From http://list.seqfan.eu/pipermail/seqfan/2017-December/018222.html
" Consider the divisors of 136 (A018299): 1, 2, 4, 8, 17, 34, 68, 136. 136 in binary is 10001000. Reading off successive bits from the left we have: 1 1 10 2 100 4 1000 8 10001 17 100010 34 1000100 68 10001000 136 This prompts the question, what other numbers have this property? i.e. Numbers whose successive binary shifted parts as decimal numbers are identical to their ordered divisors. Clearly, the powers of 2 have this property. " Then from http://list.seqfan.eu/pipermail/seqfan/2017-December/018224.html " Excluding powers of 2, the sequence begins 3, 10, 136, 32896 with no more members less than 10^7. Observation: 3= 2^0 * (2^1 +1) = 2^0 + 2^1 10= 2^1 * (2^2 + 1) = 2^1 + 2^3 136 = 2^3 * (2^4 + 1) = 2^3 + 2^7 32896 = 2^7 * (2^8 + 1) = 2^7 + 2^15 which suggests the conjecture that all integers of the form 2^(2^k-1) + 2^(2^(k+1)-1) are in the sequence. This conjecture is false. 2147516416 = 2^15 * (2^30 + 1) = 2^15 + 2^31 is in the sequence, but 9223372039002259456 = 2^15 * (2^16 + 1) = 2^15 + 2^31 is not. " Notice that 2^(2^k-1) + 2^(2^(k+1)-1) = 2^(2^k-1)* (1+2^2^k) and the second equation is wrong since -:(* >:)2^2^5x 9223372039002259456 Later this bound was doubled to "There are no further terms, other than powers of 2, up to 2 x 10^7." Puzzle: How can the power of J be used to enlarge this bound to 10^10 or more? Apart from efficient solutions should be elegant. R.E. Boss ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
