Closed form of $\sum _{k=1}^{n }\:\lfloor{n/k}\rfloor$

Asked Apr 10 '22 at 16:17

Active Apr 10 '22 at 16:50

Viewed 76 times

I came across a question in which, if I am able to calculate this sum,

$\sum_{k=1}^{n }\:\lfloor{n/k}\rfloor=$ ?

it would get solved quite easily. I have never seen any closed form for this question, (though I saw a simplification by a previous answer: $\sum_{k=1}^{n }\:\lfloor{n/k}\rfloor=2\sum_{k=1}^{\lfloor\sqrt{n}\rfloor }\:\lfloor{n/k}\rfloor-\lfloor{\sqrt{n}}\rfloor^2$).

I was wondering, does there exist any closed form or other simplified ways of calculation for this sum?

Thanks in advance!

edited Apr 10 '22 at 16:50

RobPratt

50,938

asked Apr 10 '22 at 16:17

shsh23

1,174

Your sum is not defined when $k=0$. – SacAndSac Apr 10 '22 at 16:19
My apologies, I meant $k=1$. – shsh23 Apr 10 '22 at 16:20
2

Look at https://oeis.org/search?q=1%2C3%2C5%2C8%2C10%2C14%2C16%2C20%2C23%2C27&language=french&go=Chercher – TheSilverDoe Apr 10 '22 at 16:28
@TheSilverDoe Thanks a lot! – shsh23 Apr 10 '22 at 16:35
2

This is the divisor summary function, its closed form is not known. Even its leading error bound is a famous open problem. – achille hui Apr 10 '22 at 17:09
@achillehui Thanks! – shsh23 Apr 10 '22 at 17:29

Closed form of $\sum _{k=1}^{n }\:\lfloor{n/k}\rfloor$

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