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Let $d_n=a_{n+1}-a_n$ be the difference of two consecutive terms of a sequence of natural numbers.
We can easily construct sequences of natural numbers $a_n$ using trigonometric functions or the floor function which have the property:

*(1)*For infinitely many $n, d_{n+1}<d_n$ .

For example, let $a_n=cos(\frac{n\pi}{2})$ or $a_n=\lfloor \frac{n}{3}\rfloor$ and $n\geq 1$ for both cases.

  • My question is

Can anybody give me an example of a sequence having the property (1) but which has a closed formula not containing trigonometric functions,or the floor function?

  • Note

I would like to see a closed formula of such a sequence (if possible) and not something general like the primes or a sequence whose sum of reciprosals diverges.
Thank you very much in advance.

EDIT:$(-1)^n=cos(n\pi)$ so don't try something like this. So what i would like to see is something without trigonometric functions ,floor functions or $(-1)^n$.
Everything else is acceptable.

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I’m probably missing something, but doesn’t $a_n=42+(-1)^n$ already work?

EDIT: Excluding $(-1)^n$ still leaves arbitrarily many sequences, e.g., A027642, (the absolute values of) A124449, or even A067029, but it all depends on your definition of “closed formula”. To me, something like

$$a_n = \min\left\{v_p(n) | v_p(n)>0\right\}$$

(which encodes A067029) is a closed formula, but you may of course disagree.

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No. $(-1)^n=cos(n\pi)$ – Konstantinos Gaitanas Nov 24 '13 at 11:40
I.e., you are looking for something that not only does not contain trigonometric functions, but additionally, cannot be rewritten in terms of them? Or not be rewritten in fewer than a thousand trig terms? Or … I don’t know. To me, $(-1)^n$ does not contain trig functions. – Christopher Creutzig Nov 24 '13 at 11:50
i made an edit i think that it looks much more precise now. – Konstantinos Gaitanas Nov 24 '13 at 11:56
The denominator of Bernoulli numbers looks like an answer. – Konstantinos Gaitanas Nov 24 '13 at 12:33

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