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A248134 Consider a number x as a concatenation of two integers, a and b: x = concat(a,b). Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.

%I A248134 #25 Jun 18 2015 04:07:54
%S A248134 14,19,21,28,42,47,63,84,105,126,147,149,168,189,199,298,323,497,646,
%T A248134 795,911,969,1292,1499,1822,1999,2087,2733,2998,3089,3248,3379,3644,
%U A248134 4555,4997,5411,5466,6178,6377,6496,7288,7995,8199,9161,9267,9744,10822,12356
%N A248134 Consider a number x as a concatenation of two integers, a and b: x = concat(a,b). Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.
%C A248134 If the number x is rewritten as concat(a,b), the problem is to find a value of y such that x = a*F(y) + b*F(y+1), if a < b, or x = b*F(y) + a*F(y+1), if a > b, where F(y) is a Fibonacci number (see values of x, a, b, y, for 1<x<10^6, in Links).
%C A248134 Similar to A130792 but here the minimum number is deleted since the beginning.
%C A248134 All the listed numbers admit only one concatenation, concat(a,b), that, through the addition process, leads to themselves. Is there any number that admit more than one single concatenation?
%C A248134 Sequence is infinite. Let us consider the numbers 19, 199, 1999, 19...9 and let us divide them as concat(1,9), concat(1,99), concat(1,999), concat(1,9...9). In two steps we have the initial numbers back: 1 + 9 = 10 and 9 + 10 = 19; 1 + 99 = 100 and 99 + 100 = 199, etc.
%H A248134 Paolo P. Lava, <a href="/A248134/b248134.txt">Table of n, a(n) for n = 1..410</a>
%H A248134 Paolo P. Lava, <a href="/A248134/a248134.txt">List of [x,a,b,y] in the equation x = a*F(y) + b*F(y+1), for 1< x <10^6</a>
%e A248134 Let us rewrite 5411 as 54 U 11. Then:
%e A248134 11 + 54 = 65;
%e A248134 54 + 65 =  119;
%e A248134 65 + 119 = 184;
%e A248134 119 + 184 = 303;
%e A248134 184 + 303 = 487;
%e A248134 303 + 487 = 790;
%e A248134 487 + 790 = 1277;
%e A248134 790 + 1277 = 2067;
%e A248134 1277 + 2067 = 3344;
%e A248134 2067 + 3344 = 5411, that is 11*F(10) + 54*F(11) = 11*55 + 54*89 = 605 + 4806 = 5411.
%p A248134 P:=proc(q,h) local a,b,k,n,t,v; v:=array(1..h);
%p A248134 for n from 1 to q do for k from 1 to ilog10(n) do
%p A248134 a:=n mod 10^k; b:=trunc(n/10^k); if a<b then
%p A248134 v[1]:=a; v[2]:=b; else v[1]:=b; v[2]:=a; fi; t:=3;
%p A248134 v[t]:=a+b; while v[t]<n do t:=t+1; v[t]:=v[t-2]+v[t-1]; od;
%p A248134 if v[t]=n then print(n); break; fi; od; od; end: P(10^9,1000);
%Y A248134 Cf. A000045, A007629, A130792, A246544, A247012, A247013.
%K A248134 nonn,base
%O A248134 1,1
%A A248134 _Paolo P. Lava_, Oct 02 2014