A099150 Positive integers k such that f(k)+f(k)=concatenation of k and k, where f(k)=k(k+3)/2 (A000096).

8, 98, 998, 9998, 99998, 999998, 9999998, 99999998, 999999998, 9999999998, 99999999998, 999999999998, 9999999999998, 99999999999998, 999999999999998, 9999999999999998, 99999999999999998, 999999999999999998, 9999999999999999998, 99999999999999999998

Offset

1,1

Comments

By the definition, k*(k+3) = k*10^m+k. So k+3 = 10^m+1, that is k = 10^m-2. - Seiichi Manyama, Aug 31 2019

Links

Table of n, a(n) for n=1..20.

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

Formula

a(n) = A002283(n) - 1 = 10^n - 2. - Seiichi Manyama, Aug 31 2019

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

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

G.f.: x*(10*x + 8)/((x - 1)*(10*x - 1)). (End)

Example

99998*(99998+3) = 9999899998 (concatenation of 99998 and 99998).

Prog

(PARI) for(k=1, 1e9, if(k*(k+3)==eval(Str(k, k)), print1(k", "))) \\ Seiichi Manyama, Aug 31 2019
(PARI) {a(n) = 10^n-2} \\ Seiichi Manyama, Aug 31 2019

Crossrefs

Cf. A000096, A002283, A096032, A099148, A099149.

Sequence in context: A116287 A290242 A116229 * A199029 A228329 A228794

Adjacent sequences: A099147 A099148 A099149 * A099151 A099152 A099153

Keyword

nonn,base,easy

Author

John W. Layman, Sep 30 2004

Extensions

a(9)-a(20) from Seiichi Manyama, Aug 31 2019

Status

approved