A099149 -id:A099149 - cp's OEIS frontend

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

John W. Layman, Sep 30 2004

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

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

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

Cf. A000096, A002283, A096032, A099148, A099149, A177096, A198971.

for(k=1, 1e9, if(k*(k+3)==eval(Str(k, k)), print1(k", "))) \\ Seiichi Manyama, Aug 31 2019

{a(n) = 10^n-2} \\ Seiichi Manyama, Aug 31 2019

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)
E.g.f.: 1 - 2*exp(x) + exp(10*x). - Stefano Spezia, May 02 2025
a(n) = 2*A198971(n-1) = A177096(n)/5. - Elmo R. Oliveira, May 02 2025

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

A099148 Take a <= b such that f(a)+f(b)=concatenation of a and b, where f(k)=k(k+3)/2 (A000096). Sequence gives values of a.

Original entry on oeis.org

2, 3, 8, 9, 14, 15, 20, 39, 77, 98, 99, 138, 158, 164, 177, 200, 399, 492, 614, 998, 999, 1065, 1383, 1505, 1598, 1797, 1859, 1920, 2000, 2655, 3077, 3213, 3762, 3999, 6707

Offset: 1

John W. Layman, Sep 30 2004

For values of b, see A099149.

Cf. A096032, A099149.

Edited by Charles R Greathouse IV, Apr 23 2010

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

Original entry on oeis.org

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

Offset: 1

John W. Layman, Sep 30 2004

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

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

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

Cf. A000096, A096032, A099148, A099149, A099150.

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)
E.g.f.: (2 - 5*exp(x) + 3*exp(10*x))/5. - Elmo R. Oliveira, Jun 09 2025

Edited by Charles R Greathouse IV, Apr 29 2010

More terms from Chai Wah Wu, Jun 15 2020

cp's OEIS Frontend

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

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A099148 Take a <= b such that f(a)+f(b)=concatenation of a and b, where f(k)=k(k+3)/2 (A000096). Sequence gives values of a.

Views

Author

Keywords

Comments

Crossrefs

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions