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%I #4 Sep 16 2015 19:34:48
%S 2,2,2,2,4,2,3,9,9,3,7,24,34,24,7,31,103,154,154,103,31,241,778,1055,
%T 1036,1055,778,241,3121,10127,12957,10083,10083,12957,10127,3121,
%U 65521,215148,274724,184020,117846,184020,274724,215148,65521,2227681,7378804
%N 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}].
%C Row sums are:
%C {2, 4, 8, 24, 96, 576, 5184, 72576, 1596672, 55883520, 3129477120,...}.
%C If you take:
%C with H(i) as quantum Magnetic fields:
%C Product[1+H(i)*x,{i,0,n}]
%C The sequence that results is Stirling number like.
%C Making that symmetrical:
%C p(x,n)=Product[1+H(i)*x,{i,0,n}]+x^n*Product[1+H(i)/x,{i,0,n}]
%C Now, you can put in just about any a(n) sequence and get a symmetrical
%C polynomial back.
%F p(x,n) = Product[1 + Fibonacci[i]*x, {i, 0, n}] + x^n*Product[1 + Fibonacci[i]/x, {i, 0, n}];
%F t(n,m)=coefficients(p(x,n))
%e {{2},
%e {2, 2},
%e {2, 4, 2},
%e {3, 9, 9, 3},
%e {7, 24, 34, 24, 7},
%e {31, 103, 154, 154, 103, 31},
%e {241, 778, 1055, 1036, 1055, 778, 241},
%e {3121, 10127, 12957, 10083, 10083, 12957, 10127, 3121},
%e {65521, 215148, 274724, 184020, 117846, 184020, 274724, 215148, 65521},
%e {2227681, 7378804, 9521213, 6204407, 2609655, 2609655, 6204407, 9521213, 7378804, 2227681},
%e {122522401, 408057203, 530891673, 348306220, 128955206, 52011714, 128955206, 348306220, 530891673, 408057203, 122522401}
%t 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}];
%t Flatten[%]
%Y A000045
%K nonn,tabl,uned
%O 0,1
%A _Roger L. Bagula_, Jan 16 2009
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