A095372 -id:A095372 - cp's OEIS frontend

A095373 Integers k such that A095372(k) = 1 + 90*(-1+100^k)/99 is prime.

2, 3, 8, 15, 26, 33, 146, 320, 1068, 1505

Offset: 1

Labos Elemer, Jun 07 2004

a(11) > 40000. - Michael S. Branicky, Jan 06 2025

			Corresponding primes are: 9091, 909091, 909090909090909091, 909090909090909090909090909091, ...

Cf. A095372, A002275, A095370, A095371.

{ta=Table[0, {100}], u=1}; Do[s=1+(90*(100^n-1)/99);If[PrimeQ[s], Print[{n, s}]; ta[[u]]=s;u=u+1], {n, 1, 320}] ta

a(9)-a(10) from Max Alekseyev, Jan 28 2012

A096486 Period length of continued fraction for square root of A095372(n)=1+90*(-1+100^n)/99.

Original entry on oeis.org

8, 170, 2242, 2132, 1294, 976846, 216566, 9904144, 25617930, 408928520, 25346031262, 137031675878, 89950978114

Offset: 1

Labos Elemer, Jun 25 2004

Cf. A095372, A096485.

with(numtheory): [seq(nops(cfrac((1+90*(-1+100^k)/99)^(1/2), 'periodic', 'quotients')[2]), k=2..10)];

Do[Print[Length[Last[ContinuedFraction[(1+90*(-1+100^n)/99)^(1/2)]]]], {n, 1, 9}]

a(10) from Vaclav Kotesovec, Aug 29 2019
a(11)-a(12) from Chai Wah Wu, Dec 07 2019
a(13) from Chai Wah Wu, Oct 01 2021

A116436 Numbers m which when sandwiched between two 1's give a multiple of m.

Original entry on oeis.org

1, 11, 13, 77, 91, 137, 9091, 909091, 5882353, 10989011, 12987013, 52631579, 76923077, 90909091, 4347826087, 9090909091, 13698630137, 909090909091, 3448275862069, 10989010989011, 12987012987013, 76923076923077, 90909090909091, 9090909090909091, 909090909090909091

Offset: 1

Giovanni Resta, Feb 15 2006

All k-digit numbers that divide 10^{k+1} + 1. - Franklin T. Adams-Watters, Apr 23 2008
Notice the infinite pattern m = (90..90..90)91 with 1m1/m = 21, e.g., 1911/91 = 190911/9091 = 19090911/909091 = 21 (see A095372). - Zak Seidov, Apr 22 2008
Corresponding numbers k such that k * a(n) = 1.a(n).1 where '.' stands for concatenation are in A351320. - Bernard Schott, Feb 07 2022

			77 is a member since 1771 is a multiple of 77 (77*23).

Michael S. Branicky, Table of n, a(n) for n = 1..1441 (terms 1..87 from Franklin T. Adams-Watters)

Subsequence of A116437, A116438, A116439, A116440, A116441, A116442, A116443, A116444.
Cf. A136296, A329914, A329915, A351320.
Some subsequences, M such that k*M=1M1 for: A095372 \ {1} (k=21), A331630 (k=23), A351237 (k=83), A351238 (k=87), A351239 (k=101).

f[k_, d_] := Flatten@Table[Select[Divisors[k*(10^(i + 1) + 1)], IntegerLength[ # ] == i &], {i, d}]; f[1, 14] (* Ray Chandler, May 11 2007 *)

A116436(k) = {local(l, d, lb, ub); d=divisors(10^(k+1)+1);l=[];lb=10^(k-1); ub=10*lb; for(i=1,#d, if(d[i]>=lb&&d[i]A116436(i))); l
\\ Franklin T. Adams-Watters, Apr 22 2008

from sympy import isprime
from itertools import count, islice
def agen(): # generator of terms
    yield 1
    for k in count(2):
        t = 10**(k+1) + 1
        yield from (t//i for i in range(100, 10, -1) if t%i == 0)
print(list(islice(agen(), 25))) # Michael S. Branicky, Mar 26 2023 following Franklin T. Adams-Watters but removing factorization

A351320(n) * a(n) = 1.a(n).1 where "." stands for concatenation. - Bernard Schott, Feb 07 2022

A100706 Bisection of A002275.

Original entry on oeis.org

1, 111, 11111, 1111111, 111111111, 11111111111, 1111111111111, 111111111111111, 11111111111111111, 1111111111111111111, 111111111111111111111, 11111111111111111111111

Offset: 0

N. J. A. Sloane, Nov 19 2004

Also the binary representation of the n-th iteration of the elementary cellular automaton starting with a single ON (black) cell for Rules 151, 159, 183, 191, 215, 222, 223, 247, 254 and 255. - Robert Price, Feb 21 2016

The aerated sequence 1, 0, 111, 0, 11111, 0, 1111111, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -9^2, Q = -10 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A299960. - Peter Bala, Aug 28 2019

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
S. Wolfram, A New Kind of Science
Index to Elementary Cellular Automata
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (101,-100).

