

A329663


Numbers k such that the binary reversal of k (A030101) is equal to the sum of the proper divisors of k (A001065).


0



2, 1881, 49905, 54585, 63405, 196785, 853785, 2094897, 3925449, 32480685, 1925817945, 1994453385, 961201916805
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OFFSET

1,1


COMMENTS

a(13) > 1.45*10^11.


LINKS



EXAMPLE

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.
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.


MATHEMATICA

Select[Range[10^5], DivisorSigma[1, #]  # == IntegerReverse[#, 2] &]


PROG

(PARI) isok(k) = sigma(k)  k == fromdigits(Vecrev(binary(k)), 2); \\ Michel Marcus, Feb 29 2020


CROSSREFS



KEYWORD

nonn,base,more


AUTHOR



EXTENSIONS



STATUS

approved



