# f(2013) = 2014? [closed]

I wonder if there is some "surprising" function $f(\;)$ that, when input $2013$, produces $2014$? What I have in mind is more in line with the Lewis Carroll computation involving $137$ and $992$ that Gerry Myerson highlighted in this MO answer. $f(n)=n+1$ would not satisfy. :-)

$$2013 = 3 \cdot 11 \cdot 61$$ $$2014 = 2 \cdot 19 \cdot 53$$ Hmmm...    Happy New Year!

## closed as off-topic by Fernando Muro, Steven Landsburg, Andrés E. Caicedo, Andrey Rekalo, Daniel MoskovichJan 1 '14 at 1:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Fernando Muro, Steven Landsburg, Andrés E. Caicedo, Andrey Rekalo, Daniel Moskovich
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• but, $f(x) = x+1$ is the best approximation of the reality ;-) – John Dvorak Jan 1 '14 at 0:14
• $$f(x)=\begin{cases}2014 & 2^{\aleph_0}=\aleph_1\\x+1&\text{otherwise}\end{cases}$$ – Asaf Karagila Jan 1 '14 at 0:30
• 2013, 2014, and 2015 are all squarefree products of exactly 3 primes. They form the third smallest such trio, after the ones started with 1309 and 1885. See oeis.org/A165936 and see oeis.org/A167451 for another sequence in which 2013 is followed by 2014. – Gerry Myerson Jan 1 '14 at 1:04
• Well, my historically 2nd (I think?) closed question, the other also a failed attempt at New Year's bonhomie. :-) – Joseph O'Rourke Jan 1 '14 at 1:50
• Along the lines of Gerry Myerson's comment: how about defining $f(x)$ to be the least integer $n$ such that (1) $n>x$ and (2) $n$ and $x$ have the same number of prime factors? Like Jan Dvorak's answer, this has the virtue of also working next year (though afterwards his seems to work somewhat better). – Daniel Litt Jan 1 '14 at 2:26