Consider the generalized harmonic numbers evaluated at $2^{-1}$ $$H_{n,2^{-1}} = 1+ {1 \above 1.5pt \sqrt{2}}+ \ldots+ {1 \above 1.5pt \sqrt{n}}$$ The table below lists some initial values:

$$\begin{array}{nc|cccccc} n &1&2&3&4&5&6&7 \\ \hline H_{n,2^{-1}} & 1.00 & 1.71 &2.28&2.78&3.23&3.64&4.01\\ \end{array}$$

Let $$s =min\{n\text{ }|\text{ }H_{n,2^{-1}} \geq n \}$$ For example the smallest $n$ such that $H_{n,2^{-1}}\ge 1$ is $1$. Similarly the smallest $n$ such that $H_{n,2^{-1}}\ge 2$ is $3$. We have the following sequence for $s$ $$\mathfrak a(s) =1,3,5,7,10,14,18,22,\ldots$$ I am asking if the following claim is true -

$$\mathfrak a(s) = \sum_{n\leq s}\Bigg(\Bigg\lfloor{n+2 \above 1.5pt 4}\Bigg\rfloor+\Bigg\lfloor{n+1 \above 1.5pt 4}\Bigg\rfloor\Bigg)$$

Note that $\mathfrak a(s)$ is the sequence A054040.

  • 1
    $\begingroup$ Note if the above claim is true then we can also show that $$\mathfrak a(s) = {1 \above 1.5pt 2}\sum_{n\leq s}\Bigg(n -\sin \left(\frac12 n \pi\right)\Bigg)$$ $\endgroup$ Aug 17, 2016 at 22:32

1 Answer 1


Note that $$H_{n,2^{-1}}=\int_1^{n+1}\frac{dt}{\sqrt{\lfloor t\rfloor}}$$

So $$\int_1^{n+1}\frac{dt}{\sqrt{t}}\le H_{n,2^{-1}}\le1+\int_2^{n+1}\frac{dt}{\sqrt{t-1}}$$

That is,

$$2(\sqrt{n+1}-1)\le H_{n,2^{-1}}\le 1+2(\sqrt{n}-1)$$

For example, $1998 \le H_{999,999,\,2^{-1}} < 1999$. This means that $\mathfrak a(1998)\le 999,999 < \mathfrak a(1999)$.


$$\begin{align}\sum_{n\le 1999}&\left(\left\lfloor\frac{n+2}4\right\rfloor+\left\lfloor\frac{n+1}4\right\rfloor\right)\\ &=\sum_{n\le 1999}\frac{2n+3}4\\ &-\left(\frac12+\frac34\right)\cdot499\qquad\text{for $n=4k+1$}\\ &-\frac34\cdot499\qquad\text{for }n=4k+2\\ &-\frac14\cdot499\qquad\text{for }n=4k+3\\ &-\left(\frac12+\frac14\right)\cdot498\qquad\text{for }n=4k\\ &=999,413<999,999 \end{align}$$

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    $\begingroup$ If there is any mistake, please point it. $\endgroup$
    – ajotatxe
    Aug 18, 2016 at 1:57
  • $\begingroup$ Perhaps it might be better to disprove the claim by a smallest counter example. Also I am not certain your approach is correct. It is nearly 11 pm EST I will come back tomorrow and look at this. $\endgroup$ Aug 18, 2016 at 2:25
  • $\begingroup$ You are correct. Look at the sequence $$0,1,2,2,2,3,4,4,4,5,6,6,6,\ldots$$ that is "three evens followed by one odd". Then $\mathfrak a(s)$ are the partial sums of the above sequence. A smaller counter example would be $\mathfrak a(100)$. $\endgroup$ Aug 18, 2016 at 10:57

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