A376286 n! less trailing zeros (A004154) (mod nextprime(n)).

1, 1, 2, 1, 4, 5, 2, 9, 6, 10, 10, 3, 10, 7, 13, 11, 6, 8, 11, 15, 7, 9, 14, 13, 22, 20, 27, 4, 25, 16, 17, 7, 2, 29, 24, 10, 27, 3, 32, 18, 31, 21, 22, 15, 2, 9, 38, 26, 29, 43, 48, 10, 43, 55, 20, 51, 24, 11, 48, 2, 12, 57, 50, 1, 64, 14, 53, 8, 47

Offset: 0

Author

Robert G. Wilson v, Sep 18 2024

Keywords

• easy
• base
• nonn

Crossrefs

Cf. A004154, A151800.

Programs

Mathematica

a[n_] := Mod[n!/10^IntegerExponent[n!, 10], NextPrime[n]]; Array[a, 69, 0](* Becomes quicker as n increases and it uses less resources. For me, this is around 2 million *)g[n_, p_] := Block[{s = 0, e = 1}, While[t = Floor[n/p^e]; t > 0, s += t; e++]; s];f[n_] := Block[{m = NextPrime@ n, p = 1, q = 7}, p = PowerMod[2, g[n, 2] - g[n, 5], m]; p = Mod[p*PowerMod[3, g[n, 3], m], m]; While[q < n +1, p = Mod[p*PowerMod[q, g[n, q], m], m]; q = NextPrime@ q]; p]

Python

from functools import reduce
from sympy import nextprime
from sympy.ntheory.factor_ import digits
def A376286(n): return ((p:=nextprime(n))-1)*pow(reduce(lambda i, j:i*j%p, range(n+1,p),1),-1,p)*pow(10,sum(digits(n,5)[1:])-n>>2,p)%p # Chai Wah Wu, Oct 18 2024

Formula

a(n) = A004154(n) mod A151800(n).