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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. 0
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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. LINKS EXAMPLE 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). 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 *) PROG (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))) CROSSREFS Cf. A027868, A011371, A054861. Sequence in context: A309817 A194284 A009785 * A070996 A015237 A024223 Adjacent sequences:  A135707 A135708 A135709 * A135711 A135712 A135713 KEYWORD easy,nonn AUTHOR Noam D. Elkies, Nov 25 2007 STATUS approved

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Last modified October 20 12:40 EDT 2019. Contains 328257 sequences. (Running on oeis4.)