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A091984 Number of steps required for initial p = 10^n to reach 1 in the recurrence p = floor(Li(p)). 0
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; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..71.

Andrew Booker, The Nth Prime Page.

FORMULA

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.

EXAMPLE

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.

PROG

(PARI) Li(x)=-eint1(-log(x))

pr10nLi(n) = my(c); for(x=1, n, y=10^x; c=0; p=y; while(p > 1, p = floor(Li(p)); c++; ); print1(c", "))

CROSSREFS

Sequence in context: A031977 A225668 A091985 * A228651 A045762 A320985

Adjacent sequences:  A091981 A091982 A091983 * A091985 A091986 A091987

KEYWORD

easy,nonn

AUTHOR

Cino Hilliard, Mar 16 2004

STATUS

approved

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Last modified December 4 02:40 EST 2021. Contains 349469 sequences. (Running on oeis4.)