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Sums of two powers of 4.
5

%I #15 Jun 18 2013 15:25:13

%S 2,5,8,17,20,32,65,68,80,128,257,260,272,320,512,1025,1028,1040,1088,

%T 1280,2048,4097,4100,4112,4160,4352,5120,8192,16385,16388,16400,16448,

%U 16640,17408,20480,32768,65537,65540,65552,65600,65792,66560,69632,81920,131072

%N Sums of two powers of 4.

%H T. D. Noe, <a href="/A055236/b055236.txt">Rows n = 0..100 of triangle, flattened</a>

%F a(n) = 4^(n-trinv(n))+4^trinv(n), where trinv(n) = floor((1+sqrt(1+8*n))/2) = A002262(n) and n-trinv(n) = A003056(n).

%F Regarded as a triangle T(n, k) = 4^n + 4^k, so as a sequence a(n) = 4^A002262(n) + 4^A003056(n).

%t t = 4^Range[0, 9]; Select[Union[Flatten[Table[i + j, {i, t}, {j, t}]]], # <= t[[-1]] + 1 &] (* _T. D. Noe_, Oct 09 2011 *)

%t Union[Total/@Tuples[4^Range[0,9], 2]] (* _Harvey P. Dale_, Mar 25 2012 *)

%Y Cf. A052216.

%K easy,nonn,tabl

%O 0,1

%A _Henry Bottomley_, Jun 22 2000