Cf. A002275, A099814 (other bisection), A007583, A095372, A299960.

seq((10^(2*n+1) - 1)/9,n=0..15); # C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005

Table[(10^(2*n + 1) - 1)/9, {n, 0, 100}] (* Robert Price, Feb 21 2016 *)

a(n) = (10^(2*n + 1) - 1)/9; \\ Michel Marcus, Mar 12 2023

def A100706(n): return (10**((n<<1)+1)-1)//9 # Chai Wah Wu, Nov 04 2022

Numbers composed entirely of 2*n+1 concatenated 1's for n >= 0.
O.g.f.: (1+10*x)/((-1+x)*(-1+100*x)). - R. J. Mathar, Apr 03 2008
From Klaus Purath, Sep 23 2020: (Start)
a(n) = Sum_{i = 0..2*n} 10^i.
a(n) = 101*a(n-1) - 100*a(n-2).
a(n) = 110*10^(2*n-2) + a(n-1).
a(n) = 100*a(n-1) + 11.
a(n) = (a(n-1)^2 - 1210*10^(2*n-4))/a(n-2). (End)

More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI
```
A299960(n)=4^(2*n+1)\5+1
```

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
  (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
  (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
  (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
  (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
  (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
  (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
  (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
  (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
  (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A329915 a(n) is the least M such that A329914(n) * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

91, 77, 5882353, 52631579, 4347826087, 3448275862069, 2127659574468085106383, 20408163265306122449, 1694915254237288135593220339, 16393442622950819672131147541, 137, 13, 112359550561797732809, 11, 10309278350515463917525773195876288659793814433

Offset: 1

Bernard Schott, Nov 24 2019

When M is a q-digit term, then M is a divisor of 10^(q+1) + 1.
For each term k in A329914, there exist a set of numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k. This sequence gives the smallest integer M(k) = M of each set {M_k}.
See A329914 for further information about these numbers.

			A329914(1) = 21 and 21 * 91 = 1[91]1, and there is no integer < 91 that satisfies this relation, so a(1) = 91.
A329914(2) = 23 and 23 * 77 = 1[77]1, and there is no integer < 77 that satisfies this relation, so a(2) = 77.
A329914(5) = 33 and 33 * 4347826087 = 1[4347826087]1, and there is no integer < 4347826087 that satisfies this relation, so a(5) = 4347826087.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Cf. A000533, A329914 (corresponding numbers k).

Some corresponding sets {M_k} : A095372 \ {1} = {M_21}, A331630 = {M_23}, A351237 = {M_83}, A351238 = {M_87}, A351239 = {M_101}.

A331630 Numbers M such that 23 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

77, 76923077, 76923076923077, 76923076923076923077, 76923076923076923076923077, 76923076923076923076923076923077, 76923076923076923076923076923076923077, 76923076923076923076923076923076923076923077

Offset: 1

Bernard Schott, Jan 23 2020

There are only 15 numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k; 23 is the second such integer, so 23 = A329914(2), and a(1) = A329915(2) = 77; hence, the terms of this sequence form the infinite set {M_23}.

Every term M = a(n) has q = 6*n-4 digits, and 10^(q+1)+1 that has 6*n-4 zeros in its decimal expansion is equal to 13 * M, so M is a divisor of 10^(6*n-3)+1. Example: a(2) = 76923077 has 8 digits and 13 * 76923077 = 1000000001 that has 8 zeros in its decimal expansion.

			23 * 77 = 1771, hence 77 is a term.
23 * 76923076923077 = 1(76923076923077)1, and 76923076923077 is another term.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Colin Barker, Table of n, a(n) for n = 1..150
Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).

Subsequence of A116436.

Cf. A329914, A329915, A095372 \ {1} (similar for k = 21).

Maple
```
seq((10^(6*m-3)+1)/13, m=1..15);
```

Array[(10^(6 # - 3) + 1)/13 &, 9] (* Michael De Vlieger, Jan 24 2020 *)
LinearRecurrence[{1000001,-1000000},{77,76923077},10] (* Harvey P. Dale, Mar 03 2023 *)

vector(9, n, (10^(6*n-3)+1)/13) \\ Michel Marcus, Jan 25 2020

Vec(77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)) + O(x^10)) \\ Colin Barker, Jan 25 2020

apply( {A331630(n)=10^(6*n-3)\/13}, [1..9]) \\ M. F. Hasler, Jan 26 2020, following Michel Marcus

a(n) = (10^(6*n-3)+1)/13 for n >= 1.
From Colin Barker, Jan 25 2020: (Start)
G.f.: 77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)).
a(n) = 1000001*a(n-1) - 1000000*a(n-2) for n>2.
a(n) = (1000 + 1000^(2*n))/13000 for n>0.
(End)
E.g.f.: exp(x)*(1000 + exp(999999*x))/13000 - 77/1000. - Stefano Spezia, Jan 26 2020

A351237 Numbers M such that 83 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

137, 13698630137, 1369863013698630137, 136986301369863013698630137, 13698630136986301369863013698630137, 1369863013698630136986301369863013698630137, 136986301369863013698630136986301369863013698630137

Offset: 1

Bernard Schott, Feb 05 2022

There are only 15 numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k; 83 is the eleventh such integer, so 83 = A329914(11), and a(1) = A329915(11) = 137; hence, the terms of this sequence form the infinite set {M_83}.

