A126109 -id:A126109 - cp's OEIS frontend

A133384 Numbers with n 0's between 1 and 2.

12, 102, 1002, 10002, 100002, 1000002, 10000002, 100000002, 1000000002, 10000000002, 100000000002, 1000000000002, 10000000000002, 100000000000002, 1000000000000002, 10000000000000002, 100000000000000002, 1000000000000000002, 10000000000000000002

Offset: 0

Paul Curtz, Oct 29 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Subsequence of A052148.

Cf. A004086, A126109.

[10^(n+1)+2: n in [0..20]]; // Vincenzo Librandi, Aug 10 2011

Table[FromDigits[Join[PadRight[{1},n,0],{2}]],{n,20}] (* or *) 10^Range[20]+2 (* or *) LinearRecurrence[{11,-10},{12,102},20] (* Harvey P. Dale, Oct 03 2013 *)

a(n) = 10^(n+1) + 2. - Vincenzo Librandi, Aug 10 2011
a(n) = 11*a(n-1) - 10*a(n-2); a(0)=12, a(1)=102. - Harvey P. Dale, Oct 03 2013
From Stefano Spezia, Nov 30 2023: (Start)
O.g.f.: 6*(2 - 5*x)/((1 - x)*(1 - 10*x)).
E.g.f.: 2*exp(x)*(1 + 5*exp(9*x)).
(R(a(n)) + 1)/(a(n) - 1) = 2, where R(k) = A004086(k). (End)
a(n) = 6 * A126109(n). - Alois P. Heinz, Nov 30 2023

A199685 a(n) = 5*10^n + 1.

Original entry on oeis.org

6, 51, 501, 5001, 50001, 500001, 5000001, 50000001, 500000001, 5000000001, 50000000001, 500000000001, 5000000000001, 50000000000001, 500000000000001, 5000000000000001, 50000000000000001, 500000000000000001, 5000000000000000001, 50000000000000000001, 500000000000000000001

Offset: 0

Vincenzo Librandi, Nov 09 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A126109.

Magma
```
[5*10^n+1: n in [0..30]];
```

LinearRecurrence[{11,-10},{6,51},30] (* Harvey P. Dale, Apr 06 2013 *)

a(n)=5*10^n+1 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 3*A126109(n).
a(n) = 10*a(n-1) - 9.
a(n) = 11*a(n-1) - 10*a(n-2).
G.f.: 3*(2-5*x)/((1-x)*(1-10*x)).
E.g.f.: exp(x)*(5*exp(9*x) + 1). - Elmo R. Oliveira, Sep 15 2024

A102940 Numbers k such that 10^k + 6*R_k + 1 is prime, where R_k = 11...1 is the repunit (A002275) of length k.

Original entry on oeis.org

0, 1, 2, 3, 5, 9, 11, 14, 32, 54, 55, 60, 153, 200, 461, 569, 840, 847, 1296, 1356, 2007, 2627, 2847, 3110, 6876, 9161, 17765, 33555, 59142, 65773, 280710

Offset: 1

Robert G. Wilson v, Dec 16 2004

Also numbers k such that (5*10^k + 1)/3 is prime.
Numbers k such that A126109(k) is prime.
a(32) > 3*10^5. - Robert Price, Nov 15 2014

Makoto Kamada, Prime numbers of the form 166...667.
Index entries for primes involving repunits

Cf. A000040, A002275, A102024, A126109.

A102940:=n->`if`(isprime((5*10^n+1)/3),n,NULL): seq(A102940(n),n=0..1000); # Wesley Ivan Hurt, Nov 15 2014

Do[ If[ PrimeQ[(5*10^n + 1)/3], Print[n]], {n, 0, 10000}]

a(n) = A102024(n-1) + 1.

Edited by N. J. A. Sloane, Mar 16 2007
Addition of a(27) from Kamada data by Robert Price, Dec 08 2010
a(28) from Erik Branger May 01 2013 by Ray Chandler, Aug 16 2013
a(29)-a(31) from Kamada data by Robert Price, Nov 15 2014

A350382 a(n) = 9 + 4 * 10^n.

Original entry on oeis.org

49, 409, 4009, 40009, 400009, 4000009, 40000009, 400000009, 4000000009, 40000000009, 400000000009, 4000000000009, 40000000000009, 400000000000009, 4000000000000009, 40000000000000009, 400000000000000009, 4000000000000000009, 40000000000000000009, 400000000000000000009, 4000000000000000000009

Offset: 1

Bernard Schott, Dec 28 2021

The 4th problem of 16th Tournament of Towns in 1994-1995, Spring tour 1995, 8-9 grades, Training option, asked for a proof that the number 400...009 with at least one zero is not a perfect square (see link).
Indeed, the first few squares whose digits are 0, 4 and 9 are 4900, 9409, 490000, 940900, 994009, ... (comes from A019544).
Generalization: the 4th problem of 16th Tournament of Towns in 1994-1995, Spring tour 1995, 10-11 grades, Training option, asked for a proof that the number d00...009 with at least one zero is not a perfect square, when d is a digit with 1 <= d <= 9 (see link).

			a(3) = 9 + 4 * 10^3 = 4009 = 19 * 211 is not a square.

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 (in fact, it is Problem 4) of Tournament of Towns 1995, page 301.

Tournament of Towns 1994-1995, Spring tour Problem 4, 8-9 grades, Training option & Problem 4, 10-11 grades, Training option (in Russian and English, problems in red).
Index entries for linear recurrences with constant coefficients, signature (11,-10).
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A019544, A126109, A133384, A199684.

Maple
```
Data := [seq(9 + 4*10^n,  n = 1..20)];
```

a[n_] := 9 + 4*10^n; Array[a, 20] (* Amiram Eldar, Dec 28 2021 *)

a(n) = 9 + 4*10^n = 4*A133384(n-1) + 1.
a(n) = 24*A126109(n-1) + 1 = 10*A199684(n-1) - 1. - Hugo Pfoertner, Dec 28 2021
From Stefano Spezia, Dec 28 2021: (Start)
G.f.: x*(49 - 130*x)/((1 - x)*(1 - 10*x)).
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. (End)

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A133384 Numbers with n 0's between 1 and 2.

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A199685 a(n) = 5*10^n + 1.

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A102940 Numbers k such that 10^k + 6*R_k + 1 is prime, where R_k = 11...1 is the repunit (A002275) of length k.

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A350382 a(n) = 9 + 4 * 10^n.

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