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%I #44 Feb 15 2019 18:45:35
%S 2,110,152,506,830,882,8138,13826,15878,19514,37634,93242,99002,
%T 153216,218978,576902,998978,2259758,3041798,5326106,6654278,7709006,
%U 7772762,31833002,44564438,106657202,279422306,1702668664,1774448104,2132364366,3932536504,3966201002,4954728904
%N Sequence lists numbers k > 1 such that k^3 == d(k) (mod sigma(k)), where d = A000005 and sigma = A000203.
%C From _Jinyuan Wang_, Feb 03 2019: (Start)
%C Conjecture: All terms are even.
%C a(34) > 5*10^9. (End)
%F Solutions of k^3 mod sigma(k) = d(k).
%e sigma(2) = 3 and 2^3 mod 3 = 2 = d(2).
%p with(numtheory): op(select(n->n^3 mod sigma(n)=tau(n),[$1..7772762]));
%t Select[Range[10^8], PowerMod[#1, 3, #3] == #2 & @@ Prepend[DivisorSigma[{0, 1}, #], #] &] (* _Michael De Vlieger_, Jan 18 2019 *)
%o (PARI) for(k=1,10^8,x=sigma(k);if(Mod(k,x)^3==Mod(numdiv(k),x),print1(k,", "))) \\ _Jinyuan Wang_, Feb 02 2019
%Y Cf. A000005, A000203, A323249, A323251.
%K nonn
%O 1,1
%A _Paolo P. Lava_, Jan 08 2019
%E a(24)-a(25) from _Michael De Vlieger_, Jan 18 2019
%E a(26)-a(33) from _Jinyuan Wang_, Feb 02 2019
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