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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. 1


%S 12,45,80,90,144,180,189,240,360,378,448,637,720,756,945,1274,1344,

%T 1512,1625,1728,1890,1911,2025,2240,2548,2673,3024,3185,3250,3780,

%U 3822,4032,4050,4875,5096,5346,5733,6048,6125,6370,6400,6500,6517,6720,7007,7560,7644

%N 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.

%C 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).

%C 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.

%C 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.

%D 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.

%H Alois P. Heinz, <a href="/A135710/b135710.txt">Table of n, a(n) for n = 1..2000</a>

%e Example: 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).

%p a:= proc(n) option remember; local k; for k from 1+

%p `if`(n=1, 1, a(n-1)) while (s-> nops(s)<2 or (l->

%p l[-2]<l[-1])(sort(map(p-> (p-1)*padic[ordp](k, p),

%p s))))([numtheory[factorset](k)[]]) do od; k

%p end:

%p seq(a(n), n=1..50); # _Alois P. Heinz_, Oct 23 2020

%t 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 *)

%o (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)))

%Y Cf. A027868, A011371, A054861.

%K easy,nonn

%O 1,1

%A _Noam D. Elkies_, Nov 25 2007

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