A341232 Numerator of the expected fraction of guests without a napkin in Conway's napkin problem with n guests.

0, 0, 1, 11, 39, 473, 19897, 63683, 5731597, 22926439, 280212089, 20175270749, 224810160067, 6294684482461, 1321883741325001, 1208579420640469, 68486167169628137, 17258514126746312369, 178860964586279976467, 6053755724458706915971, 3305350625554453976644453

Offset: 1

Author

Pontus von Brömssen, Feb 07 2021

Keywords

• nonn
• frac

Examples

			0, 0, 1/12, 11/96, 39/320, 473/3840, 19897/161280, 63683/516096, 5731597/46448640
		

References

• Peter Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004, p. 122.

Links

• Anders Claesson and T. Kyle Petersen, Conway’s napkin problem, arXiv:math/0505080 [math.CO], 2005.
• Anders Claesson and T. Kyle Petersen, Conway's napkin problem, Amer. Math. Monthly 114 (No. 3, 2007), 217-231.
• Niklas Eriksen, The freshman's approach to Conway's napkin problem, Amer. Math. Monthly 115 (No. 6, 2008), 492-498.
• Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020, p. 2.

Crossrefs

Cf. A248788, A341233 (denominators).

Programs

Python

from sympy import numer, S, factorial
def A341232(n):
  return numer(sum((1-S(2)**(2-k))/factorial(k) for k in range(2,n+1)))

Python

from math import factorial
from fractions import Fraction
def a(n):
  s = sum(Fraction(2**k-4, 2**k*factorial(k)) for k in range(2, n+1))
  return s.numerator
print([a(n) for n in range(1, 22)]) # Michael S. Branicky, Feb 07 2021

Formula

a(n)/A341233(n) = Sum_{k=2..n} (1-2^(2-k))/k!.

Lim_{n->oo} a(n)/A341233(n) = (2-sqrt(e))^2 (A248788).