

A171462


Number of hands a bartender needs to have in order to win at the blind bartender's problem with n glasses in a cycle.


24



0, 1, 2, 2, 4, 4, 6, 4, 6, 8, 10, 8, 12, 12, 12, 8, 16, 12, 18, 16, 18, 20, 22, 16, 20, 24, 18, 24, 28, 24, 30, 16, 30, 32, 30, 24, 36, 36, 36, 32, 40, 36, 42, 40, 36, 44, 46, 32, 42, 40, 48, 48, 52, 36, 50, 48, 54, 56, 58, 48, 60, 60, 54, 32, 60, 60, 66, 64, 66, 60, 70, 48, 72
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OFFSET

1,3


COMMENTS

For n greater than 1, the nth entry is given by n*(11/p) where p is largest prime dividing n.


REFERENCES

W. T. Laaser and L. Ramshaw, Probing the Rotating Table, The Mathematical Gardner (edited by David A. Klarner), Prindle, Weber & Schmidt, Boston, Massachusetts, 1981, pages 285307.


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
R. Ehrenborg and C. Skinner, The blind bartender's problem, Journal of Combinatorial Theory, Series A 70 (1995), 249266.
T. Lewis and S. Williard, The rotating table, Mathematics Magazine, vol. 53, no. 3 (May 1980) pages 174179.


FORMULA

Conjecture: n > 1: k=1..n: a(n) = n*min(A191898(n, k)/k). Verified up to n=10000.  Mats Granvik, Apr 19 2021


EXAMPLE

The fourth entry is two since is the classical problem with four glasses on a tray, the blind bartender needs 2 hands.


MATHEMATICA

{0}~Join~Array[# (1  1/FactorInteger[#][[1, 1]]) &, 72, 2] (* Michael De Vlieger, Jul 08 2020 *)


PROG

(PARI) a(n) = {if (n == 1, return (0)); f = factor(n); p = f[#f~, 1]; return (n * (p  1)/p); } \\ Michel Marcus, Jun 09 2013
(Haskell)
a171462 n = div n p * (p  1) where p = a006530 n
 Reinhard Zumkeller, Apr 06 2015
(Python)
from sympy import primefactors
def a(n): return 0 if n == 1 else n  n//(primefactors(n)[1])
print([a(n) for n in range(1, 74)]) # Michael S. Branicky, Apr 19 2021


CROSSREFS

Cf. A006530, A060681.
Sequence in context: A128923 A153835 A338098 * A075857 A023817 A300754
Adjacent sequences: A171459 A171460 A171461 * A171463 A171464 A171465


KEYWORD

easy,nonn


AUTHOR

Richard Ehrenborg, Dec 09 2009


STATUS

approved



