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%I #10 May 30 2024 19:45:13
%S 0,1,2,3,4,5,7,8,9,10,14,15,31,32,33,34,35,37,38,39,40,43,44,63,64,65,
%T 68,69,73,76,79,80,83,86,88,96,116,118,119,120,124,125,128,140,267,
%U 426,440,445,446,447,460,474,604,733,774,775,777,778,779,785,797,818,819
%N Slowest increasing sequence: the sum of three consecutive terms shares no digit with any of the summands.
%C The sequence is finite and has 112 terms. _Max Alekseyev_ (who computed the sequence together with Peter Pein) proved that 175414854, 415748410, and 1631058958 are the last three terms: (Quoting _Max Alekseyev_) Suppose that the next term is x, then the sum s = 415748410 + 1631058958 + x = 2046807368 + x may contain only decimal digit 2. Therefore all solutions are given by the formula x(k) = 2*(10^k-1)/9 - 2046807368 where k>=10. It is easy to see that while x(10)=175414854 is smaller than 1631058958 (hence it cannot be an element of our sequence), all other x(k) contain a decimal digit 2 which is not allowed: x(11) = 20175414854, x(12) = 220175414854, x(13) = 2220175414854, ... Therefore there is no next term in this sequence. QED. (End of quote)
%H Eric Angelini, May 25 2007, <a href="/A129268/b129268.txt">Table of n, a(n) for n = 0..111</a>
%K base,easy,fini,full,nonn
%O 0,3
%A _Eric Angelini_, May 25 2007