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

0,6

Links

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

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)

a(n) = A246817(floor(n/5)+1). - Jason Bard, Sep 06 2025

Mathematica

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

Prog

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

Crossrefs

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

Sequence in context: A245357 A001735 A245101 * A077307 A181943 A243757

Adjacent sequences: A246836 A246837 A246838 * A246840 A246841 A246842

Keyword

nonn,base

Author

Chai Wah Wu, Sep 04 2014

Status

approved