A099150 -id:A099150 - cp's OEIS frontend

A198971 a(n) = 5*10^n - 1.

4, 49, 499, 4999, 49999, 499999, 4999999, 49999999, 499999999, 4999999999, 49999999999, 499999999999, 4999999999999, 49999999999999, 499999999999999, 4999999999999999, 49999999999999999, 499999999999999999, 4999999999999999999, 49999999999999999999, 499999999999999999999

Offset: 0

Vincenzo Librandi, Nov 02 2011

Also maximal value of GCD of 2 distinct (n+1)-digit numbers (compare with A126687). - Michel Marcus, Jun 24 2013
Also, a(n) is the largest obtained remainder when an (n+1)-digit number m is divided by any k with 1 <= k <= m. This remainder is obtained when 10^(n+1)-1 is divided by 5*10^n, example: 999 = 500 * 1 + 499, and a(2) = 499. - Bernard Schott, Nov 23 2021
Also numbers k whose digital reversal equals 2*(k - 2). - Stefano Spezia, Sep 15 2024

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

Cf. A002283, A086940, A086942, A099150, A126687.

Magma
```
[5*10^n-1 : n in [0..20]];
```

CoefficientList[Series[(4 + 5*x)/(1 - 11*x + 10*x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 03 2013 *)
LinearRecurrence[{11,-10},{4,49},20] (* Harvey P. Dale, Dec 30 2018 *)

a(n)=5*10^n-1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 10*a(n-1) + 9.
a(n) = 11*a(n-1) - 10*a(n-2), n>1.
G.f.: (4 + 5*x)/(1 - 11*x + 10*x^2). - Vincenzo Librandi, Jan 03 2013
E.g.f.: exp(x)*(5*exp(9*x) - 1). - Stefano Spezia, Nov 17 2022
a(n) = A086942(n+1)/8 = A086940(n+1)/4 = A099150(n+1)/2. - Elmo R. Oliveira, May 02 2025

A086942 Integers k such that R(k+8) = 4.

Original entry on oeis.org

32, 392, 3992, 39992, 399992, 3999992, 39999992, 399999992, 3999999992, 39999999992, 399999999992, 3999999999992, 39999999999992, 399999999999992, 3999999999999992, 39999999999999992, 399999999999999992, 3999999999999999992, 39999999999999999992

Offset: 1

Ray Chandler, Jul 24 2003

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A004086, A086940, A086943, A099150, A198971.

[4*10^n-8: n in [1..25]]; // Vincenzo Librandi, Aug 22 2011

4*10^Range[20] - 8 (* Paolo Xausa, Sep 21 2024 *)

a(n) = 4*10^n - 8.
R(a(n)) = A086943(n).
G.f.: 8*x*(5*x+4)/((10*x-1)*(x-1)).
a(n) = 8*A198971(n-1).
From Elmo R. Oliveira, May 01 2025: (Start)
E.g.f.: 4*(1 - 2*exp(x) + exp(10*x)).
a(n) = 4*A099150(n) = 2*A086940(n).
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. (End)

A116229 Numbers k such that k*(k+6) gives the concatenation of two numbers m and m-9.

Original entry on oeis.org

8, 98, 998, 9998, 99998, 999998, 9999998, 99999998, 999999998, 9999999998, 36363636361, 45454545452, 54545454543, 63636363634, 72727272725, 81818181816, 90909090907, 99999999998, 999999999998, 9999999999998

Offset: 1

Giovanni Resta, Feb 06 2006

			99999998 * 100000004 = 100000001//99999992, where // denotes concatenation.

Cf. A099150, A116098, A116228, A116230, A116237.

A255398 Numbers k such that k^2 lacks the digit 1 in its decimal expansion.

Original entry on oeis.org

0, 2, 3, 5, 6, 7, 8, 15, 16, 17, 18, 20, 22, 23, 24, 25, 26, 27, 28, 30, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 64, 65, 66, 67, 68, 70, 73, 74, 75, 76, 77, 78, 80, 82, 83, 84, 85, 86, 87, 88, 92, 93, 94, 95, 97, 98, 143, 144, 150, 153, 155, 156, 157, 158

Offset: 1

Vincenzo Librandi, Feb 22 2015

			98 is in this sequence because 98^2 = 9604.
99 is not in this sequence because 99^2 = 9801.

Mohammed Yaseen, Table of n, a(n) for n = 1..10000

Cf. A052383, A052040, A099150.

[n: n in [0..200] | not 1 in Intseq(n^2)];

filter:= n -> not member(1, convert(n^2,base,10)):
select(filter, [$0..200]); # Robert Israel, Apr 27 2023

Select[Range[0, 200], DigitCount[#^2, 10, 1]==0 &]

isok(k) = !vecsearch(Set(digits(k^2)), 1); \\ Michel Marcus, Apr 29 2023

def ok(k): return "1" not in str(k**2)
print([k for k in range(160) if ok(k)]) # Michael S. Branicky, Apr 27 2023

From Mohammed Yaseen, Apr 18 2023: (Start)
The smallest n-digit term ~ sqrt(2) * 10^(n-1).
The largest n-digit term = 10^n - 2 (see A099150). (End)

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

A086941 a(n) = k where R(k+6) = 3.

Original entry on oeis.org

24, 294, 2994, 29994, 299994, 2999994, 29999994, 299999994, 2999999994, 29999999994, 299999999994, 2999999999994, 29999999999994, 299999999999994, 2999999999999994, 29999999999999994, 299999999999999994, 2999999999999999994, 29999999999999999994, 299999999999999999994

Offset: 1

Ray Chandler, Jul 24 2003

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A004086, A086944, A099150.

