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a(n) = 15*2^(n+1) + 1.
12

%I #69 Apr 02 2024 11:07:13

%S 31,61,121,241,481,961,1921,3841,7681,15361,30721,61441,122881,245761,

%T 491521,983041,1966081,3932161,7864321,15728641,31457281,62914561,

%U 125829121,251658241,503316481,1006632961,2013265921,4026531841,8053063681,16106127361

%N a(n) = 15*2^(n+1) + 1.

%C Binary numbers of form 1111(0^n)1 where n is the index and number of 0's.

%C Base 10 numbers of this sequence always end in 1.

%C An Engel expansion of 1/15 to the base 2 as defined in A181565, with the associated series expansion 1/15 = 2/31 + 2^2/(31*61) + 2^3/(31*61*121) + 2^4/(31*61*121*241) + ... . - _Peter Bala_, Oct 29 2013

%C The only squares in this sequence are 121 = 11^2 and 961 = 31^2. - _Antti Karttunen_, Sep 24 2023

%H Vincenzo Librandi, <a href="/A195744/b195744.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2).

%F a(n) = A052996(n+3) + A164094(n+2).

%F From _Bruno Berselli_, Sep 23 2011: (Start)

%F G.f.: (31-32*x)/(1-3*x+2*x^2).

%F a(n) = 2*a(n-1)-1.

%F a(n) = A110286(n+1)+1 = A128470(2^n). (End)

%F E.g.f.: exp(x)*(1 + 30*exp(x)). - _Stefano Spezia_, Oct 08 2022

%F For n >= 0, A005940(a(n)) = A030514(2+n). - _Antti Karttunen_, Sep 24 2023

%e First few terms in binary are 11111, 111101, 1111001, 11110001, 111100001.

%t 15*2^Range[50] + 1 (* _Paolo Xausa_, Apr 02 2024 *)

%o (Magma) [15*2^(n+1) + 1: n in [0..30]]; // _Vincenzo Librandi_, Sep 24 2011

%o (PARI) a(n)=30*2^n+1 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A052996, A164094.

%Y Cf. A020737, A083575, A083683, A083686, A083705, A168596, A181565.

%Y Cf. A110286, A128470.

%Y Cf. A005940, A030514.

%K easy,nonn

%O 0,1

%A _Brad Clardy_, Sep 23 2011

%E Corrected by _Arkadiusz Wesolowski_, Sep 23 2011