A301474 - OEIS

%I #8 Mar 29 2018 10:13:59

%S 1,12,1214,221418,1213241618,3223242618,121334153618,10322314361718,

%T 10322334361718,10222334461718,10222344361819,10222334261738,

%U 1022134415361718,20323334261728,10121334255618,1021421314153618,20522324261728,1011121354252618,3062131425161718

%N Summarize the double of the previous term (digits in increasing order), starting with a(1) = 1.

%C From 56th term the sequence goes into a cycle of 4 terms: 10222334362819, 1022133415361728, 2032233415361718, 20123344461718.

%e a(1) = 1 and 2*1 = 2 ('one 2') then a(2) = 12;

%e 2*12 = 24 ('one 2, one 4') then a(3) = 1214. And so on.

%p P:=proc(q,h) local a,b,c,j,k,n; a:=h; print(a);

%p for n from 1 to q do a:=convert(2*a,base,10);

%p b:=0; for k from 0 to 9 do c:=0; for j from 1 to nops(a) do

%p if a[j]=k then c:=c+1; fi; od;

%p if c>0 then b:=b*10^(ilog10(c*10+k)+1)+c*10+k; fi; od;

%p a:=b; print(a); od; end: P(10,1); # _Paolo P. Lava_, Mar 22 2018

%Y Cf. A005151.

%K nonn,base,easy

%O 1,2

%A _Paolo P. Lava_, Mar 22 2018