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.

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

Offset: 1

Author

Richard Ehrenborg, Dec 09 2009

Keywords

• easy
• nonn

Comments

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

Examples

			a(4) = 2 since in the classical problem with 4 glasses on a tray, the blind bartender needs 2 hands.
		

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 285-307.

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), 249-266.
• T. Lewis and S. Williard, The rotating table, Mathematics Magazine, vol. 53, no. 3 (May 1980) pages 174-179.

Crossrefs

Cf. A006530, A052126, A060681.

Programs

Haskell

a171462 n = div n p * (p - 1) where p = a006530 n
-- Reinhard Zumkeller, Apr 06 2015

Mathematica

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

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

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

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

a(n) = n - A052126(n) = n - n/A006530(n). - Antti Karttunen, Jan 03 2024