A048764 Largest factorial <= n.
1, 2, 2, 2, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24
Offset: 1
References
- J. Castillo, Other Smarandache Type Functions: Inferior/Superior Smarandache f-part of x, Smarandache Notions Journal, Vol. 10, No. 1-2-3 (1999), 202-204.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
- Krassimir T. Atanassov, On Some of Smarandache's Problems.
- Vassia K. Atanassova and Krassimir T. Atanassov, On the 43rd and 44th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5, No. 2, (1999), 86-88.
- Li Jie, On the inferior and superior factorial part sequences, in Zhang Wenpeng (ed.), Research on Smarandache Problems in Number Theory (collected papers), 2004, pp. 47-48.
- Florentin Smarandache, Only Problems, Not Solutions!.
Programs
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Mathematica
Table[k = 1; While[(k + 1)! <= n, k++]; k!, {n, 80}] (* Michael De Vlieger, Aug 30 2016 *)
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PARI
a(n)=my(t=1,k=1);while(t<=n,t*=k++);t/k \\ Charles R Greathouse IV, Sep 19 2012
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Python
from sympy import factorial as f def a(n): k=1 while f(k + 1)<=n: k+=1 return f(k) print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 21 2017, after Mathematica code
Formula
n log log n / log n << a(n) <= n. - Charles R Greathouse IV, Sep 19 2012
From Amiram Eldar, Aug 02 2022: (Start)
Sum_{n>=1} 1/a(n)^m = Sum_{k>=1} k/k!^m (Li Jie, 2004).
In particular:
Sum_{n>=1} 1/a(n)^2 = e (A001113).
Sum_{n>=1} 1/a(n)^3 = BesselI(1,2) (A096789). (End)