OFFSET
1,2
COMMENTS
See A001744 for the other version.
If n-1 is represented as a base-4 number (see A007090) according to n-1 = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=0,6,8,9 for k=0..3. - Hieronymus Fischer, May 30 2012
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 1..10000
FORMULA
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = ((b_m(n)+6) mod 9 + floor((b_m(n)+2)/3) - floor(b_m(n)/3))*10^m + Sum_{j=0..m-1} (b_j(n) mod 4 +5*floor((b_j(n)+3)/4) +floor((b_j(n)+2)/4)- 6*floor(b_j(n)/4)))*10^j, where n>1, b_j(n)) = floor((n-1-4^m)/4^j), m = floor(log_4(n-1)).
a(1*4^n+1) = 6*10^n.
a(2*4^n+1) = 8*10^n.
a(3*4^n+1) = 9*10^n.
a(n) = 6*10^log_4(n-1) for n=4^k+1,
a(n) < 6*10^log_4(n-1), otherwise.
a(n) > 10^log_4(n-1) for n>1.
G.f.: g(x) = (x/(1-x))*Sum_{j>=0} 10^j*x^4^j *(1-x^4^j)* (6 + 8x^4^j + 9(x^2)^4^j)/(1-x^4^(j+1)).
Also: g(x) = (x/(1-x))*(6*h_(4,1)(x) + 2*h_(4,2)(x) + h_(4,3)(x) - 9*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*(x^4^j)^k/(1-(x^4^j)^4). (End)
EXAMPLE
a(1000) = 99896.
a(10^4) = 8690099.
a(10^5) = 680688699.
MATHEMATICA
Union[Flatten[Table[FromDigits/@Tuples[{0, 6, 8, 9}, n], {n, 3}]]] (* Harvey P. Dale, Sep 04 2013 *)
PROG
(PARI) is(n) = #setintersect(vecsort(digits(n), , 8), [1, 2, 3, 4, 5, 7])==0 \\ Felix Fröhlich, Sep 09 2019
CROSSREFS
KEYWORD
base,nonn,easy
AUTHOR
EXTENSIONS
Examples added by Hieronymus Fischer, May 30 2012
STATUS
approved