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Search: id:A091984
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| A091984 |
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Number of steps required for initial p = 10^n to reach 1 in the recurrence p = floor(Li(p)). |
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+0
1
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| 4, 6, 8, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 38, 38, 39, 39, 40, 40, 40, 41, 41, 42, 42, 43, 43, 44, 44, 45, 45, 46, 46
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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The number of steps for the recurrence in this sequence stopping at 1 compares closely to the steps in the pi(n) recurrence stopping at 0. If we define Li(1) = 0 and allow that step then A(using Li seq#) - A(using pi seq#) = 1 for n <=13. So one may conjecture the steps in the Li method is always 1 greater than the steps in the pi method. Question is can the difference be greater than 1? For the largest value allowed in the link (3*10^13) A(Li) = 17 (assuming L1(1)= 0) and n=13 A(pi) = 16 from Booker so the difference = 1 as before.
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LINKS
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Andrew Booker, The Nth Prime Page.
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FORMULA
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Li(n) = logarithmic integral = integral(x=2..n, dx/log(x)). This gives a very good approximation to the number of primes less than or equal to n. By repeating n=Li(n), n will reach 1 in a finite number of steps.
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EXAMPLE
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Li(100) = 30
Li(30) = 13
Li(13) = 7
Li(7) = 4
Li(4) = 2
Li(2) = 1
Total steps to reach 1 = 6. Thus 6 is the 2nd entry in the sequence corresponding to n=2.
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PROGRAM
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(PARI) pr10nLi(n) = { local c; for(x=1, n, y=10^x; c=0; p=y; while(p > 1, p = floor(Li(p)); c++; ); print1(c", ") ) } Li(x) = \Logarithmic integral in PARI { -eint1(log(1/x)) }
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CROSSREFS
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Sequence in context: A019516 A031977 A091985 this_sequence A045762 A047748 A085558
Adjacent sequences: A091981 A091982 A091983 this_sequence A091985 A091986 A091987
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KEYWORD
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easy,nonn
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AUTHOR
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Cino Hilliard (hillcino368(AT)gmail.com), Mar 16 2004
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