symmetrical:

p(x,n)=Product[1+H(i)*x,{i,0,n}]+x^n*Product[1+H(i)/x,{i,0,n}]

Now, you can put in just about any a(n) sequence and get a symmetrical

polynomial back.

p(x,n) = Product[1 + Fibonacci[i]*x, {i, 0, n}] + x^n*Product[1 + Fibonacci[i]/x, {i, 0, n}];

t(n,m)=coefficients(p(x,n))

{{2},

{2, 2},

{2, 4, 2},

{3, 9, 9, 3},

{7, 24, 34, 24, 7},

{31, 103, 154, 154, 103, 31},

{241, 778, 1055, 1036, 1055, 778, 241},

{3121, 10127, 12957, 10083, 10083, 12957, 10127, 3121},

{65521, 215148, 274724, 184020, 117846, 184020, 274724, 215148, 65521},

{2227681, 7378804, 9521213, 6204407, 2609655, 2609655, 6204407, 9521213, 7378804, 2227681},

{122522401, 408057203, 530891673, 348306220, 128955206, 52011714, 128955206, 348306220, 530891673, 408057203, 122522401}

Clear[p, x, n]; p[x_, n_] = Product[1 + Fibonacci[i]*x, {i, 0, n}] + x^n*Product[1 + Fibonacci[i]/x, {i, 0, n}]; \! Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];

Flatten[%]

A000045

Sequence in context: A036555 A046927 A084718 this_sequence A037445 A003036 A089818

Adjacent sequences: A154848 A154849 A154850 this_sequence A154852 A154853 A154854

nonn,tabl,uned,tabl

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 16 2009