A228829 a(n) = (m+n-k) mod (m-n+k) where k = BigOmega(n) and m is the next larger integer after n with the same k = BigOmega(m) as n.

0, 1, 0, 2, 3, 2, 3, 2, 4, 2, 0, 4, 0, 2, 0, 2, 0, 1, 4, 2, 0, 2, 8, 1, 3, 0, 0, 2, 9, 4, 12, 2, 1, 1, 0, 2, 0, 2, 0, 2, 3, 4, 2, 4, 3, 1, 28, 2, 4, 2, 0, 6, 4, 2, 0, 2, 4, 2, 12, 1, 0, 0, 2, 0, 1, 2, 0, 1, 6, 2, 4, 4, 4, 0, 1, 3, 14, 1, 18, 0, 0, 3, 0, 1, 0, 2, 0, 5, 4, 2, 7, 2, 1, 4, 18, 2

Offset

2,4

Comments

Let k = A001222(n) be the number of prime divisors of n and let m>n be the smallest number larger than n with the same number of prime divisors, k=A001222(m). Then a(n) = (m+n-k) mod (m-n+k).

Links

Table of n, a(n) for n=2..97.

Example

a(1) is undefined because there is only 1 0-almost prime (the 1 itself).
a(2) = 0 because (3 + 2 - 1 mod 3 - 2 + 1) = (4 mod 2) = 0 where 1 < 2 < 3 and 2, 3 are consecutive 1-almost primes,
a(3) = 1 because (5 + 3 - 1 mod 5 - 3 + 1) = (7 mod 3) = 1 where 1 < 3 < 5 and 3, 5 are consecutive 1-almost primes,
a(4) = 0 because (6 + 4 - 2 mod 6 - 4 + 2) = (8 mod 4) = 0 where 1 < 4 < 6 and 4, 6 because consecutive 2-almost primes,
a(5) = 2 because (7 + 5 - 1 mod 7 - 5 + 1) = (11 mod 3) = 2 where 1 < 5 < 7 and 5, 7 are consecutive 1-almost primes,
a(6) = 3 because (9 + 6 - 2 mod 9 - 6 + 2) = (13 mod 5) = 3 where 1 < 6 < 9 and 6, 9 are consecutive 2-almost primes.

Maple

A228829 := proc(n)
    local k, m ;
    k := numtheory[bigomega](n) ;
    for m from n+1 do
        if numtheory[bigomega](m) = k then
            return modp(m+n-k, m-n+k)
        end if;
     end do:
end proc: # R. J. Mathar, Sep 13 2013

Crossrefs

Cf. A226534.

Sequence in context: A069898 A245511 A259940 * A341982 A337686 A007978

Adjacent sequences: A228826 A228827 A228828 * A228830 A228831 A228832

Keyword

nonn

Author

Juri-Stepan Gerasimov, Sep 04 2013

Extensions

Corrected by R. J. Mathar, Sep 13 2013

Status

approved