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Subset of A077436. Proof: Since a(n) is of the form (2^i-1))*2^j, i,j>= >= 0, a(n)^2 = [ = (2^(2i)-) - 2^(i+1)]))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004
Numbers n such thatwhose binary representation contains no "01". - Benoit Cloitre, May 23 2004
Powers of 2 represented in bases which are membersterms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017
Numbers nk such that nk = 2^(1 + A000523(nk)) - 2^A007814(nk). - Daniel Starodubtsev, Aug 05 2021
If v is the dyadic valuation (i.e. ., A007814), then A002260(n) = v(a(n)/2^v(a(n))+1) and A002024(n) = A002260(n) + v(a(n)). - Lorenzo Sauras Altuzarra, Feb 01 2023
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