OFFSET
1,3
COMMENTS
If n-1 is represented as a base-6 number (see A007092) 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)=0,1,2,3,5,7 for k=0..5. - Hieronymus Fischer, May 30 2012
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
FORMULA
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = (b_m(n)+1) mod 10 + floor((b_m(n)+2)/5) + floor((b_m(n)+1)/5) - 2*floor(b_m(n)/5))*10^m + sum_{j=0..m-1} (b_j(n) mod 6 + floor((b_j(n)+1)/6) + floor((b_j(n)+2)/6) - 2*floor(b_j(n)/6)))*10^j, where n>1, b_j(n)) = floor((n-1-6^m)/6^j), m = floor(log_6(n-1)).
a(1*6^n+1) = 1*10^n.
a(2*6^n+1) = 2*10^n.
a(3*6^n+1) = 3*10^n.
a(4*6^n+1) = 5*10^n.
a(5*6^n+1) = 7*10^n.
a(n) = 10^log_6(n-1) for n=6^k+1, k>0,
a(n) < 10^log_6(n-1) else.
a(n) <= A084984(n), equality holds if the representation of n-1 as a base-6 number only has digits 0 or 1.
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^6^j *(1-x^6^j)* (1 + 2x^6^j + 3(x^2)^6^j + 5(x^3)^6^j + 7(x^4)^6^j)/(1-x^6^(j+1)).
Also: g(x) = (x/(1-x))*(h_(6,1)(x) + h_(6,2)(x) + h_(6,3)(x) + 2*h_(6,4)(x) + 2*h_(6,5)(x) - 7*h_(6,6)(x)), where h_(6,k)(x) = sum_{j>=0} 10^j*x^(k*6^j)/(1-x^6^(j+1)). (End)
Sum_{n>=2} 1/a(n) = 4.945325883472729555972742252181522711968119529132581193614012706741310832798... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024
EXAMPLE
a(1000) = 5353.
a(10^4) = 115153
a(10^5) = 2070753.
a(10^6) = 33233353.
MATHEMATICA
Union[Flatten[FromDigits/@Tuples[{0, 1, 2, 3, 5, 7}, 3]]] (* Harvey P. Dale, Mar 11 2015 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Jaroslav Krizek, Dec 25 2011
EXTENSIONS
Examples added by Hieronymus Fischer, May 30 2012
STATUS
approved