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 A154851 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}]. 0
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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. LINKS FORMULA 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)) EXAMPLE {{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} MATHEMATICA 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[%] CROSSREFS Sequence in context: A036555 A046927 A084718 * A281854 A037445 A318307 Adjacent sequences:  A154848 A154849 A154850 * A154852 A154853 A154854 KEYWORD nonn,tabl,uned AUTHOR Roger L. Bagula, Jan 16 2009 STATUS approved

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Last modified February 21 05:23 EST 2020. Contains 332086 sequences. (Running on oeis4.)