A135543 Record number of steps under iterations of "map n to n - (largest prime <= n)" (A064722) until reaching the limiting value 0 or 1. Also, places where A121561 reaches a new record.

1, 2, 9, 122, 1357323

Offset: 0

Author

Sergio Pimentel, Feb 22 2008

Keywords

• hard
• more
• nonn

Comments

a(5) must be very large (> 100000000). Can anyone extend the sequence?

Conjecture: there exist positive values of n for which a(n) != A175079(n) - 1. - Jaroslav Krizek, Feb 05 2010

From Thomas R. Nicely's data (see link) it seems that the smallest known prime with following prime gap of length a(4)+1 or more is 90823#/510510 - 1065962 (39279 digits), so a(5) = A104138(a(4)) + a(4) <= 90823#/510510 - 1065962 + 1357323 = A002110(8787)/510510 + 291361. (The bounding primes of this prime gap are only known to be probable primes, but if either of them were not prime, the gap would only be larger and the bound on a(5) would still hold.) - Pontus von Brömssen, Jul 31 2022

Examples

			a(4) = 1357323 because after iterating n - (largest prime <= n) we get:
  1357323 - 1357201 = 122 =>
  122 - 113 = 9 =>
  9 - 7 = 2 =>
  2 - 2 = 0,
which takes 4 steps.
		

Links

• Thomas R. Nicely, First known occurrence prime gaps (1000000 to 999999998).

Crossrefs

Cf. A002110, A064722, A104138, A121559, A121560, A121561, A175078, A175079.

Programs

Mathematica

LrgstPrm[n_] := Block[{k = n}, While[ !PrimeQ@ k, k-- ]; k]; f[n_] := Block[{c = 0, d = n}, While[d > 1, d = d - LrgstPrm@d; c++ ]; c]; lst = {}; record = -1; Do[ a = f@n; If[a > record, record = a; AppendTo[lst, a]; Print@ n], {n, 100}] (* Robert G. Wilson v *)

Python

from sympy import prevprime
from functools import lru_cache
from itertools import count, islice
@lru_cache(maxsize=None)
def f(n): return 0 if n == 0 or n == 1 else 1 + f(n - prevprime(n+1))
def agen(record=-1):
    for k in count(1):
        if f(k) > record: record = f(k); yield k
print(list(islice(agen(), 4))) # Michael S. Branicky, Jul 26 2022

Formula

Iterate n - (largest prime <= n) until reaching 0 or 1. Count the iterations required to reach 0 or 1 and determine if it is a new record.

From Pontus von Brömssen, Jul 31 2022: (Start)

a(n) = A104138(a(n-1)) + a(n-1) for n >= 2.

A121561(a(n)) = n.

a(n) = A175079(n) - 1 for n >= 1, i.e., the conjecture in the Comments is false. This follows from the result that A175078(n) = A121561(n-1) for n >= 2.

(End)