A121559 - OEIS

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

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OFFSET

1,1

COMMENTS

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

LINKS

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

FORMULA

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

EXAMPLE

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.

MATHEMATICA

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

PROG

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