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Search: id:A114551
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| A114551 |
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Continued fraction expansion of the constant (A114550) equal to the sum Sum_{n>=0} 1/A112373(n), such that the partial quotients satisfy: a(2n) = A112373(n) for n>0 and a(2n+1) = A112373(n+1)/A112373(n) for n>=0. |
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3
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| 2, 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742, 342022190843338960032, 1710009514450915230711940280907486, 584861200495456320274313200204390612579749188443599552
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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A112373 is defined by the recurrence: let b(n) = A112373(n), then
b(n) =(b(n-1)^3 + b(n-1)^2)/b(n-2) for n>=2 with b(0)=b(1)=1.
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FORMULA
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a(2n) = a(2n-1)*a(2n-2) for n>=2, a(2n+1) = a(2n)*a(2n-1) + a(2n-1) for n>=1, with a(0)=2, a(1)=a(2)=1. - Jeffrey Shallit (shallit(AT)graceland.uwaterloo.ca).
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EXAMPLE
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2.584401724019776724812076147153331342112382090467969...
= Sum_{n>=0} 1/A112373(n) = 1/1 +1/1 +1/2 +1/12 +1/936 +1/68408496 +...
= [2;1,1,2,2,6,12,78,936,73086,68408496,...] (continued fraction).
The recurrence of partial quotients is demonstrated by:
(odd-index) a(7) = 78 = a(6)*a(5) + a(5) = 12*6 + 6;
(even-index) a(8) = 936 = a(7)*a(6) = 78*12.
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PROGRAM
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(PARI) a(n)=if(n<0, 0, if(n<3, [2, 1, 1][n+1], a(n-1)*a(n-2)+(n%2)*a(n-2)))
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CROSSREFS
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Cf. A112373, A114550 (constant), A114552 (bisection).
Adjacent sequences: A114548 A114549 A114550 this_sequence A114552 A114553 A114554
Sequence in context: A083698 A128976 A046772 this_sequence A136256 A090996 A089309
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KEYWORD
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cofr,nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Dec 08 2005
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