A135710 Positive integers b such that more than one prime factor p of b attains the maximum of (p-1)*v_p(b) where v_p(b) is the valuation of b at p.

12, 45, 80, 90, 144, 180, 189, 240, 360, 378, 448, 637, 720, 756, 945, 1274, 1344, 1512, 1625, 1728, 1890, 1911, 2025, 2240, 2548, 2673, 3024, 3185, 3250, 3780, 3822, 4032, 4050, 4875, 5096, 5346, 5733, 6048, 6125, 6370, 6400, 6500, 6517, 6720, 7007, 7560, 7644

Offset: 1

Author

Noam D. Elkies, Nov 25 2007

Keywords

• easy
• nonn

Comments

Given b, the number of trailing zeros at the end of the base-b representation of x! is asymptotic to x/M where M is the maximum over p|b of (p-1)*v_p(b).

Usually only one prime p attains the maximum and then the number is v_p(x!)/v_p(b) for all but finitely many x.

But for b=12,45,80,90,..., at least two v_p(x!) must be computed. For example: if b=12 then for x=2006 there are 998 trailing zeros due to v_3 but for x=2007 there are 999 due to v_2.

Examples

			For b=90 we have (p-1)*v_p(b) = 1, 4, 4 for p = 2, 3, 5 respectively so the maximum of 4 is attained twice (p=3 and p=5).
		

References

• Eryk LIPKA, Automaticity of the sequence of the last nonzero digits of n! in a fixed base, Journal de Théorie des Nombres de Bordeaux 31 (2019), 283-291. [See Theorem 3.7 on page 290, and consider the complementary sequence.] - Jean-Paul Allouche and Don Reble, Oct 22 2020.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..2000
• J.-P. Allouche, J. Shallit, and R. Yassawi, How to prove that a sequence is not automatic, arXiv:2104.13072 [math.NT], 2021.

Crossrefs

Cf. A027868, A011371, A054861.

Programs

Maple

a:= proc(n) option remember; local k; for k from 1+
     `if`(n=1, 1, a(n-1)) while (s-> nops(s)<2 or (l->
      l[-2] (p-1)*padic[ordp](k, p),
       s))))([numtheory[factorset](k)[]]) do od; k
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Oct 23 2020

Mathematica

F[n_] := Module[{f, p, v, vmax}, f = FactorInteger[n]; p = f[[All, 1]]; v = Table[ f[[i, 2]]*(p[[i]]-1), {i, 1, Length[p]}]; vmax = Max[v]; Sum[Boole[v[[i]] == vmax], {i, 1, Length[v]}]]; Reap[For[n = 1, n <= 6400, n++, If[F[n] > 1, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jan 09 2014, translated from PARI *)

PARI

{ F(n, f,p,v,vmax)= f=factor(n); p=f[,1]; v=vector(length(p),i,f[i,2]*(p[i]-1)); vmax=vecmax(v); sum(i=1,length(v),v[i]==vmax) } for(n=1,6400,if(F(n)>1,print(n)))