[seqfan] Re: A202211

[Moshe Levin]

Here are three different periods:

P=3: 21,25,25,21,25
P=25: in your message
P=51: 93, 97, 257, 709, 1933, 5285, 4169, 1061, 2881, 2257, 257, 637, 541, 1109, 
3301, 8821, 24245, 18389, 889, 397, 1049, 2893, 2625, 541, 1097, 3277, 2421, 
765, 573, 417, 661, 1601, 4525, 3565, 425, 97, 229, 653, 1765, 2013, 829, 
1781, 1933, 4141, 4069, 829, 2285, 2573, 1081, 261, 153, 93, 97

Should be other (longer) periods.
Seems there are no sequences without period.

ML 


15 декабря 2011, 07:04 от Vladimir Shevelev <shevelev at bgu.ac.il>:
> Dear SeqFans,
> 
>  I  submitted the following sequence A202211: 
> a(1)=2, a(2)=3, for n>=3, a(n)=2*(gpd(a(n-1)+gpd(a(n-2))+1,
>  where gpd(n) is the great prime divisor of n:
> 2, 3, 11, 29, 81, 65, 33, 49, 37, 89, 253, 225, 57, 49, 53, 121, 129, 109, 305, 341, 185, 137, 349, 973, 977, 2233, 2013, 181, 485, 557, 1309, 1149, 801, 945, 193, 401, 1189, 885, 201, 253, 181, 409, 1181, 3181, 8725, 7061, 1313, 817,289
> Here we have an alternative: either the sequence is unbounded or it has a period, beginning with some place. D.S.McNeil found that the sequence has period of length 25: a(61)=a(85)=85 and a(62)=a(86)=73 and, consequently, is bounded.
> A problem is the following: a) do exist initials a(1) and a(2) depending on a given N for which the sequence has the least period of length>=N? b) do exist initials a(1) and a(2) for which the sequence has not any period? 
> 
> Best regards,
> Vladimir
> 
>  Shevelev Vladimir‎
> 
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