A246839 Number of trailing zeros in A002109(n).

0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 15, 15, 15, 15, 15, 30, 30, 30, 30, 30, 50, 50, 50, 50, 50, 100, 100, 100, 100, 100, 130, 130, 130, 130, 130, 165, 165, 165, 165, 165, 205, 205, 205, 205, 205, 250, 250, 250, 250, 250, 350, 350, 350, 350, 350, 405, 405, 405, 405

Offset: 0

Author

Chai Wah Wu, Sep 04 2014

Keywords

• nonn
• base

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..999

Crossrefs

Cf. A002109, A112765, A122840, A246817, A027868, A249152.

Programs

Mathematica

(n=#;k=0;While[Mod[n,10]==0,n=n/10;k++];k)&/@Hyperfactorial@Range[0,60] (* Giorgos Kalogeropoulos, Sep 14 2021 *)

PARI

a(n) = sum(i=1, n, i*valuation(i, 5)); \\ Michel Marcus, Sep 14 2021

Python

def a(n):
  s = 1
  for k in range(n+1):
    s *= k**k
  i = 1
  while not s % 10**i:
    i += 1
  return i-1
n = 1
while n < 100:
  print(a(n),end=', ')
  n += 1 # Derek Orr, Sep 04 2014

Python

from sympy import multiplicity
A246839, p5 = [0,0,0,0,0], 0
for n in range(5,10**3,5):
    p5 += multiplicity(5,n)*n
    A246839.extend([p5]*5)
# Chai Wah Wu, Sep 05 2014

Formula

From Michel Marcus, Sep 14 2021: (Start)

a(n) = A122840(A002109(n)), but also,

a(n) = A112765(A002109(n)), see explanation in A002109; so

a(n) = Sum_{i=1..n} i*v_5(i), where v_5(i) = A112765(i) is the exponent of the highest power of 5 dividing i. After a similar formula in A249152. (End)