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Numbers k > 1 for which floor(b(k)) = floor(b(k-1)), where b(m) = Sum_{j=1..m} (j/(j+2)).
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%I #18 Jul 05 2021 09:11:08

%S 2,5,9,17,29,49,81,135,225,371,614,1013,1672,2757,4548,7499,12365,

%T 20388,33615,55423,91378,150659,248395,409536,675212,1113237,1835419,

%U 3026095,4989189,8225783,13562025,22360001,36865410,60780788,100210579,165219314,272400598,449112661

%N Numbers k > 1 for which floor(b(k)) = floor(b(k-1)), where b(m) = Sum_{j=1..m} (j/(j+2)).

%C I conjecture that lim_{n->infinity} a(n)/a(n-1) = sqrt(e). For integers not in the sequence, b(m) = 1 + b(m-1).

%e floor(b(1)) = floor(1/3) = 0;

%e floor(b(2)) = floor(1/3 + 2/4) = 0;

%e hence 2 is a term.

%o (PARI) A=0;for(n=1,1000000,B=A;A=B+(n/(n+2));if(floor(A)-floor(B)-1,print1(n,", ")))

%K easy,nonn

%O 1,1

%A Philippe LALLOUET (philip.lallouet(AT)orange.fr), Nov 28 2007

%E More terms from _Hugo Pfoertner_, Jul 04 2021