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
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.
Programs
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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 *)
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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)))
Comments