A147553 Numbers k such that k^2 divides k.k where dot "." means concatenation.

1, 143, 142857143, 142857142857143, 142857142857142857143, 142857142857142857142857143, 142857142857142857142857142857143, 142857142857142857142857142857142857143, 142857142857142857142857142857142857142857143, 142857142857142857142857142857142857142857142857143

Offset

1,2

Comments

I proved that for n > 0, a(n+1) = (10^(6*n-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).

Formula

From Elmo R. Oliveira, Mar 24 2026: (Start)

G.f.: x*(857000*x^2 - 999858*x + 1)/((x - 1)*(1000000*x - 1)).

a(n) = 1000001*a(n-1) - 1000000*a(n-2) for n > 3. (End)

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

Cf. A147554, A243162 (k^2 divides k.k.k).

Sequence in context: A204683 A205159 A205308 * A030122 A057404 A173713

Adjacent sequences: A147550 A147551 A147552 * A147554 A147555 A147556

Keyword

base,easy,nice,nonn

Author

Farideh Firoozbakht, Dec 23 2008

Status

approved