A268839 a(n) = Sum_{j=1..10^n-1} 2^f(j) where f(j) is the number of zero digits in the decimal representation of j.
9, 108, 1197, 13176, 144945, 1594404, 17538453, 192922992, 2122152921, 23343682140, 256780503549, 2824585539048, 31070440929537, 341774850224916, 3759523352474085, 41354756877214944, 454902325649364393, 5003925582143008332, 55043181403573091661
Offset: 1
Examples
a(1) = 9 because 2^f(1) + 2^f(2) + ... + 2^f(9) = 2^0 + 2^0 + ... + 2^0 = 9; a(2) = 108 because 2^f(1) + 2^f(2) + ... + 2^f(99) = 9*10 + 2*9 = 108, where f(10) = f(20) = ... = f(90) = 1 and f(i) = 0 otherwise.
Links
- Colin Barker, Table of n, a(n) for n = 1..950
- Index entries for linear recurrences with constant coefficients, signature (12,-11).
Programs
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Magma
[(9/10)*(11^n-1): n in [1..20]]; // Vincenzo Librandi, Feb 15 2016
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Maple
for n from 1 to 100 do: x:=(9/10)*(11^n-1):printf(‘%d, ‘,x):od:
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Mathematica
Table[Table[(9/10) (11^n - 1), {n, 1, 20}]] (* Bruno Berselli, Feb 15 2016 *) CoefficientList[Series[9/((1 - 11 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Feb 15 2016 *)
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PARI
Vec(9*x/((1-11*x)*(1-x)) + O(x^30)) \\ Colin Barker, Feb 22 2016
Formula
a(n) = (9/10)*(11^n-1) = 9*A016123(n-1).
From Vincenzo Librandi, Feb 15 2016: (Start)
G.f.: (9*x)/((1-11*x)*(1-x)).
a(n) = 11*a(n-1) + 9. (End)
E.g.f.: 9*exp(x)*(exp(10*x) - 1)/10. - Stefano Spezia, Sep 13 2023
Extensions
Name edited by Jon E. Schoenfield, Sep 13 2017
Comments