# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a135710 Showing 1-1 of 1 %I A135710 #20 Jul 28 2021 12:58:47 %S A135710 12,45,80,90,144,180,189,240,360,378,448,637,720,756,945,1274,1344, %T A135710 1512,1625,1728,1890,1911,2025,2240,2548,2673,3024,3185,3250,3780, %U A135710 3822,4032,4050,4875,5096,5346,5733,6048,6125,6370,6400,6500,6517,6720,7007,7560,7644 %N 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. %C A135710 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 A135710 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 A135710 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 A135710 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 A135710 Alois P. Heinz, Table of n, a(n) for n = 1..2000 %H A135710 J.-P. Allouche, J. Shallit, and R. Yassawi, How to prove that a sequence is not automatic, arXiv:2104.13072 [math.NT], 2021. %e A135710 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 A135710 a:= proc(n) option remember; local k; for k from 1+ %p A135710 `if`(n=1, 1, a(n-1)) while (s-> nops(s)<2 or (l-> %p A135710 l[-2] (p-1)*padic[ordp](k, p), %p A135710 s))))([numtheory[factorset](k)[]]) do od; k %p A135710 end: %p A135710 seq(a(n), n=1..50); # _Alois P. Heinz_, Oct 23 2020 %t A135710 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 A135710 (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 A135710 Cf. A027868, A011371, A054861. %K A135710 easy,nonn %O A135710 1,1 %A A135710 _Noam D. Elkies_, Nov 25 2007 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE