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a(1)=2, a(2)=3; a(n) is the smallest k > a(n-1) such that k + a(n-1) is a multiple of a(n-2).
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%I #16 Nov 01 2019 15:02:26

%S 2,3,5,7,8,13,19,20,37,43,68,104,168,248,256,488,536,928,1216,1568,

%T 2080,2624,3616,4256,6592,10432,15936,25792,37952,39424,74432,83264,

%U 140032,193024,227072,352000,556288,851712,1373440,2033408,2086912,4013312

%N a(1)=2, a(2)=3; a(n) is the smallest k > a(n-1) such that k + a(n-1) is a multiple of a(n-2).

%C (a(n+1)+a(n+2))/a(n) gives the sequence 4, 4, 3, 3, 4, 3, 3, 4, 3, 4, 4, 4, 3, 3, 4, 3, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 3, 3, 4, 3, 4, 3, 3, 4, 4, 4, 4, 3, 3, 4, 3, 4, 3, 3, 3, 4, 3, ...

%H Robert Israel, <a href="/A328724/b328724.txt">Table of n, a(n) for n = 1..6866</a>

%e a(7)=19; a(8)=20. 37 is the smallest number, larger than 20, that can be added to 20 and the result (57) is divisible by 19. So, a(9)=37.

%p A[1]:= 2: A[2]:= 3:

%p for n from 3 to 50 do

%p k:= -A[n-1] mod A[n-2];

%p A[n]:= k + A[n-2]*(1+floor((A[n-1]-k)/A[n-2]));

%p od:

%p seq(A[i],i=1..50); # _Robert Israel_, Oct 27 2019

%K nonn

%O 1,1

%A _Ali Sada_, Oct 26 2019