Still thinking about the puzzle.
Some tools to explore the properties of the conjecture:
NB. pretty print
toca =: '0123456789' {~ ]
tobi =: 3 : '(2#~1+>.2^.>./y)#:y'
todc =: 3 : '(10#~>.10^.>./y)#:y'
todcc =: 3 : '(1+>.10^.>./y)":,.y'
NB. divisor formula from Roger Hui,
http://code.jsoftware.com/wiki/Essays/Divisors
check =: 3 : 'y, (2&#.\2#. inv y) -: /:~ @: , @: > @: (*/&.>/) @:
((^ i.@>:)&.>/)
@: (__&q:) y'"0
NB. from the article (2^_1+2^k) + 2^_1+2^k+1
gen =: 3 : '+/"1 (2) 2&^\x:<:2^i.y'
check gen 7
3 1
10 1
136 1
32896 1
2147516416 1
9223372039002259456 0
check gen 8
3 1
10 1
136 1
32896 1
2147516416 1
9223372039002259456 0
170141183460469231740910675752738881536 0
(toca&tobi,.todcc) gen 7
00000000000000000000000000000000000000000000000000000000000000011
3
00000000000000000000000000000000000000000000000000000000000001010
10
00000000000000000000000000000000000000000000000000000000010001000
136
00000000000000000000000000000000000000000000000001000000010000000
32896
00000000000000000000000000000000010000000000000001000000000000000
2147516416
01000000000000000000000000000000010000000000000000000000000000000
9223372039002259456
On Sat, Dec 30, 2017 at 2:29 PM, R.E. Boss <[email protected]> wrote:
> 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
----------------------------------------------------------------------
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