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A007456 Number of days required to spread gossip to n people. 7
0, 1, 3, 2, 4, 3, 4, 3, 5, 4, 5, 4, 5, 4, 5, 4, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
On the first day, each gossip has his own tidbit. On each successive day, disjoint pairs of gossips may share tidbits (over the phone). After a(n) days, all gossips have all tidbits.
a(A240277(n)) = n and a(m) < n for m < A240277(n). - Reinhard Zumkeller, Apr 03 2014
REFERENCES
D. Shasha, Gossiping Defenders, The Puzzling Adventures of Dr. Ecco, pp. 62-4;156 W. H. Freeman NY 1988.
LINKS
C. Kenneth Fan, Bjorn Poonen and George Poonen, How to spread rumors fast, Mathematics Magazine 70 (Feb, 1997), pp. 40-42.
I. Peterson, Spreading Rumors, MathLand, March 17, 1997.
FORMULA
a(1) = 0; for n >= 2, a(n) = floor(log_2(n-1)) + ((n-2) mod 2) + 1.
G.f.: -1 + (1/(1-z))*(1/(1+z) + Sum_{k>=0} z^(2^k)). - Ralf Stephan, Apr 06 2003
MATHEMATICA
Join[{0}, Table[Floor[Log[2, n - 1]] + Mod[n - 2, 2] + 1, {n, 2, 100}]] (* T. D. Noe, Mar 16 2012 *)
PROG
(Haskell)
a007456 1 = 0
a007456 n = a000523 (n - 1) + mod n 2 + 1
-- Reinhard Zumkeller, Apr 03 2014
CROSSREFS
Cf. A000523.
Sequence in context: A243289 A134559 A333773 * A316141 A119707 A354679
KEYWORD
nonn,nice,easy
AUTHOR
Alex Graesser (AlexG(AT)sni.co.za)
EXTENSIONS
More terms from David W. Wilson
Formulae corrected by Johannes W. Meijer, May 15 2009
STATUS
approved

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Last modified March 18 22:56 EDT 2024. Contains 370952 sequences. (Running on oeis4.)