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# User talk:Giovanni Resta

Remarkable. Just rediscovered your 11853735811. Thought I had successfully already searched and obtained a negative to the prior term, and had planned to submit.James G. Merickel 17:56, 8 July 2013 (UTC) The problem of finding the next term is somewhat interesting. It requires a less exhaustive approach by using (something like) the modern methods of efficiently obtaining the prime-counting function's values.James G. Merickel 21:06, 30 May 2015 (UTC)

Thanks for adding the Mathematica to A216208. --Susanne Bobzien 18:41, 13 March 2013 (UTC)

@Giovanni Resta - Hi, Re : A217254 -could you please verify your comment to see where it stands after round was changed to floor Specifically, the Floor function is used - in the formula section. Alexander R. Povolotsky 00:05, 21 March 2013 (UTC)

https://oeis.org/A181392/b181392.txt How did you compute this sequence A181392 so fast and why is "The last term is 999999999786876856487355368576387875784644". Did you use bruteforce? Patrick Wieschollek 21:13, 20 May 2013 (UTC)

- Yes, is there some chance of you re-wording your comment in any case? I like the result, and am interested in its basis also and think some brief remark on it would be good to include. It's also the case that the 43-digit terms mentioned fail to exist entirely, so at least a rewording. I myself would say 45-digit numbers in the sequence are not square because [reason or method: Exhaustive with a 0 always appearing, by a modulus other than 3, how?], 44-digit numbers fail modulo 3, and there are no 43-digit numbers, plus perhaps something about the specific finding. It's a nice but unfounded result as is, with a minutely bad wording.James G. Merickel 05:42, 26 July 2015 (UTC)
- Problem here mostly mine. Nice improvement, in any case.James G. Merickel 18:51, 26 July 2015 (UTC)

- With only its statement as fact, I can't guess why you are sure of your comment (on an absence of 4th and higher power terms) at the originating sequence, A108571, either. I trust the assertion is firmly based, but would like a little clue upon what.James G. Merickel 05:52, 26 July 2015 (UTC)

@Giovanni Resta - Hi Giovanni, Re A023212: - as far as submitted PARI program is concerned, did you notice that it does not explicitly test (n^n-1)mod (4*n+1)= 0 for *any* n (as you seems to imply in your pinky note) - it only does it for primes (note that the loop construction "forprime" is used). Alexander R. Povolotsky 01:15, 24 May 2013 (UTC)

## Contents

## A240563

Thanks for the assist & patience on this. I imagine a family of curves could be generated, from various starting points (2, then 3, then 5, etc). These are the types of sequences I need for my work ...--Bill McEachen 14:26, 10 April 2014 (UTC)

## New @ A239412

I decided for an entirely different reason than yours--that of confusing sourcing on those ranks (of island sizes)--to change the sequence (still in editing) to that which is more mathematical and derived from the original idea (with date change). Just to inform you. I don't think we should have every list here, I agree. The consecutive nature of four of the rankings does have a freakish nature, though; and the first 7 rank values, according to the source I have (Hammond 2002), produces an interesting result.James G. Merickel 18:52, 22 April 2014 (UTC)

## A103514

Another editor sent back up this one you proposed for review with a claim I did something wrong I should know better about already. My last experience with him was to have a different editor approve something he didn't seem to grasp at all. Could you please see who's lost in this case? Maybe I'm totally wrong. It will be ridiculous if he just means something about dates, but I doubt he means that.James G. Merickel 12:23, 8 August 2015 (UTC) Jon resolved the issue. I needed to hyphenate.James G. Merickel 21:51, 8 August 2015 (UTC)

## A261752

Hi Giovanni. Thanks for extending (and, I assume, verifying) the sequence. Cheers! Matthew Conroy 19:11, 31 August 2015 (UTC)