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Distance from 2^n to the nearest triangular number.
3

%I #22 Feb 28 2014 11:12:26

%S 0,1,1,2,1,4,2,8,3,16,11,32,1,64,87,128,167,256,306,512,500,1024,552,

%T 2048,688,4096,3041,8192,579,16384,20854,32768,37075,65536,55618,

%U 131072,37108,262144,222296,524288,147729,1048576,891994,2097152,602155,4194304,3523022

%N Distance from 2^n to the nearest triangular number.

%F a(2*k+1) = 2^k.

%F Specifically, both the nearest triangular number below: A006516(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) and the nearest triangular number above: A007582(n) = A000217(2^n) = 2^(2n-1) + 2^(n-1) are at the same distance from 2^(2n-1). - _Antti Karttunen_, Feb 26 2014

%e Triangular number nearest to 2^8=256 is 253, so a(8)=256-253=3.

%t a[n_] := Module[{k, k0}, k0 = k /. FindRoot[2^n == k*(k+1)/2, {k, 2^(n/2)}, WorkingPrecision -> 20] // Round; Abs[2^n-k0*(k0+1)/2]]; Table[a[n], {n, 0, 46}] (* _Jean-François Alcover_, Feb 27 2014 *)

%Y Bisections: a(2n) = abs(A238455(n)), a(2n+1) = A000079(n).

%Y Cf. A000217, A006516, A007582.

%K nonn,easy

%O 0,4

%A _Alex Ratushnyak_, Feb 23 2014