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%I #21 Sep 05 2017 06:10:26
%S 1,2,2,2,4,2,2,4,2,2,4,2,2,4,2,4,2,4,4,4,2,4,4,4,2,4,6,4,6,2,4,4,4,4,
%T 4,4,4,4,4,4,4,2,2,4,4,4,2,4,4,2,4,2,4,4,4,2,2,4,4,2,4,4,4,2,4,2,4,4,
%U 2,4,4,4,4,4,4,4,4,4,4,4,4,2,2,4,2,4,4,2,4,4,4,4,4,4,4,4,2,4,2,4,4,2,4,4,2
%N a(n) = the number of divisors of A160689(n) = the number of divisors of A160690(n).
%C From _Farideh Firoozbakht_, May 28 2009: (Start)
%C For the first 200000 natural numbers n, a(n) is in the set {1,2,4,6,8,12}
%C and in fact we have:
%C For one number n, a(n)=1.
%C For 13 numbers n, a(n)=12 (see the sequence A158963).
%C For 4785 numbers n, a(n)=6.
%C For 6706 numbers n, a(n)=8.
%C For 26790 numbers n, a(n)=2.
%C For 161705 numbers n, a(n)=4.
%C Also n=2 is the only number (less than 200000) such that a(n) = a(n+1) = a(n+2) = 2.
%C And for the 53 consecutive numbers 64833, 64834, ..., 64885 we have a(n)=4. (End)
%C a(n)=10 for n=271532 and n=424519 (up to 5*10^5). - _Michel Marcus_, Sep 05 2017
%H Michel Marcus, <a href="/A160691/b160691.txt">Table of n, a(n) for n = 1..5000</a>
%t c[1] = 1; c[n_] := c[n] = (s = Sum[c[k], {k, n - 1}]; For[m = 1, DivisorSigma[0, m] != DivisorSigma[0, s + m], m++]; m); a[n_] := a[n] = DivisorSigma[0, c[n]]; Table[a[n], {n, 105}] (* _Farideh Firoozbakht_, May 28 2009 *)
%o (PARI) lista(nn) = {k = 1; print1(numdiv(k), ", "); last = k; for (n=2, nn, k = last+1; while(numdiv(k) != numdiv(k - last), k++); print1(numdiv(k), ", "); s += k; last = k;);} \\ _Michel Marcus_, Sep 05 2017
%Y Cf. A160689, A160690.
%Y Cf. A158963, A158964. [_Farideh Firoozbakht_, May 28 2009]
%K nonn
%O 1,2
%A _Leroy Quet_, May 24 2009
%E More terms from _Farideh Firoozbakht_, May 28 2009
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