[seqfan] Re: A202211

[Harvey P. Dale]

	Here is a Mathematica program that will calculate the terms of the sequence rapidly:

nxt[{a_,b_}]:={b,2(FactorInteger[a][[-1,1]]+FactorInteger[b][[-1,1]])+1}; Transpose[NestList[nxt,{2,3},120]][[1]]

To test other initial terms, i.e., different terms for a(1) and a(2), merely insert them inside the parentheses in the NestList statement.

	Best,

	Harvey
 

-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Vladimir Shevelev
Sent: Wednesday, December 14, 2011 5:48 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] A202211

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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