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

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, 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, 32, 4, 5, 6, 4

Offset: 4

Author

Lekraj Beedassy, Nov 27 2003

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 4..10000

Crossrefs

Cf. A090368, A090369.

Programs

Maple

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

Mathematica

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 *)

PARI

a(n) = {m = 4; while (gcd(n-1,m*n - 1) != m-1, m++); return (m);} \\ Michel Marcus, Jul 27 2013

Formula

a(n) = 1 + A090368(k) for n=2k. [corrected by Søren Eilers, Aug 09 2018]

a(n) = 1 + A090369(k) for n=2k+1.

Extensions

a(46) and a(49) corrected by Søren Eilers, Aug 09 2018