OFFSET
1,2
COMMENTS
I proved that for n > 0, a(n+1) = (10^(6n-3) + 1)/7. Namely for n > 1, a(n) is of the form 142857.142857. ... .142857.143. Except for a(1), 11 divides all terms, so there is no prime p such that p^2 divides p.p. For n > 1, a(n).a(n)/(a(n)*a(n))=7.
LINKS
Robert Israel, Table of n, a(n) for n = 1..167
Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).
EXAMPLE
143*143|143.143 (143143/(143*143)=7) so 143 is in the sequence.
MAPLE
1, seq((10^(6*n-3)+1)/7, n=1..20); # Robert Israel, Sep 26 2016
MATHEMATICA
a[0]=1; a[n_]:=(10^(6n-3)+1)/7; Table[a[k], {k, 0, 8}]
Do[d=Divisors[10^i+1]; s=Select[d, Length[IntegerDigits[#]]==i&]; If[Length[s]>0, Do[Print[s[[j]]], {j, Length[s]}]], {i, 69}] (* Hans Havermann, May 31 2014 *)
LinearRecurrence[{1000001, -1000000}, {1, 143, 142857143}, 20] (* Harvey P. Dale, Apr 02 2018 *)
CROSSREFS
KEYWORD
base,easy,nice,nonn
AUTHOR
Farideh Firoozbakht, Dec 23 2008
STATUS
approved