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a(n) = reverse(2^n) mod 2^n.
1

%I #9 Dec 12 2021 19:47:43

%S 0,0,0,13,23,46,53,140,215,105,210,2808,2918,15593,21187,63556,7987,

%T 179118,358137,466945,420750,4034914,8068838,10946113,23445533,

%U 46880176,22406063,117663950,219078635,1060248229,2021396468,2632727628,2954399858,13837158803

%N a(n) = reverse(2^n) mod 2^n.

%C Remainder if (2^n written backwards) is divided by 2^n.

%e a(4) = reverse(2^4) mod 2^4 = reverse(16) mod 16 = 61 mod 16 = 13.

%t Table[Mod[FromDigits[Reverse[IntegerDigits[2^n]]], 2^n], {n, 1, 256}]

%o (Python)

%o def a(n): t = 2**n; return int(str(t)[::-1])%t

%o print([a(n) for n in range(1, 35)]) # _Michael S. Branicky_, Dec 12 2021

%Y Cf. A002113, A071590, A103164-A103167.

%K base,nonn

%O 1,4

%A _Labos Elemer_, Jan 28 2005