A099151 Positive integers a such that f(3a)+f(a)=concatenation of 3a and a, where f(k)=k(k+3)/2 (A000096).

5, 59, 599, 5999, 59999, 599999, 5999999, 59999999, 599999999, 5999999999, 59999999999, 599999999999, 5999999999999, 59999999999999, 599999999999999, 5999999999999999, 59999999999999999, 599999999999999999, 5999999999999999999, 59999999999999999999, 599999999999999999999

Offset

1,1

Comments

Is it difficult to prove that the sequence continues in the expected way?

Links

Table of n, a(n) for n=1..21.

Index entries for linear recurrences with constant coefficients, signature (11, -10).

Formula

From Chai Wah Wu, Jun 15 2020: (Start)

a(n) = 6*10^(n-1) - 1.

a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.

G.f.: x*(4*x + 5)/((x - 1)*(10*x - 1)).

Proof: let m be a term and r be the number of decimal digits of m. Then m satisfies the equation 3m(3m+3)/2 + m(m+3)/2 = 3m*10^r + m = m(3*10^r+1). Solving for m we get m = 6*10^(r-1) - 1 and it is clear that m indeed has r decimal digits.

(End)

Example

599 is in the sequence because (3*599)(3*599+3)/2 + 599(602)/2 = 1797*1800/2 + 599*602/2 = 1797599.

Crossrefs

Cf. A096032, A099148, A099149, A099150.

Sequence in context: A106105 A178003 A036947 * A243219 A113055 A020468

Adjacent sequences: A099148 A099149 A099150 * A099152 A099153 A099154

Keyword

nonn,base,easy

Author

John W. Layman, Sep 30 2004

Extensions

Edited by Charles R Greathouse IV, Apr 29 2010

More terms from Chai Wah Wu, Jun 15 2020

Status

approved