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I am looking for a function that for any natural number n returns a natural number m. Inverse function should exist for any $m=F(n), n \in \mathbb{N}$. Sequence should be simple enough for a person to calculate first 1000 members on paper in a day. Sequence should have little apparent order.

Examples not matching criteria
m=0*n+14 No inverse function
m=34*n*n-42: Easy to guess this law
m= mod(a^n, d) : modular arithmetic is periodic and inverse function returns values (1,n)
https://oeis.org/A000055 : Close, but too complex to compute
Prime numbers: Best example so far. Pros

  • Not trivial
  • Defined for all n
  • Easily calculatable

Cons

  • Well known

Which sequence behaves as nice as prime counting, but is relatively less studied?

Update: some cellular automata sequences seem promissing, but they

  • Grow fast (nearly exponential growth)
  • Hard to remember (error prone procedure)
Stepan
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  • Just to be clear...you want the sequence to be difficult to discern, yes? I mean, $F(n)=n$ won't work for you. Of course, "difficult to discern" is subjective. – lulu Sep 29 '18 at 18:35
  • Correct. Any polinomial P1(n) to P10(n) is easy to guess and wouldn't work. Any high degree polynomial is hard to compute on paper. – Stepan Sep 29 '18 at 18:39
  • Well, what about something like A000928, the sequence of irregular primes? It is known that there are infinitely many of these and they defy simple description. Certainly, most people would fail to spot the pattern. Might not be that easy to compute though. – lulu Sep 29 '18 at 18:44
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    Maybe something like Mertens Function. – Hazem Orabi Sep 29 '18 at 18:47
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    I fear the question is too subjective. if $p_n$ is the $n^{th}$ prime, then take some increasing polynomial $P(n)$ and define $F(n)=P(p_n)$. Easy to compute, and it seems a safe bet that nobody would spot it. Too artificial? – lulu Sep 29 '18 at 18:48
  • A000928 (a subset of primes): very promising. But all members are primes. Statistical analysis will indicate which pruning mechanism was used. – Stepan Sep 29 '18 at 18:53
  • Mertens Function (sequence A028442) is promising. Please post it as an answer so that I can accept it. – Stepan Sep 29 '18 at 18:57
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    Merten's function is not injective, not by a long shot. The sequence you referenced is not the values of Merten's functions...rather it is the list of zeroes of the Merten's function (Not sure that is known to be infinite, but of course it may be). – lulu Sep 29 '18 at 18:59
  • P(F(some number)) of high order (square or above) would grow too fast. (some_number)^4 will quickly exceed 4.2 Billion. m=a*(n)+b is a good obfuscation technique on top of A000928 – Stepan Sep 29 '18 at 18:59
  • this question claims to show that there are infinitely many zeroes for the Merten function. I didn't study the proof carefully, but it looks easy to follow. – lulu Sep 29 '18 at 19:01

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