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A160691
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a(n) = the number of divisors of A160689(n) = the number of divisors of A160690(n).
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5
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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, 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, 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
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OFFSET
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1,2
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COMMENTS
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For the first 200000 natural numbers n, a(n) is in the set {1,2,4,6,8,12}
and in fact we have:
For one number n, a(n)=1.
For 13 numbers n, a(n)=12 (see the sequence A158963).
For 4785 numbers n, a(n)=6.
For 6706 numbers n, a(n)=8.
For 26790 numbers n, a(n)=2.
For 161705 numbers n, a(n)=4.
Also n=2 is the only number (less than 200000) such that a(n) = a(n+1) = a(n+2) = 2.
And for the 53 consecutive numbers 64833, 64834, ..., 64885 we have a(n)=4. (End)
a(n)=10 for n=271532 and n=424519 (up to 5*10^5). - Michel Marcus, Sep 05 2017
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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