|
| |
|
|
A090370
|
|
Least m>3 such that gcd(n-1,m*n - 1)=m-1.
|
|
0
| |
|
|
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, 6, 24, 48, 5, 8, 6, 4, 5, 54, 4, 6, 5, 4, 30, 60, 4, 62, 32, 4, 5, 6, 4
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 4,1
|
|
|
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.
|
|
|
FORMULA
| a(n)=1+A090368(k-1) for n=2k. a(n)=1+A090369(k) for n=2k+1.
|
|
|
EXAMPLE
| We have a(50)=8 because 50*8=400 is the least multiple of 50 such that (50-1,400-1)=8-1=7.
|
|
|
CROSSREFS
| Cf. A090368, A090369.
Sequence in context: A201933 A016719 A196999 * A002129 A113184 A136004
Adjacent sequences: A090367 A090368 A090369 * A090371 A090372 A090373
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Lekraj Beedassy (blekraj(AT)yahoo.com), Nov 27 2003
|
| |
|
|
