A086223 Every integer can be represented uniquely as m = k*2^(j+1)+2^j-1. Sequence gives values of k for m = repunit(n).

0, 1, 3, 69, 694, 6944, 69444, 694444, 6944444, 69444444, 694444444, 6944444444, 69444444444, 694444444444, 6944444444444, 69444444444444, 694444444444444, 6944444444444444, 69444444444444444, 694444444444444444

Offset

1,3

Comments

j = A007814(m+1).

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (11,-10).

Formula

a(n) = A025480(A002275(n)).

G.f.: -x^2*(35*x^3-46*x^2+8*x-1) / ((x-1)*(10*x-1)). - Colin Barker, Apr 29 2015

a(n) = (125*10^(n-3)-8)/18 for n >= 4. - Robert Israel, Apr 29 2015

a(n) = 11*a(n-1)-10*a(n-2) for n>5.

Example

1 = 0*4+1; 11 = 1*8+3; 111 = 3*32+15.
For n > 3, repunit(n) = [69*10^(n-4)+(10^(n-4)-1)*4/9]*16+7.

Mathematica

CoefficientList[Series[x (35 x^3 - 46 x^2 + 8 x - 1)/((1 - x)(10 x - 1)), {x, 0, 20}], x] (* Vincenzo Librandi, Apr 30 2015 *)

Prog

(PARI) concat(0, Vec(-x^2*(35*x^3-46*x^2+8*x-1)/((x-1)*(10*x-1)) + O(x^100))) \\ Colin Barker, Apr 30 2015
(Magma) [0, 1, 3] cat [(125*10^(n-3)-8)/18: n in [4..25]]; // Vincenzo Librandi, Apr 30 2015
(Python)
def A086223(n): return (-(k:=(m:=(10**n-1)//9)&~(m+1))+m+1)//((k+1)<<1) # Chai Wah Wu, Jul 07 2022

Crossrefs

Cf. A002275, A007814, A025480.

Sequence in context: A166835 A166806 A219070 * A089455 A012201 A284058

Adjacent sequences: A086220 A086221 A086222 * A086224 A086225 A086226

Keyword

nonn,easy

Author

Marco Matosic, Jul 27 2003

Extensions

Edited and extended by David Wasserman, Feb 17 2005

Status

approved