A309891 a(n) is the total number of trailing zeros in the representations of n over all bases b >= 2.

0, 1, 1, 3, 1, 3, 1, 5, 3, 3, 1, 6, 1, 3, 3, 8, 1, 6, 1, 6, 3, 3, 1, 9, 3, 3, 5, 6, 1, 7, 1, 10, 3, 3, 3, 11, 1, 3, 3, 9, 1, 7, 1, 6, 6, 3, 1, 13, 3, 6, 3, 6, 1, 9, 3, 9, 3, 3, 1, 12, 1, 3, 6, 14, 3, 7, 1, 6, 3, 7, 1, 15, 1, 3, 6, 6, 3, 7, 1, 13, 8, 3, 1, 12

Offset: 1

Author

Rémy Sigrist, Aug 21 2019

Keywords

• nonn
• base
• changed

Comments

a(n) depends only on the prime signature of n.

a(n) is the sum of the k-adic valuations of n for k >= 2. - Friedjof Tellkamp, Jan 25 2025

Examples

			For n = 12: 12 has 2 trailing zeros in base 2 (1100), 1 trailing zero in bases 3, 4, 6 and 12 (110, 30, 20, 10) and no trailing zero in other bases, hence a(12) = 1*2 + 4*1 = 6.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A001221, A006218, A007814, A007949, A033093, A056239, A112765, A122841, A169594, A214411, A235127, A244413, A286561, A293514, A369180.

Cf. A111003.

First differences of A078632.

Programs

Mathematica

Table[DivisorSum[n, IntegerExponent[n, #] &, # > 1 &], {n, 84}] (* Jon Maiga, Aug 25 2019 *)

PARI

a(n) = sumdiv(n, d, if (d>1, valuation(n,d), 0))

PARI

a(n) = {if(n == 1, return(0)); my(f = factor(n)[, 2], res = 0, t = 2, of = f, nf = f >> 1, nd(v) = prod(i = 1, #v, v[i] + 1)); while(Set(of) != [0], res += (nd(of) - nd(nf)) * (t-1); of = nf; t++; nf = f \ t); res} \\ David A. Corneth, Aug 22 2019

Formula

a(n) = Sum_{d|n, d>1} A286561(n,d), where A286561 gives the d-valuation of n.

a(p) = 1 for any prime number p.

a(p^k) = A006218(k) for any k >= 0 and any prime number p.

a(n) = 2^A001221(n) - 1 for any squarefree number n.

a(n) = 3 for any semiprime number n.

a(m*n) >= a(m) + a(n).

a(n) >= A007814(n) + A007949(n) + A235127(n) + A112765(n) + A122841(n) + A214411(n) + A244413(n).

a(n) = A056239(A293514(n)). - Antti Karttunen, Aug 22 2019

a(n) <= A033093(n). - Michel Marcus, Aug 22 2019

a(n) = A169594(n) - 1. - Jon Maiga, Aug 25 2019

From Friedjof Tellkamp, Feb 27 2024: (Start)

G.f.: Sum_{k>=2, j>=1} x^(k^j)/(1-x^(k^j)).

Dirichlet g.f.: zeta(s) * Sum_{k>=1} (zeta(k*s) - 1).

Sum_{n>=1} a(n)/n^2 = Pi^2/8 (A111003). (End)