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Let f(p) = greatest prime divisor of p-1. Sequence gives smallest prime which takes at least n steps to reach 2 when f is iterated.
+10
6
2, 3, 7, 23, 47, 283, 719, 1439, 2879, 34549, 138197, 1266767, 14619833, 36449279, 377982107, 1432349099, 22111003847
OFFSET
0,1
COMMENTS
There is a remarkable and unexplained agreement: if 3 and 7 are replaced by 11 and 14619833 is replaced by 14920303, the result is sequence A056637 (least prime of class n-, according to the Erdős-Selfridge classification of primes).
From David A. Corneth, Oct 18 2016: (Start):
If a(n) * k + 1 is prime then a(n + 1) <= a(n) * k + 1.
a(18), a(19), ..., a(23) <= 309554053859, 619108107719, 19811459447009, 433142367554861, 866284735109723, 22523403112852799 respectively. (End)
Conjecture: a(n) is the smallest prime p such that b(p) = n, where f(2) = 0 and for an odd prime p, f(p) = 1 + max{q|(p-1), q prime} f(q). In other words, a(n) is the smallest prime p such that A364332(primepi(p)) = n. Verified for n <= 13. - Jianing Song, Apr 28 2024
REFERENCES
Steven G. Johnson, Postings to Number Theory List, Apr 23 and Apr 25, 2003.
EXAMPLE
a(2) = 7 since 7 -> 3 -> 2 takes two steps, and smaller primes require less than 2 steps.
For p = 2879, 8 steps are needed (2879 -> 1439 -> 719 -> 359 -> 179 -> 89 -> 11 -> 5 -> 2), so a(8) = 2879, since smaller primes require less than 8 steps.
MATHEMATICA
(* Assuming a(n) > 2 a(n-1) if n>1 *) Clear[a, f]; f[p_] := FactorInteger[p - 1][[-1, 1]]; f[2] = 2; a[n_] := a[n] = For[p = NextPrime[2 a[n-1]], True, p = NextPrime[p], k = 0; If[Length[FixedPointList[f, p]] == n+2, Return[p]]]; a[0]=2; a[1]=3; Table[Print[a[n]]; a[n], {n, 0, 16}] (* Jean-François Alcover, Oct 18 2016 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
N. J. A. Sloane, Apr 25 2003
EXTENSIONS
Edited by Klaus Brockhaus, May 01 2003
a(16) from Donovan Johnson, Nov 17 2008
STATUS
approved
a(2) = 0; a(n) = a(n-1) + 1 if n is an odd prime; otherwise a(n) = max{a(k) : k is divisor of n, 1 < k < n}.
+10
1
0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 1, 0, 1, 1, 2, 1, 2, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 0, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 1, 3, 4, 1, 2, 1, 1, 2, 3, 1, 2, 2, 2, 3, 4, 1, 2, 2, 2, 0, 2, 2, 3, 1, 3, 2, 3, 1, 2, 2, 1, 2, 2, 2, 3, 1, 1, 2, 3, 2, 1, 3, 3, 2, 3, 1, 2, 3, 2, 4, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 3, 4, 1, 2, 2, 2, 2
OFFSET
2,6
COMMENTS
This sequence is a kind of measure of the "amount of information" in an integer. The post at Zhihu wonders whether one can calculate this sequence without using prime decomposition.
FORMULA
a(2) = 0,
a(n) = a(n-1) + 1 if n is an odd prime,
a(n) = max{a(k) : k|n, 1<k<n} otherwise.
EXAMPLE
a(238)=2, since a(2)=0, a(7)=2, a(14)=2, a(17)=1, a(34)=1, a(119)=2, and the largest among them is 2.
And a(239)=3, since 239 is a prime number, and a(238)=2.
MATHEMATICA
Nest[Function[list,
Module[{n = Length[list] + 1},
Append[list,
If[PrimeQ[n], Last[list] + 1,
Max[(list[[First[#]]]) & /@ FactorInteger[n]]]]]], {0, 0}, 110]//Rest
CROSSREFS
For values at primes, see A364332.
KEYWORD
nonn
AUTHOR
Steven Lu, Jul 18 2023
STATUS
approved

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