OFFSET
0,2
COMMENTS
For corresponding numbers m of pairs (m, n) that generate matching times using the given equation, see A121578. [The pair (94, 127) thus implies that a(94)=75317 which stands for 7hr53min17s on a clock will read exactly a(127)=103926, i.e., 10hr39min26s when the hour and minute hands come to merely interchange positions.]
REFERENCES
J. D. E. Konhauser et al., Which Way Did The Bicycle Go?, Problem 171 pp. 54;207 MAA Washington DC 1996.
S. G. Krantz, Techniques of Problem-Solving, Problem 4.11 pp. 421-2 AMS Providence RI 1997.
H. Steinhaus, One Hundred Problems In Elementary Mathematics, pp. 123-4 Dover NY 1979.
A. Latto, "A Clock Puzzle" in 'Puzzler's Tribute' Ed. D. Wolfe and T. Rodgers pp. 41-2 A. K. Peters 2002.
J. Roberts, Lure Of The Integers, "Clockhands:Entry 143" pp. 223 MAA Washington DC 1992.
P. Tougne, "Montre de mathématicien", Réponse Au Jeu-Concours No. 68 in 'Pour La Science' (French edition of 'Scientific American') pp. 7 No. 270 April 2000 Paris.
LINKS
B. Ford et al., Confusing Clocks, Mathematics Magazine, pp. 190-1 vol. 71 No. 3 1998.
FORMULA
a(n)=round(43200*n/143) expressed in doubly-spaced sexagesimal scale. In other words, the required times occur at 143 successive positions after every (12/143)hr, i.e., 5min2s and 14/143s from noon or midnight.
EXAMPLE
2008 and 30116, for instance, in the sequence stand respectively for 00:20:08 and 3:01:16 or 0hr20min08s and 3hr01min16s.
We have a(25)=7552 to base 60 (doubly-spaced), i.e. 7552 = 2*60^2 + 05*60 + 52*1 to base 60, which is 20552.
CROSSREFS
KEYWORD
fini,nonn
AUTHOR
Lekraj Beedassy, Aug 08 2006
STATUS
approved