Every term M = a(n) has q = 8*n-5 digits, and 10^(q+1)+1 that has q = 8*n-5 zeros in its decimal expansion is equal to 73 * M, so a(n) = M is a divisor of 10^(8*n-4)+1. Example: a(2) = 13698630137 has 11 digits and 73 * 13698630137 = 1000000000001 that has 11 zeros in its decimal expansion.

			83 * 137 = 1[137]1, hence 137 is a term.
83 * 13698630137 = 1[13698630137]1, and 13698630137 is another term.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Index entries for linear recurrences with constant coefficients, signature (100000001,-100000000).

Subsequence of A116436.
Cf. A329914, A329915.
Similar for: A095372 \ {1} (k=21), A331630 (k=23), this sequence (k=83), A351238 (k=87), A351239 (k=101).

Maple
```
seq((10^(8*n-4)+1)/73, n=1..15);
```

Table[(10^(8*n-4)+1)/73, {n, 1, 7}] (* Amiram Eldar, Feb 06 2022 *)
LinearRecurrence[{100000001,-100000000},{137,13698630137},20] (* Harvey P. Dale, Nov 01 2022 *)

a(n) = (10^(8*n-4)+1)/73 for n >= 1.

A351238 Numbers M such that 87 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

13, 12987013, 12987012987013, 12987012987012987013, 12987012987012987012987013, 12987012987012987012987012987013, 12987012987012987012987012987012987013, 12987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987012987013

Offset: 1

Bernard Schott, Feb 05 2022

There are only 15 numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k; 87 is the twelfth such integer, so 87 = A329914(12), and a(1) = A329915(12) = 13; hence, the terms of this sequence form the infinite set {M_87}.

Every term M = a(n) has q = 6*n-4 digits, and 10^(q+1)+1 that has q = 6*n-4 zeros in its decimal expansion is equal to 77 * M, so a(n) = M is a divisor of 10^(6*n-3)+1. Example: a(2) = 12987013 has 8 digits and 77 * 12987013 = 1000000001 that has 8 zeros in its decimal expansion.

			87 * 13 = 1[13]1, hence 13 is a term.
87 * 12987013 = 1[12987013]1, and 12987013 is a term.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).

Subsequence of A116436.
Cf. A329914, A329915.
Similar for: A095372 \ {1} (k=21), A331630 (k=23), A351237 (k=83), this sequence (k=87), A351239 (k=101).

Maple
```
seq((10^(6*n-3)+1)/77, n=1..15);
```

Table[(10^(6*n - 3) + 1)/77, {n, 1, 10}] (* Amiram Eldar, Feb 06 2022 *)

a(n) = (10^(6*n-3)+1)/77 for n >= 1.

A351239 Numbers M such that 101 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

11, 10989011, 10989010989011, 10989010989010989011, 10989010989010989010989011, 10989010989010989010989010989011, 10989010989010989010989010989010989011, 10989010989010989010989010989010989010989011, 10989010989010989010989010989010989010989010989011

Offset: 1

Bernard Schott, Feb 05 2022

There are only 15 numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k; 101 is the fourteenth such integer, so 101 = A329914(14), and a(1) = A329915(14) = 11; hence, the terms of this sequence form the infinite set {M_101}.

Every term M = a(n) has q = 6*n-4 digits, and 10^(q+1)+1 that has q = 6*n-4 zeros in its decimal expansion is equal to 91 * M, so a(n) = M is a divisor of 10^(6*n-3)+1. Example: a(2) = 10989011 has 8 digits and 91 * 10989011 = 1000000001 that has 8 zeros in its decimal expansion.

			101 * 11 = 1[11]1, hence 11 is a term.
101 * 10989011 = 1[10989011]1 and 10989011 is another term.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).

Subsequence of A116436.
Cf. A329914, A329915.
Similar for: A095372 \ {1} (k = 21), A331630 (k = 23), A351237 (k = 83), A351238 (k = 87), this sequence (k = 101).

Maple
```
seq((10^(6*n-3)+1)/91, n=1..15);
```

Table[(10^(6*n - 3) + 1)/91, {n, 1, 9}] (* Amiram Eldar, Feb 06 2022 *)
LinearRecurrence[{1000001,-1000000},{11,10989011},10] (* Harvey P. Dale, Sep 12 2022 *)

a(n) = (10^(6*n-3)+1)/91 for n >= 1.

cp's OEIS Frontend

A095373 Integers k such that A095372(k) = 1 + 90*(-1+100^k)/99 is prime.

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A096486 Period length of continued fraction for square root of A095372(n)=1+90*(-1+100^n)/99.

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A116436 Numbers m which when sandwiched between two 1's give a multiple of m.

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A100706 Bisection of A002275.

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A299960 a(n) = (4^(2*n+1) + 1) / 5.

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A329915 a(n) is the least M such that A329914(n) * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

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A331630 Numbers M such that 23 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

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