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a(1)=1; thereafter a(n) = 2^(number of bits in binary expansion of a(n-1)) + 1 + a(n-1).
2

%I #15 May 05 2020 13:32:12

%S 1,4,13,30,63,128,385,898,1923,3972,8069,16262,32647,65416,130953,

%T 262026,524171,1048460,2097037,4194190,8388495,16777104,33554321,

%U 67108754,134217619,268435348,536870805,1073741718,2147483543,4294967192,8589934489,17179869082,34359738267,68719476636

%N a(1)=1; thereafter a(n) = 2^(number of bits in binary expansion of a(n-1)) + 1 + a(n-1).

%C An infinite subsequence of A010061.

%H Harvey P. Dale, <a href="/A232228/b232228.txt">Table of n, a(n) for n = 1..1001</a>

%H D. W. Bange, <a href="http://www.jstor.org/stable/2319017">Solution to Problem E 2408</a>, Amer. Math. Monthly, 81 (1974), 407.

%H <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>

%p f:=proc(n) option remember; if n=1 then 1 else 2^(nops(convert(f(n-1),base,2)))+1+f(n-1); fi; end;

%p [seq(f(n),n=1..40)];

%t NestList[2^IntegerLength[#,2]+1+#&,1,40] (* _Harvey P. Dale_, May 05 2020 *)

%Y Cf. A010061.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Nov 22 2013