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A228829
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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.
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1
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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
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OFFSET
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2,4
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COMMENTS
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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).
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LINKS
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EXAMPLE
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a(1) is undefined because there is only 1 0-almost prime (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.
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MAPLE
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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:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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