cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A308900 An explicit example of an infinite sequence with a(1)=1 and, for n >= 2, a(n) and S(n) = Sum_{i=1..n} a(i) have no digit in common.

Original entry on oeis.org

1, 6, 4, 66, 34, 666, 334, 6666, 3334, 66666, 33334, 666666, 333334, 6666666, 3333334, 66666666, 33333334, 666666666, 333333334, 6666666666, 3333333334, 66666666666, 33333333334, 666666666666, 333333333334, 6666666666666, 3333333333334, 66666666666666, 33333333333334
Offset: 1

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Author

N. J. A. Sloane, Jul 15 2019

Keywords

Comments

Used in a proof that the initial terms of A309151 are correct.
The S(n) sequence is 1, 7, 11, 77, 111, 777, 1111, 7777, 11111, 77777, ...
A093137 interleaved with positive terms of A002280. - Felix Fröhlich, Jul 15 2019

Crossrefs

Programs

  • Magma
    I:=[1,6,4]; [n le 3 select I[n] else - Self(n-1) + 10*Self(n-2) + 10*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 20 2019
  • Maple
    1, seq(op([6*(10^i-1)/9, 3*(10^i-1)/9+1]), i=1..30); # Robert Israel, Jul 15 2019
  • Mathematica
    CoefficientList[Series[(1 + 7 x)/((1 + x) (1 - 10 x^2)), {x, 0, 26}], x] (* Michael De Vlieger, Jul 18 2019 *)
    LinearRecurrence[{-1,10,10},{1,6,4},30] (* Harvey P. Dale, Jan 02 2022 *)
  • PARI
    Vec((1+7*x)/((1+x)*(1-10*x^2)) + O(x^20)) \\ Felix Fröhlich, Jul 15 2019
    
  • PARI
    a(n) = if(n==1, 1, if(n%2==0, 6*(10^(n/2)-1)/9, 3*(10^((n-1)/2)-1)/9+1)) \\ Felix Fröhlich, Jul 15 2019
    

Formula

For even n >= 2, a(n) = 6666...66 (with n/2 6's). For odd n >= 5, a(n) = 3333...334 (with (n-3)/2 3's and a single 4).
From Robert Israel, Jul 15 2019: (Start)
G.f. (1+7*x)/((1+x)*(1-10*x^2)).
a(n) = -a(n - 1) + 10*a(n - 2) + 10*a(n - 3). (End)
a(-n) = a(n+1). - Paul Curtz, Jul 18 2019
a(n) = (1/60)*(-40*(-1)^n + (1 + (-1)^n)*(2^(2+n/2)*5^(1+n/2)) + (1 + (-1)^(n+1))*10^((1+n)/2)). - Stefano Spezia, Jul 20 2019