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%I #8 Apr 03 2023 10:36:10
%S 4,6,8,9,10,11,12,12,13,14,15,15,16,17,17,18,19,19,20,20,21,22,22,23,
%T 23,24,24,25,25,26,27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35,
%U 35,36,36,37,37,38,38,39,39,40,40,40,41,41,42,42,43,43,44,44,45,45,46,46
%N Number of steps required for initial p = 10^n to reach 1 in the recurrence p = floor(Li(p)).
%C 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.
%H Andrew Booker, <a href="https://t5k.org/nthprime/index.php#piofx">The Nth Prime Page</a>.
%F 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.
%e Li(100) = 30
%e Li(30) = 13
%e Li(13) = 7
%e Li(7) = 4
%e Li(4) = 2
%e Li(2) = 1
%e Total steps to reach 1 = 6. Thus 6 is the 2nd entry in the sequence corresponding to n=2.
%o (PARI) Li(x)=-eint1(-log(x))
%o 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","))
%K easy,nonn
%O 1,1
%A _Cino Hilliard_, Mar 16 2004
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