A073095 Numbers k such that the final nonzero digit of k! is the same as the last digit of binomial(2k,k) (in base 10).

5, 12, 26, 31, 35, 51, 56, 60, 136, 152, 157, 177, 182, 252, 257, 275, 280, 287, 300, 305, 312, 627, 632, 650, 655, 662, 675, 680, 687, 751, 756, 760, 786, 811, 886, 902, 907, 927, 932, 1251, 1256, 1260, 1286, 1311, 1377, 1382, 1400, 1405, 1412, 1425

Offset: 1

Author

Benoit Cloitre, Aug 18 2002

Keywords

• base
• easy
• nonn

Examples

			12! = 479001600, binomial(24,12) = 2704156, and the last nonzero digit of 12! is the same as the last digit of binomial(24,12), hence 12 is in the sequence.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Programs

Mathematica

Select[Range[1500],Mod[#!/10^IntegerExponent[#!,10],10]==Mod[Binomial[2 #,#],10]&] (* Harvey P. Dale, Sep 13 2022 *)

Python

from math import comb
from functools import reduce
from itertools import count, zip_longest, islice
from sympy.ntheory.factor_ import digits
from sympy.ntheory.modular import crt
def A073095_gen(startvalue=2): # generator of terms >= startvalue
    for n in count(max(startvalue,2)):
        s, s2 = digits(n,5)[-1:0:-1], digits(n<<1,5)[-1:0:-1]
        if reduce(lambda x,y:x*y%10,(((6,2,4,8,6,2,4,8,2,4,8,6,6,2,4,8,4,8,6,2)[(a<<2)|(i*a&3)] if i*a else (1,1,2,6,4)[a]) for i, a in enumerate(s)),6)==crt([2,5],[0,reduce(lambda x,y:x*y%5,(comb(a, b) for a, b in zip_longest(s2,s,fillvalue=0)))])[0]:
            yield n
A073095_list = list(islice(A073095_gen(),50)) # Chai Wah Wu, Dec 07 2023

Formula

k such that A008904(k) = binomial(2k, k) reduced (mod 10).