A182765 - OEIS

%I #17 Jun 24 2017 00:49:06

%S 1,3,5,7,9,11,12,14,16,18,20,22,24,25,27,29,31,33,35,37,38,40,42,44,

%T 46,48,50,51,53,55,57,59,61,63,64,66,68,70,72,74,75,77,79,81,83,85,87,

%U 88,90,92,94,96,98,100,101,103,105,107,109,111,113,114,116,118,120,122,124,126,127

%N Beatty sequence for (6 + sqrt(2))/4.

%C Let u=(1+sqrt(2))/2 and v=sqrt(2). Jointly rank {ju} and {kv} as in the first comment at A182760; a(n) is the position of nu.

%F a(n) = floor(r*n), where r = (6 + sqrt(2))/4.

%F a(n) = 2*n - 1 - floor(n/7) for n < 41, but this fails for a(41) = 75 onwards. - _M. F. Hasler_, Jun 23 2016

%t Table[Floor[(6 + Sqrt@ 2) n/4], {n, 70}] (* _Michael De Vlieger_, Jun 23 2016 *)

%o (PARI) A182765(n)=n*(6+sqrt(2))\4 \\ Requires sufficient realprecision (but the 64-bit default is enough up to n = 10^38). - _M. F. Hasler_, Jun 23 2016

%Y Cf. A182766.

%K nonn

%O 1,2

%A _Clark Kimberling_, Nov 29 2010