[seqfan] Re: 1,2,3,5,8,13,21,4,25,29,6,...

[Éric Angelini]

Well done, thank you Hans!
Best,
É.

> Le 13 sept. 2019 à 16:40, Hans Havermann <gladhobo at bell.net> a écrit :
> 
> EA: "Start S with a(1) = 1 and a(2) = 2. Now if a(n) and a(n+1) don't share any digit, then a(n+2) = a(n) + a(n+1). Else a(n+2) = the smallest integer not yet in S."
> 
> Almost all adjacent terms share a digit so let's focus on those terms that are sums (term#,sum). From (3,3) to (107,239) there are 43 such. From here on to term 10^3 are 16 more; from 10^3 to 10^4, 16 more; from 10^4 to 10^5, 16 more:
> 
> (125,198)    (1013,1998)    (10015,19998)
> (212,399)    (2007,3999)    (20009,39999)
> (213,599)    (2008,5999)    (20010,59999)
> (215,800)    (2010,8000)    (20012,80000)
> (310,599)    (3009,5999)    (30011,59999)
> (311,899)    (3010,8999)    (30012,89999)
> (313,1200)   (3012,12000)   (30014,120000)
> (411,798)    (4011,7998)    (40013,79998)
> (412,1198)   (4012,11998)   (40014,119998)
> (513,999)    (5013,9999)    (50015,99999)
> (514,1499)   (5014,14999)   (50016,149999)
> (614,1198)   (6014,11998)   (60016,119998)
> (615,1798)   (6015,17998)   (60017,179998)
> (715,1399)   (7016,13999)   (70018,139999)
> (716,2099)   (7017,20999)   (70019,209999)
> (815,1600)   (8017,16000)   (80019,160000)
> 
> Perhaps this continues forever.
> 
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