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A095116
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a(n) = prime(n) + n - 1.
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9
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2, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, 53, 56, 61, 68, 75, 78, 85, 90, 93, 100, 105, 112, 121, 126, 129, 134, 137, 142, 157, 162, 169, 172, 183, 186, 193, 200, 205, 212, 219, 222, 233, 236, 241, 244, 257, 270, 275, 278, 283, 290, 293, 304, 311, 318, 325
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OFFSET
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1,1
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COMMENTS
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a(n) = b(n)-th highest positive integer not equal to any a(k), 1 <= k <= n-1, where b(n) = primes = A000040(n). a(1) = 2, a(n) = a(n-1) + A000040(n) - A000040(n-1) + 1 for n >= 2. a(1) = 2, a(n) = a(n-1) + A001223(n-1) + 1 for n >= 2. a(n) = A014688(n) - 1. [Jaroslav Krizek, Oct 28 2009]
On March 23 2024, Davide Rotondo sent me an email with the following conjecture. (I've simplified it a bit.)
For a positive integer n, define a sequence b by b(0) = n; b(i) = n - pi(b(i-1)) for i >= 1, where pi(x) = number of primes <= x.
The conjecture is that after some initial terms, b becomes periodic with period length 1 or 2, and the n for which the period is 2 are 3 together with the present sequence, that is, 2, 3, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, ... (End)
This is simply a consequence of the fact that if x < y, 0 <= pi(y) - pi(x) <= y - x and the inequality on the right is strict if y-x > 1 except for the case of 1 and 3.
Thus we start with b(0) - b(1) = pi(n). While |b(i) - b(i+1)| > 2 we get |b(i+1) - b(i+2)| = |pi(b(i+1)) - pi(b(i+2))| < |b(i) - b(i+1)|.
Eventually we must either reach |b(j+1) - b(j)| = 0 or |b(j+1) - b(j)| = 1.
If we reach 0, i.e. b(j+1) = b(j), then clearly b(k) = b(j) for all k > j.
If b(j+1) = b(j) + 1 = n - pi(b(j)), then b(j+2) = n - pi(b(j)+1) = b(j+1) or b(j+1)-1.
If b(j+1) = b(j) - 1, then b(j+2) = n - pi(b(j)-1) = b(j+1) or b(j+1)+1.
Thus from this point on we either get a 2-cycle or a 1-cycle. (End)
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LINKS
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FORMULA
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MAPLE
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with (numtheory):seq(n+ithprime(n+1), n=0..56); # Zerinvary Lajos, Aug 24 2008
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MATHEMATICA
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PROG
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(Haskell)
a095116 n = a000040 n + toInteger n - 1
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CROSSREFS
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Essentially the same sequence as A014690.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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