[seqfan] Re: An algorithm for multiplicative order of 2 mod 2n+1 (A002326)

[Vladimir Shevelev]

Dear Richard,
thanks, but I did not find 
any so close, at least, 
direct connection...

In addition, see also the comment 
in a submitted by me and Antti
new sequence A292720.
There it is proved that the
number of steps in the algorithm 
for the calculation of A002326(n)
required to reach  (the first) 1 
does not exceed n. Sometimes
n is reached. For example, for 
n=9, 2*n+1=19, we have exactly
9 steps with the advent of a 
permutation of all odd residues
(which are partial fractions)
 <= 2*n-1= 17 modulo 19:
5,3,11,15,17,9,7,13,1.

Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Richard J. Mathar [mathar at mpia-hd.mpg.de]
Sent: 04 October 2017 14:19
To: seqfan at seqfan.eu
Subject: [seqfan] Re: An algorithm for multiplicative order of 2 mod 2n+1 (A002326)

The algorithm in http://list.seqfan.eu/pipermail/seqfan/2017-October/017983.html
seems to be closely related to the association between terminating binary
expansions and the multiplicative orders (xi, Haupt-exponent) in Cunningham's
article "on binal fractions" of 1908. The article is cited in A002326:
http://dx.doi.org/10.2307/3602595 .

Richard

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