A168624 a(n) = 1 - 10^n + 100^n.

1, 91, 9901, 999001, 99990001, 9999900001, 999999000001, 99999990000001, 9999999900000001, 999999999000000001, 99999999990000000001, 9999999999900000000001, 999999999999000000000001, 99999999999990000000000001, 9999999999999900000000000001, 999999999999999000000000000001

Offset: 0

Author

Jason Earls, Dec 01 2009

Keywords

• easy
• nonn

Comments

Prime values for n = 2,4,6,8, with no others up to n = 3400. Beiler mentions this pattern in the reference.

From Peter Bala, Sep 27 2015: (Start)

Calculation suggests the continued fraction expansion of sqrt(a(n)), for n >= 1, begins [10^n - 1, 1, 1, 1/3*(2*10^n - 5), 1, 5, 1/9*(2*10^n - 11), 1, 17, (2*10^n - 20 - 9*(1 - MOD(n, 3)))/27, ...]. Note the large partial quotients early in the expansion.

A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions. Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n) for m = 2, 3, 4,... as n increases. Some examples are given below. Cf. A000533, A002283, A066138. (End)

Examples

			Simple continued fraction expansions showing large partial quotients:
sqrt(a(10)) = [9999999999; 1, 1, 6666666665, 1, 5, 2222222221, 1, 17, 740740740, 1, 1, 1, 5, 2, 1, 246913579, 1, 1, 4, 1, 1, 3, 1, 1, ...].
a(18)^(1/3) = [999999999999; 1, 2999999, 499999999999, 1, 1439999, 2582644628099, 5, 1, 3, 4, 1, 58, 1, 1, 1, 8, ...].
a(30)^(1/5) = [999999999999; 1, 4999999999999999999, 333333333333, 3, 217391304347826086, 1, 1, 1, 1, 1, 8, 2398081534, 1, 1, 1, 9, 1, 98, 1, 125052522059263, 1, 9, 7, 1, ...]. - _Peter Bala_, Sep 27 2015
		

References

• A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 85.

Links

• Colin Barker, Table of n, a(n) for n = 0..499
• P. Bala, A168624 and some empirical continued fraction expansions.
• Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Crossrefs

Cf. A000533, A002283, A066138, A187868.

Programs

Mathematica

Table[1-10^n+100^n,{n,0,20}] (* Harvey P. Dale, Dec 01 2013 *)

PARI

Vec(-(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

Formula

From Colin Barker, Sep 27 2015: (Start)

a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.

G.f.: -(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)

E.g.f.: exp(x)*(exp(99*x) - exp(9*x) + 1). - Elmo R. Oliveira, Sep 12 2024