A sequence

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Antreas Hatzipolakis

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Jan 31, 2025, 4:50:14 AMJan 31
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  7, 13, 25. 43, 67, 97, 133, 175,. .

cf.  A102724

1884, p. 111

APH
JdME 1884.png

Antti Karttunen

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Jan 31, 2025, 5:17:36 AMJan 31
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On Fri, Jan 31, 2025 at 11:50 AM Antreas Hatzipolakis <anopo...@gmail.com> wrote:

  7, 13, 25. 43, 67, 97, 133, 175,. .

cf.  A102724

Well, clearly it's not the same sequence.

A102724(9) = 227, while a(8) = 223, where a(n) = 3*(n^2 + n)+7, the polynomial formula given in your scan of that journal page.

(It would have been quite astonishing if it had been... Or maybe I missed the actual point of your mail?)

I wonder whether the latter does merit an entry in OEIS, (with a reference to that book)?
Best regards,

Antti
 

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Antreas Hatzipolakis

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Jan 31, 2025, 5:25:05 AMJan 31
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On Fri, Jan 31, 2025 at 12:17 PM Antti Karttunen <antti.k...@gmail.com> wrote:


On Fri, Jan 31, 2025 at 11:50 AM Antreas Hatzipolakis <anopo...@gmail.com> wrote:

  7, 13, 25. 43, 67, 97, 133, 175,. .

cf.  A102724

Well, clearly it's not the same sequence.

I did not say it is the same

I said "cf" = compare 

APH

Antreas Hatzipolakis

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Jan 31, 2025, 5:40:30 AMJan 31
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I mentioned it, since OEIS gave it after my search.

Also it is interesting what OEIS says.

Did not find it !

APH
Seq.png

Daniel Mondot

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Jan 31, 2025, 8:01:54 AMJan 31
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For those interested and who can't read French,
Here is a translation of the original attachment:

Solution by Mr Paul Bourgarel, from Antibes

Journal of elementary mathematics, 1884 page 111

Given the unlimited series: 7, 13, 25, 43, 67, 97, 133, 175, ...
which follows a(n) = 3(n^2 + n) + 7,
demonstrate the following propositions:
1. among any 5 consecutive terms in the series, one is divisible by 5.
2. among any 7 consecutive terms in the series, two are divisible by 7.
3. among any 13 consecutive terms in the series, two are divisible by 13.
4. no term in the series is a cube
5. an infinite number of terms are squares divisible by 25. e.g. a(2) = 25, a(37) = 4225, etc...


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