

A180055


In binary expansion, number of 1's in 5n is less than in n.


2



13, 26, 29, 52, 53, 55, 58, 61, 77, 103, 104, 106, 109, 110, 111, 116, 117, 119, 122, 125, 154, 157, 205, 206, 207, 208, 212, 213, 215, 218, 219, 220, 221, 222, 223, 231, 232, 234, 237, 238, 239, 244, 245, 247, 250, 253, 308, 309, 311, 314, 317, 333, 359, 365
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OFFSET

1,1


COMMENTS

Or, binary weight of 5n is less than binary weight of n.
Also called the 5flimsy numbers; see the Stolarsky reference.
If m is here, 2m is too. Hence the "primitive solutions" are all odd ones:
13,29,53,55,61,77,103,109,111,117,119,125,157,205,207,213,215,219,221,223,231.


LINKS



FORMULA



MATHEMATICA

Select[Range[1000], Count[IntegerDigits[5#, 2], 1]<Count[IntegerDigits[ #, 2], 1]&]


PROG

(PARI) for(k=1, 370, if(hammingweight(5*k) < hammingweight(k), print1(k, ", "))) \\ Hugo Pfoertner, Dec 27 2019


CROSSREFS



KEYWORD

base,nonn


AUTHOR



STATUS

approved



