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A154851
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Symmetrical triangular sequence of Fibonacci numbers (A000045): p(x,n) = Product[1 + Fibonacci[i]*x, {i, 0, n}] + x^n*Product[1 + Fibonacci[i]/x, {i, 0, n}].
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0
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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
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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))
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EXAMPLE
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{{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}
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MATHEMATICA
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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[%]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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