A218723 a(n) = (256^n - 1)/255.
0, 1, 257, 65793, 16843009, 4311810305, 1103823438081, 282578800148737, 72340172838076673, 18519084246547628289, 4740885567116192841985, 1213666705181745367548161, 310698676526526814092329217, 79538861190790864407636279553, 20361948464842461288354887565569
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..400
- Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. Cites this sequence.
- Index entries related to partial sums.
- Index entries related to q-numbers.
- Index entries for linear recurrences with constant coefficients, signature (257,-256).
Programs
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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
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Mathematica
LinearRecurrence[{257, -256}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
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PARI
A218723(n)=256^n\255
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Python
def A218723(n): return (1<<(n<<3))//255 # Chai Wah Wu, Nov 10 2022
Formula
a(n) = floor(256^n/255).
From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 256*x)).
a(n) = 257*a(n-1) - 256*a(n-2). (End)
E.g.f.: exp(x)*(exp(255*x) - 1)/255. - Stefano Spezia, Mar 23 2023
Comments