[3*(10^n-2): n in [1..20]]; // Vincenzo Librandi, Aug 22 2011

3(10^Range[20]-2) (* or  *) 10*Table[FromDigits[PadRight[{2},n,9]],{n,20}]+4 (* Harvey P. Dale, Jul 25 2023 *)

a(n) = 3*(10^n - 2).
R(a(n)) = A086944(n).
From Elmo R. Oliveira, Apr 30 2025: (Start)
G.f.: 6*x*(5*x+4)/((x-1)*(10*x-1)).
E.g.f.: 3*(1 - 2*exp(x) + exp(10*x)).
a(n) = 3*A099150(n).
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. (End)

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

A177096 a(n) = 5*(10^n - 2).

Original entry on oeis.org

40, 490, 4990, 49990, 499990, 4999990, 49999990, 499999990, 4999999990, 49999999990, 499999999990, 4999999999990, 49999999999990, 499999999999990, 4999999999999990, 49999999999999990, 499999999999999990, 4999999999999999990, 49999999999999999990, 499999999999999999990

Offset: 1

Vincenzo Librandi

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

Cf. A099150, A198971.

Vec(10*x*(4+5*x)/(10*x^2-11*x+1) + O(x^21)) \\ Elmo R. Oliveira, May 02 2025

G.f.: 10*x*(4+5*x)/((10*x-1)*(x-1)). - R. J. Mathar, Jan 06 2011
From Elmo R. Oliveira, May 02 2025: (Start)
E.g.f.: 5*(1 + exp(10*x) - 2*exp(x)).
a(n) = 5*A099150(n) = 10*A198971(n-1).
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. (End)

More terms from Elmo R. Oliveira, May 02 2025

A342214 Primes formed by the concatenation of exactly three successive composite numbers.

Original entry on oeis.org

138140141, 180182183, 240242243, 250252253, 330332333, 400402403, 408410411, 478480481, 546548549, 570572573, 600602603, 646648649, 660662663, 676678679, 768770771, 838840841, 876878879, 928930931, 940942943, 970972973, 996998999, 100810101011, 109610981099

Offset: 1

Bernard Schott, Mar 05 2021

The primes that are obtained by the concatenation of exactly three successive composite numbers are always of the form c||c+2||c+3, with c+1 prime and c+3 odd <> 5, hence c must necessary ends with 0, 6, 8 (see examples).
No such primes can be obtained with the two other possible configurations of 3 successive composite numbers: c||c+1||c+2 or c||c+1||c+3.
The number of digits in each term is a multiple of 3. If a term existed for which this were not true, then c would necessarily be of the form 10^k - 2 (A099150), but then c+1 = 10^k - 1 would not be prime.

			a(1) = 138140141 because 138, 140, 141 are 3 successive composite numbers, then concat(138, 140, 141) = 138140141 is prime and is the least prime with this property (see link Prime Curios!).
The smallest such primes whose first composite ends respectively with 0, 6, 8 are: a(2) = 180182183, a(9) = 546548549, a(1) = 138140141.
If (3,q) is the smallest term formed by the concatenation of 3 successive composite numbers with each q digits: (3,3) = a(1) = 138140141, (3,4) = a(22) = 100810101011.

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 138140141.

Cf. A087341, A281684, A342049 (similar, with 2 consecutive composites).

nextc[n_] := Module[{k = n + 1}, While[PrimeQ[k], k++]; k]; seq = {}; n1 = 4; n2 = nextc[n1]; Do[n3 = nextc[n2]; c = FromDigits @ Flatten @ Join[IntegerDigits /@ {n1, n2, n3}]; If[PrimeQ[c], AppendTo[seq, c]]; n1 = n2; n2 = n3, {1000}]; seq (* Amiram Eldar, Mar 05 2021 *)

lista(nn) = {my(ca=4, cb=6); forcomposite(c=7, nn, if (isprime(x=eval(concat(Str(ca), concat(Str(cb), Str(c))))), print1(x, ", ")); ca = cb; cb = c;);} \\ Michel Marcus, Mar 05 2021

from sympy import isprime
def aupto(limit):
  c, t, alst = 6, 689, []
  while t < limit:
    t = int("".join(map(str, [c, c+2, c+3])))
    if isprime(c+1) and not isprime(c+3) and isprime(t): alst.append(t)
    c += [6, 4, 2, 2, 2][(c%10)//2]
  return alst
print(aupto(109610981099)) # Michael S. Branicky, Mar 05 2021

A292552 Nontotients of the form 10^k - 2.

Original entry on oeis.org

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

Offset: 1

Torlach Rush, Sep 18 2017

nonn

There are no k for which (2^n)*(5^n)[p1*p2*...*pk]-2[p1*p2*...*pk]=m[(p1-1)*(p2-1)*...*(pk-1)].
Up to k = 60, the only totient of the form 10^k-2 is obtained for k=1. - Giovanni Resta, Sep 20 2017
For 10^k-2 with k > 1 to be a totient, it would have to be of the form (p-1)*p^m for some odd prime p and m >= 2. - Robert Israel, Sep 20 2017

			a(1) = A011557(2) - 2 = A005277(13);
a(2) = A011557(3) - 2 = A005277(210);
a(3) = A011557(4) - 2 = A005277(2627);
a(4) = A011557(5) - 2 = A005277(29747).

Cf. A005277, A011557, A099150.

More terms from Giovanni Resta, Sep 20 2017

cp's OEIS Frontend

A198971 a(n) = 5*10^n - 1.

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A086942 Integers k such that R(k+8) = 4.

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A116229 Numbers k such that k*(k+6) gives the concatenation of two numbers m and m-9.

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A255398 Numbers k such that k^2 lacks the digit 1 in its decimal expansion.

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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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A086941 a(n) = k where R(k+6) = 3.

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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).

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A177096 a(n) = 5*(10^n - 2).

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