I was considering the number of consecutive terms of the integer reciprocal sequence (1/1, 1/2, 1/3...) which could be summed without exceeding 1, starting from the nth reciprocal. If we call this number of consecutive terms s(n) then, for example, s(2) = 2 because 1/2 + 1/3 < 1 but 1/2 + 1/3 + 1/4 > 1. The sequence begins:

1, 2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 24, 26, 28, 30, 31, 33...

A quick search on OEIS found that it formed the first column of this table which, as the page says, is a permutation of of the natural numbers. I'm not clear where the "x" in the given formula comes from, but anyway...

I also looked at the sequence t(n) = s(n) + n - 1:

1, 3, 6, 9, 11, 14, 17, 19, 22, 25, 28, 30, 33, 36, 38, 41, 44, 47, 49, 52...

My reason for doing this is best explained with an example: starting with 1/5, s(5)=7 consecutive integer reciprocals can be summed without exceeding 1. The last of these reciprocals is 1/t(5)=1/11.

Searching OEIS found this, which has a simple formula and moreover is precisely the set of "numbers n [not to be confused with the index, n, of the terms of the sequence] such that the first digit in ternary expansion on 2^n is 2" and also "appears in connection with the 3x+1 problem". I found this highly curious, as I can see no particular reason why the first digit of a power of two in ternary or, indeed, the 3x+1 problem should have anything to do with the index of the last reciprocal in a sum. Can anyone shed any light on this?

## A curious connection

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### Re: A curious connection

That is curious. Are you positive that the sequence you've generated is identical to the one on OEIS? I encourage you to check another twenty or thirty terms to see. Coincidences happen a lot in sequences. t(n), for instance, agrees in all but three of the displayed terms with u(n) = ceiling(n*e). Clearly it's not that, but the point is that initial terms can be deceiving.

either

a(x) <= t(x) + k and a(z) >= t(z) + k - 1

or

a(x) <= t(x) + k+1 and a(z) >= t(z) + k

for all x<y<z.

In other words, the amount by which t(n) is underestimated by a(n) appears to be a rounded approximation of a strictly-(and slowly)-increasing function.

Edit: ironic discovery. Approximating 1/n + 1/(n+1) + ... + 1/(t(n)) as the integral from n to t(n) of 1/n, we obtain 1 - 1/(t(n)+1) < ln(t(n)) - ln(n) < 1 [using the bounds already defined]. Taking exponentials both sides, e^LHS < t(n) / n < e. Since the integral is an approximation which improves for large n, it is fair to say that t(n) approximately equals n*e for large n.

More accurately, it is easy to show that

ln(n+1) < 1/1 + 1/2 + ... + 1/n < 1+ln(n).

Therefore,

ln(t(n)+1) - (1+ln(n-1)) < 1/n + 1/(n+1) + ... + 1/(t(n)) < 1+ln(t(n)) - ln(n)

We already know that

1 - 1/(t(n)+1) < 1/n + 1/(n+1) + ... + 1/(t(n)) < 1

Therefore,

ln(t(n)+1) - (1+ln(n-1)) < 1 and 1 - 1/(t(n)+1) < 1+ln(t(n)) - ln(n)

ln(t(n)+1) - ln(n-1) < 2 and -1/(t(n)+1) < ln(t(n)) - ln(n)

ln( (t(n)+1) / (n-1) ) < 2 and -1/(t(n)+1) < ln( t(n) / n )

t(n)+1 < (n-1)e^2 and t(n) > n / e^(1/(t(n)+1)

The second statement tells us nothing that wasn't obvious, but the first statement sets an upper bound on t(n) of (n-2)e^2, and the approximation above suggests that in fact t(n) stays close to n*e.

### Re: A curious connection

OEIS A083088 gives explicit formula for this sequence

a(n) = floor(n/sqrt(2)) + n + 1

a(n) = floor(n/sqrt(2)) + n + 1

### Re: A curious connection

Holy "five years late" batman!

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