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a(n) = (256^n - 1)/255.
35

%I #45 Mar 23 2023 17:23:53

%S 0,1,257,65793,16843009,4311810305,1103823438081,282578800148737,

%T 72340172838076673,18519084246547628289,4740885567116192841985,

%U 1213666705181745367548161,310698676526526814092329217,79538861190790864407636279553,20361948464842461288354887565569

%N a(n) = (256^n - 1)/255.

%C Partial sums of powers of 256 (A133752), q-integers for q=256.

%H Vincenzo Librandi, <a href="/A218723/b218723.txt">Table of n, a(n) for n = 0..400</a>

%H Quynh Nguyen, Jean Pedersen, and Hien T. Vu, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Pedersen/pedersen2.html">New Integer Sequences Arising From 3-Period Folding Numbers</a>, Vol. 19 (2016), Article 16.3.1. Cites this sequence.

%H <a href="/index/Par#partial">Index entries related to partial sums</a>.

%H <a href="/index/Q#q-numbers">Index entries related to q-numbers</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (257,-256).

%F a(n) = floor(256^n/255).

%F From _Vincenzo Librandi_, Nov 07 2012: (Start)

%F G.f.: x/((1 - x)*(1 - 256*x)).

%F a(n) = 257*a(n-1) - 256*a(n-2). (End)

%F E.g.f.: exp(x)*(exp(255*x) - 1)/255. - _Stefano Spezia_, Mar 23 2023

%t LinearRecurrence[{257, -256}, {0, 1}, 30] (* _Vincenzo Librandi_, Nov 07 2012 *)

%o (PARI) A218723(n)=256^n\255

%o (Magma) [n le 2 select n-1 else 257*Self(n-1) - 256*Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Nov 07 2012

%o (Python)

%o def A218723(n): return (1<<(n<<3))//255 # _Chai Wah Wu_, Nov 10 2022

%Y Cf. A133752.

%Y Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

%K nonn,easy

%O 0,3

%A _M. F. Hasler_, Nov 04 2012