A121560 -id:A121560 - cp's OEIS frontend

A121559 Final result (0 or 1) under iterations of {r mod (max prime p <= r)} starting at r = n.

1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0

Offset: 1

Kerry Mitchell, Aug 07 2006

Previous name: Find r1 = n modulo p1, where p1 is the largest prime not greater than n. Then find r2 = r1 modulo p2, where p2 is the largest prime not greater than r1. Repeat until the last r is either 1 or 0; a(n) is the last r value.
The sequence has the form of blocks of 0's between 1's. See sequence A121560 for the lengths of the blocks of zeros.
The function r mod (max prime p <= r), which appears in the definition, equals r - (max prime p <= r) = A064722(r), because p <= r < 2*p by Bertrand's postulate, where p is the largest prime less than or equal to r. - Pontus von Brömssen, Jul 31 2022

			a(9) = 0 because 7 is the largest prime not larger than 9, 9 mod 7 = 2, 2 is the largest prime not greater than 2 and 2 mod 2 = 0.

Kerry Mitchell, Table of n, a(n) for n = 1..7919

Cf. A007917 and A064722 (both for the iterations).

Cf. A121560, A121561, A121562, A175077, A200947.

Abs[Table[FixedPoint[Mod[#,NextPrime[#+1,-1]]&,n],{n,110}]] (* Harvey P. Dale, Mar 17 2023 *)

a(n) = if (n==1, return (1)); na = n; while((nb = (na % precprime(na))) > 1, na = nb); return(nb); \\ Michel Marcus, Aug 22 2014

a(p) = 0 when p is prime. - Michel Marcus, Aug 22 2014
a(n) = A175077(n+1) - 1. - Pontus von Brömssen, Jul 31 2022
a(n) = A200947(n) mod 2. - Alois P. Heinz, Jun 12 2023

New name from Michel Marcus, Aug 22 2014

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.

Original entry on oeis.org

1, 2, 9, 122, 1357323

Offset: 0

Sergio Pimentel, Feb 22 2008

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

			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.

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

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

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 *)

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

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)

cp's OEIS Frontend

A121559 Final result (0 or 1) under iterations of {r mod (max prime p <= r)} starting at r = n.

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