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A090370 Least m > 3 such that gcd(n-1, m*n - 1) = m-1.

%I A090370 #23 Aug 10 2018 12:49:03
%S A090370 4,5,6,4,8,5,4,6,12,4,14,8,4,5,18,4,20,5,4,12,24,4,6,14,4,5,30,4,32,5,
%T A090370 4,18,6,4,38,20,4,5,42,4,44,5,4,24,48,4,8,6,4,5,54,4,6,5,4,30,60,4,62,
%U A090370 32,4,5,6,4
%N A090370 Least m > 3 such that gcd(n-1, m*n - 1) = m-1.
%C A090370 Choosing a pair (m, n) so as to redefine 1 hour = m*n minutes and 1 minute = m*n seconds, then the three hands of a fictitious n-hour clock coincide in exactly m-1 equally spaced positions, including that of the n o'clock position. For instance, in the cases where we select (m, n) as (6, 11), (8, 15), (4, 25), with m*n respectively equal to 66, 120, 100 (implying 1 hour = 66 minutes, 1 minute = 66 seconds; 1 hour = 120 minutes, 1 minute = 120 seconds; 1 hour = 100 minutes, 1 minute = 100 seconds), the hands coincide in exactly 6-1=5, 8-1=7, 4-1=3 equally spaced positions on a 11-hour, 15-hour, 25-hour clock respectively.
%H A090370 Giovanni Resta, <a href="/A090370/b090370.txt">Table of n, a(n) for n = 4..10000</a>
%F A090370 a(n) = 1 + A090368(k) for n=2k. [corrected by _Søren Eilers_, Aug 09 2018]
%F A090370 a(n) = 1 + A090369(k) for n=2k+1.
%e A090370 We have a(50)=8 because 50*8 = 400 is the least multiple of 50 such that gcd(50-1, 400-1) = 8 - 1 = 7.
%p A090370 A090370:=proc(n) local m; m:=4; while  (gcd(n-1, m*n - 1) <> m-1) do m:=m+1; end;  return m; end; # _Søren Eilers_, Aug 09 2018
%t A090370 a[n_] := Block[{m=4}, While[GCD[n-1, n*m-1] != m-1, m++]; m]; Table[a[k], {k, 4, 67}] (* _Giovanni Resta_, Aug 09 2018 *)
%o A090370 (PARI) a(n) = {m = 4; while (gcd(n-1,m*n - 1) != m-1, m++); return (m);} \\ _Michel Marcus_, Jul 27 2013
%Y A090370 Cf. A090368, A090369.
%K A090370 nonn
%O A090370 4,1
%A A090370 _Lekraj Beedassy_, Nov 27 2003
%E A090370 a(46) and a(49) corrected by _Søren Eilers_, Aug 09 2018