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Integers in (2+A191113)/4; contains A191113 as a proper subsequence.
3

%I #4 Mar 30 2012 18:57:30

%S 1,2,3,4,6,10,12,14,16,21,22,28,30,38,40,46,48,54,62,64,82,84,86,102,

%T 110,112,118,120,136,138,150,158,160,182,183,184,190,192,214,244,246,

%U 254,256,264,326,328,334,336,342,352,354,406,408,426,438,446,448,470,472,478,480,542,544,550,552,568,570,598,630,638,640,726,730

%N Integers in (2+A191113)/4; contains A191113 as a proper subsequence.

%C See A191113 and A191146.

%e A191113=(1,2,4,6,10,...), whence (2+A191113)/4=(3/4,1,6/4,2,3,...), so that, retaining only integers, A191149=(1,2,3,...).

%t (See A191113.)

%Y Cf. A191113, A191146.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 27 2011