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A063938
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Numbers k that divide tau(k), where tau(k)=A000594(k) is Ramanujan's tau function.
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17
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 50, 54, 56, 60, 63, 64, 70, 72, 75, 80, 81, 84, 88, 90, 91, 92, 96, 98, 100, 105, 108, 112, 115, 120, 125, 126, 128, 135, 140, 144, 147, 150, 160, 161, 162, 168
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OFFSET
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1,2
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COMMENTS
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Although most small numbers are in the sequence, it becomes sparser for larger values; e.g., only 504 numbers up to 10000 and only 184 numbers from 10001 to 20000 are in the sequence.
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LINKS
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MATHEMATICA
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(* First do <<NumberTheory`Ramanujan` *) test[n_] := Mod[RamanujanTau[n], n]==0; Select[Range[200], test]
(* Second program: *)
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PROG
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(PARI 2.8) for (n=1, 1000, if(Mod(ramanujantau(n), n)==0, print1(n", "))) \\ Dana Jacobsen, Sep 06 2015
(Perl) use ntheory ":all"; my @p = grep { !(ramanujan_tau($_) % $_) } 1..1000; say "@p"; # Dana Jacobsen, Sep 06 2015
(Python)
from itertools import count, islice
from sympy import divisor_sigma
def A063938_gen(startvalue=1): # generator of terms >= startvalue
return filter(lambda n: not -840*(pow(m:=n+1>>1, 2, n)*(0 if n&1 else pow(m*divisor_sigma(m), 2, n))+(sum(pow(i, 4, n)*divisor_sigma(i)*divisor_sigma(n-i) for i in range(1, m))<<1)) % n, count(max(startvalue, 1)))
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CROSSREFS
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For the sequence when n is prime see A007659.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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