A099148 -id:A099148 - 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

A099149 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 b.

Original entry on oeis.org

5, 6, 8, 8, 6, 5, 59, 78, 384, 98, 98, 506, 78, 1803, 59, 599, 798, 960, 921, 998, 998, 14555, 921, 860, 798, 599, 506, 384, 5999, 6785, 24615, 7343, 27170, 7998, 36005

Offset: 1

John W. Layman, Sep 30 2004

For values of a, see A099148.

			From _R. J. Mathar_, Oct 10 2017 (Start)
Values of a, then b, then f(a)+f(b) and concatenations (first a, then b or first b, then a), sorted on b, are:
0 0 0+0=0
2 5 5+20=25
3 6 9+27=36
8 8 44+44=88
8 9 44+54=98
6 14 27+119=146
5 15 20+135=155
0 17 0+170=170
20 59 230+1829=2059
39 78 819+3159=3978
98 98 4949+4949=9898
98 99 4949+5049=9998
78 158 3159+12719=15878
59 177 1829+15930=17759
77 384 3080+74304=77384
138 506 9729+128777=138506
200 599 20300+180299=200599
399 798 80199+319599=399798
492 860 121770+371090=492860
614 921 189419+425502=614921
998 998 499499+499499=998998
998 999 499499+500499=999998
921 1383 425502+958419=1383921
860 1505 371090+1134770=1505860
798 1598 319599+1279199=1598798
599 1797 180299+1617300=1797599
164 1803 13694+1628109=1641803
506 1859 128777+1730729=1859506
384 1920 74304+1846080=1920384
2000 5999 2003000+18002999=20005999
2655 6785 3528495+23028290=26556785
 3213 7343 5166504+26970839=32137343
 3999 7998 8001999+31995999=39997998
(End)

Cf. A096032, A099148, A099150.

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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A099149 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 b.

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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