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%I #12 Feb 29 2020 16:04:48
%S 2,1881,49905,54585,63405,196785,853785,2094897,3925449,32480685,
%T 1925817945,1994453385,961201916805
%N Numbers k such that the binary reversal of k (A030101) is equal to the sum of the proper divisors of k (A001065).
%C a(13) > 1.45*10^11.
%C a(14) > 5*10^12, if it exists. - _Giovanni Resta_, Feb 29 2020
%e 2 is a term since its binary representation is 10, its binary reversal is 01 = 1 which is equal to the sum of the proper divisors of 2.
%e 1881 is a term since its binary representation is 11101011001, its binary reversal is 10011010111 which is equal to 1239, which is also the sum of the proper divisors of 1881: 1 + 3 + 9 + 11 + 19 + 33 + 57 + 99 + 171 + 209 + 627 = 1239.
%t Select[Range[10^5], DivisorSigma[1, #] - # == IntegerReverse[#, 2] &]
%o (PARI) isok(k) = sigma(k) - k == fromdigits(Vecrev(binary(k)), 2); \\ _Michel Marcus_, Feb 29 2020
%Y Cf. A001065, A030101, A072234.
%K nonn,base,more
%O 1,1
%A _Amiram Eldar_, Feb 28 2020
%E a(13) from _Giovanni Resta_, Feb 29 2020
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