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Integers n > 0 such that n XOR 62*n = 63*n.
6

%I #12 Oct 31 2021 15:27:41

%S 1,2,4,8,16,32,64,65,128,129,130,256,257,258,260,512,513,514,516,520,

%T 1024,1025,1026,1028,1032,1040,2048,2049,2050,2052,2056,2064,2080,

%U 4096,4097,4098,4100,4104,4112,4128,4160,4161,8192,8193,8194,8196,8200,8208

%N Integers n > 0 such that n XOR 62*n = 63*n.

%H Ivan Neretin, <a href="/A115424/b115424.txt">Table of n, a(n) for n = 1..10000</a>

%F This sequence also seems to satisfy:

%F 3*a(n) XOR 41*a(n) = 42*a(n);

%F 5*a(n) XOR 35*a(n) = 38*a(n);

%F 6*a(n) XOR 35*a(n) = 37*a(n);

%F 7*a(n) XOR 35*a(n) = 36*a(n); etc.

%t Select[Range[8300],BitXor[#,62 #]==63 #&] (* _Harvey P. Dale_, Oct 31 2021 *)

%o (Perl)

%o $cnt=0;

%o foreach(1..8_000){

%o print ++$cnt," $_\n" if ((62*$_)^$_)==63*$_;

%o }

%o # _Ivan Neretin_, Nov 11 2016

%o (PARI) isok(n) = bitxor(n, 62*n) == 63*n; \\ _Michel Marcus_, Nov 11 2016

%Y Cf. A003714 (Fibbinary numbers), A048715, A048718, A115422, A115423.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Jan 22 2006