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%I #25 Jun 18 2015 04:07:54
%S 14,19,21,28,42,47,63,84,105,126,147,149,168,189,199,298,323,497,646,
%T 795,911,969,1292,1499,1822,1999,2087,2733,2998,3089,3248,3379,3644,
%U 4555,4997,5411,5466,6178,6377,6496,7288,7995,8199,9161,9267,9744,10822,12356
%N 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 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 Similar to A130792 but here the minimum number is deleted since the beginning.
%C 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 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 Paolo P. Lava, <a href="/A248134/b248134.txt">Table of n, a(n) for n = 1..410</a>
%H 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 Let us rewrite 5411 as 54 U 11. Then:
%e 11 + 54 = 65;
%e 54 + 65 = 119;
%e 65 + 119 = 184;
%e 119 + 184 = 303;
%e 184 + 303 = 487;
%e 303 + 487 = 790;
%e 487 + 790 = 1277;
%e 790 + 1277 = 2067;
%e 1277 + 2067 = 3344;
%e 2067 + 3344 = 5411, that is 11*F(10) + 54*F(11) = 11*55 + 54*89 = 605 + 4806 = 5411.
%p P:=proc(q,h) local a,b,k,n,t,v; v:=array(1..h);
%p for n from 1 to q do for k from 1 to ilog10(n) do
%p a:=n mod 10^k; b:=trunc(n/10^k); if a<b then
%p v[1]:=a; v[2]:=b; else v[1]:=b; v[2]:=a; fi; t:=3;
%p v[t]:=a+b; while v[t]<n do t:=t+1; v[t]:=v[t-2]+v[t-1]; od;
%p if v[t]=n then print(n); break; fi; od; od; end: P(10^9,1000);
%Y Cf. A000045, A007629, A130792, A246544, A247012, A247013.
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, Oct 02 2014
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