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Search: id:A025172
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| A025172 |
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Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n. |
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3
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| 1, 1, -7, -23, 17, 241, 329, -1511, -5983, 1633, 57113, 99529, -314959, -1525679, -216727, 13297657, 28545857, -62587199, -382087111, -200889431, 3037005137, 7882015153, -11569015927, -94076168231, -84031193119, 678623127841, 2113526993753
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Used when showing that the regular simplex is not "scisssors-dissectable" to a cube, thus answering Hilbert's third problem.
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REFERENCES
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J. L. Dupont, Scissors Congruences, Group Homology and Characteristic Classes, World Scientific, 2001. See p. 4.
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FORMULA
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a(0) = 1, a(1) = 1; for n >= 2, a(n) = 2*a(n-1) - 9*a(n-2). - Warut Roonguthai (warut822(AT)yahoo.com), Oct 11 2005
a(n) = (1/2)*(1-2*i*2^(1/2))^(n+1)+(1/2)*(1+2*i*2^(1/2))^(n+1), where i=sqrt(-1). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Apr 19 2003
a(n) is the permanent of the matrix M^n, where M = [i, 2; 1, i]. - Simone Severini (simoseve(AT)gmail.com), Apr 27 2007
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MAPLE
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f:=proc(n) option remember; if n <= 1 then RETURN(1); fi; 2*f(n-1)-9*f(n-2); end;
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MATHEMATICA
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Table[ n/2 3^n GegenbauerC[ n, 1/3 ], {n, 24} ]
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PROGRAM
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(PARI) {a(n)= if(n<0, 0, 3^(n-1)* subst(3* poltchebi(abs(n)), x, 1/3))} /* Michael Somos Mar 14 2007 */
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CROSSREFS
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Sequence in context: A067959 A070411 A121815 this_sequence A115023 A009228 A031450
Adjacent sequences: A025169 A025170 A025171 this_sequence A025173 A025174 A025175
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KEYWORD
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sign
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AUTHOR
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Wouter Meeussen (wouter.meeussen(AT)pandora.be)
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EXTENSIONS
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Better description from Vladeta Jovovic (vladeta(AT)Eunet.yu), Apr 19 2003
Edited by njas, Feb 22 2007. Among other things, I changed the offset and the beginning of the sequence, so some of the formulae may need to be adjusted slightly.
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