A374781 Numbers whose divisors have a record value of the mean abundancy index.
1, 2, 4, 6, 8, 12, 24, 48, 72, 96, 120, 144, 240, 288, 360, 480, 576, 720, 1440, 2160, 2880, 4320, 5760, 8640, 12960, 15120, 17280, 20160, 25920, 30240, 40320, 51840, 60480, 90720, 120960, 181440, 241920, 302400, 362880, 483840, 604800, 725760, 907200, 1209600
Offset: 1
Keywords
Examples
The values of A374777(k)/A374778(k) for k = 1..6 are {1, 5/4, 7/6, 17/12, 11/10, 35/24} = {1, 1.25, 1.166..., 1.416..., 1.1, 1.458...}. The records values, {1, 1.25, 1.416..., 1.458...} occur at k = 1, 2, 4, and 6, the first 4 terms of this sequence.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..778
Programs
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Mathematica
f[p_, e_] := ((e+1)*p^2 - (e+2)*p + p^(-e))/((e+1)*(p-1)^2); s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; seq[kmax_] := Module[{v = {}, smax = 0, s1}, Do[s1 = s[k]; If[s1 > smax, AppendTo[v, k]; smax = s1], {k, 1, kmax}]; v]; seq[10^5]
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PARI
s(n) = {my(f = factor(n)); prod(i = 1, #f~, p=f[i,1]; e=f[i,2]; (-2*p - e*p + p^2 + e*p^2 + p^(-e))/((e + 1)*(p - 1)^2));} lista(kmax) = {my(smax = 0, s1); for(k = 1, kmax, s1 = s(k); if(s1 > smax, print1(k, ", "); smax = s1));}
Comments