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Search: id:A074700
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| A074700 |
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tau(F(2^n)) where tau(x) is the number of divisors of x (A00005(x)) and F(k) the k-th Fibonacci number (A00045(k)). |
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+0
1
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| 1, 2, 4, 8, 16, 64, 256, 1024, 8192, 131072, 1048576
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Is there any pattern in this sequence ? It seems also that tau(F(m^n)) is a power of 2 for any m, any n >0.
F(2^n)=L(2)L(4)L(8)...L(2^(n-1)) where L(x) is the Lucas numbers. This greatly reduces the difficulty of factoring these numbers. To find a(9) one needs the factorization of F(512); this was done long ago: F(2^9)=3 * 7 * 47 * 127 * 1087 * 2207 * 4481 * 34303 * 119809 * 73327699969 * 186812208641 * 4698167634523379875583 * 125960894984050328038716298487435392001; hence a(9) = 2^13 = 8192. Since L(512), L(1024) are completed factored the next few terms are also known. L(2048) has 1 known factor and a C411, thus the next term is at least 2^23. - Sean A. Irvine (sairvin(AT)xtra.co.nz), Jun 02 2005
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LINKS
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Author?, Lucas number factorizations
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CROSSREFS
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Cf. A063375.
Sequence in context: A041013 A018754 A076086 this_sequence A013133 A012997 A013184
Adjacent sequences: A074697 A074698 A074699 this_sequence A074701 A074702 A074703
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KEYWORD
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more,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 03 2002
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EXTENSIONS
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More terms from Sean A. Irvine (sairvin(AT)xtra.co.nz), Jun 02 2005
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