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k such that A000203(k) mod A000005(k) = 1. [Paolo P. Lava, Jan 31 2013]
with(numtheory);
A110878:=proc(q)
local n;
for n from 1 to q do if (sigma(n) mod tau(n))=1 then print(n);
fi; od; end:
A110878(100000000); # Paolo P. Lava, Jan 31 2013
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Select[Range[20000], Divisible[DivisorSigma[1, #]-1, DivisorSigma[0, #]]&] (* Harvey P. Dale, Dec 23 2020 *)
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Positive integers n k such that the sum of the divisors of n, k, excluding 1, is a multiple of the number of divisors of nk.
n k such that A000203(nk) mod A000005(nk) = 1. [Paolo P. Lava, Jan 31 2013]
The 9 divisors of 36 are {1,2,3,4,6,9,12,18,36}, giving sigma(36)-1=90 , which is a multiple of 9. Thus 36 is a term of the sequence.
A110878(100000000); # Paolo P. Lava, Jan 31 2013.
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n such that A000203(n) mod A000005(n) = 1. [Paolo P. Lava, Jan 31 2013]
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