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%I #31 Jul 08 2022 08:23:14
%S 2,6,2,14,2,6,2,30,2,6,2,14,2,6,2,62,2,6,2,14,2,6,2,30,2,6,2,14,2,6,2,
%T 126,2,6,2,14,2,6,2,30,2,6,2,14,2,6,2,62,2,6,2,14,2,6,2,30,2,6,2,14,2,
%U 6,2,254,2,6,2,14,2,6,2,30,2,6,2,14,2,6,2,62,2,6,2,14,2,6,2,30,2,6,2,14,2
%N a(n) = (2*n-1) XOR (2*n+1), bitwise.
%H Reinhard Zumkeller, <a href="/A100892/b100892.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 2 * ((n-1) XOR n) = 2*A038712(n).
%F a(n) = 4*2^A007814(n) - 2.
%F Recurrence: a(2n) = 2a(n) + 2, a(2n+1) = 2. - _Ralf Stephan_, Aug 21 2013
%F a(n) = A088837(n) - 1. - _Filip Zaludek_, Dec 10 2016
%F a(n) = A074400(n)/A000593(n) = 2*A000203(n)/A000593(n). - _Ivan N. Ianakiev_, Jul 04 2019
%t a[n_]:=BitXor[2*n-1,2*n+1]; a/@Range[100] (* _Ivan N. Ianakiev_, Jul 04 2019 *)
%o (PARI) a(n)=4*2^valuation(n,2)-2; \\ _Ralf Stephan_, Aug 21 2013
%o (Haskell)
%o a100892 n = (2 * n - 1) `xor` (2 * n + 1)
%o a100892_list = zipWith xor (tail a005408_list) a005408_list
%o -- _Reinhard Zumkeller_, Sep 03 2013
%o (Python)
%o def A100892(n): return ((~n& n-1)<<2)+2 # _Chai Wah Wu_, Jul 07 2022
%Y Cf. A006519, A099820, A007088.
%Y Cf. A005408.
%K nonn
%O 1,1
%A _Reinhard Zumkeller_, Jan 10 2005
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