

A164917


Smallest number of steps to reach prime(n) by applying the map x>A060308(x) starting from any member of A164368.


5



0, 1, 2, 3, 0, 4, 0, 1, 5, 0, 1, 2, 0, 6, 0, 1, 0, 2, 0, 0, 3, 1, 7, 1, 0, 0, 2, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 8, 0, 2, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 1, 2, 0, 0, 1, 0, 0, 3, 9, 1, 3, 0, 0, 1, 1, 0, 0, 1, 2, 1, 2, 0, 0, 0, 2, 0, 0, 0, 3, 1, 1, 0, 1, 1, 1, 0, 0, 2, 0, 3, 0, 1, 2, 3, 1, 1, 0, 0, 2
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OFFSET

1,3


COMMENTS

Starting from some prime, iterated application of A060308 (or of the equivalent A059788) generates a chain of increasing prime numbers.
The nature of these chains is to reach higher in the list of primes, sometimes "oversatisfying" Bertrand's postulate by skipping some nearer primes, almost doubling of possible. On the other hand, A164368 contains the primes that would be skipped by a chain which contains the prime slightly above half of their value. The sequence shows how far up in chains starting from some member of A164368 we find prime(n), or equivalently, how many inverse applications of the map we need to hit a member of A164368 if starting at prime(n).
Note that by construction A164368(k) starts with the smallest prime that is not member of any chain started from any previous A164368. So each prime exists at some place in one of these chains, and the number of steps a(n) to reach it from the start of its chain is well defined.


LINKS

Table of n, a(n) for n=1..106.
V. Shevelev, On critical small intervals containing primes, arXiv:0908.2319 [math.NT], 2009.


EXAMPLE

The first prime chains of the mapping with A060308 initialized with members of A164368 are
2>3>5>7>13>23>43>83>163>317>631>1259>2503>..
11>19>37>73>139>277>547>1093>2179>4357>8713>17419>..
17>31>61>113>223>443>883>1759>3517>7027>14051>28099>..
29>53>103>199>397>787>1571>3137>6271>12541>25073>..
41>79>157>313>619>1237>2473>4943>9883>19763>39521>..
47>89>173>337>673>1327>2647>5281>10559>21107>..
The a(1) to a(4) representing the first 4 primes are all on the first chain, and need 0 to 3 steps to be reached from 2 = A164368(1). a(5) asks for the number of steps for A000040(5)=11 which is on the second chain, and needs 0 steps.


MAPLE

A060308 := proc(n) prevprime(2*n+1) ; end:
isA164368 := proc(p) local q ; q := nextprime(floor(p/2)) ; RETURN(numtheory[pi](2*q) numtheory[pi](p) >= 1); end:
A164368 := proc(n) option remember; local a; if n = 1 then 2; else a := nextprime( procname(n1)) ; while not isA164368(a) do a := nextprime(a) ; od: RETURN(a) ; fi; end:
A164917 := proc(n) local p, a, j, q, itr ; p := ithprime(n) ; a := 1000000000000000 ; for j from 1 do q := A164368(j) ; if q > p then break; fi; itr := 0 ; while q < p do q := A060308(q) ; itr := itr+1 ; od; if q = p then if itr < a then a := itr; fi; fi; od: a ; end:
seq(A164917(n), n=1..120) ; # R. J. Mathar, Sep 24 2009


MATHEMATICA

A060308[n_] := NextPrime[2*n + 1, 1];
isA164368[p_] := Module[{q}, q = NextPrime[Floor[p/2]]; Return[PrimePi[2*q]  PrimePi[p] >= 1]];
A164368[n_] := A164368[n] = Module[{a}, If[n == 1, 2, a = NextPrime[ A164368[n1]]; While[Not @ isA164368[a], a = NextPrime[a]]; Return[a]]];
A164917[n_] := Module[{p, a, j, q, itr}, p = Prime[n]; a = 10^15; For[j = 1 , True, j++, q = A164368[j]; If[q > p, Break[]]; itr = 0; While[q < p, q = A060308[q]; itr++]; If[q == p, If[itr < a, a = itr]]]; a];
Table[A164917[n], {n, 1, 120}] (* JeanFrançois Alcover, Dec 14 2017, after R. J. Mathar *)


CROSSREFS

Cf. A006992, A104272, A164368, A164288.
Sequence in context: A244119 A122059 A324906 * A166238 A293275 A014197
Adjacent sequences: A164914 A164915 A164916 * A164918 A164919 A164920


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Aug 31 2009


EXTENSIONS

Edited, examples added and extended by R. J. Mathar, Sep 24 2009


STATUS

approved



