A329915 -id:A329915 - cp's OEIS frontend

A095372 1+integers repeating "90" decimal digit pattern.

1, 91, 9091, 909091, 90909091, 9090909091, 909090909091, 90909090909091, 9090909090909091, 909090909090909091, 90909090909090909091, 9090909090909090909091, 909090909090909090909091

Offset: 0

Labos Elemer, Jun 07 2004

These numbers arise for example as divisors of several repunits (A002275).
The aerated sequence A(n) = [1, 0, 91, 0, 9091, 0, 909091,...] is a divisibility sequence, i.e., A(n) divides A(m) whenever n divides m. It is the case P1 = 0, P2 = -11^2, Q = 10 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 22 2019
Except for a(0) = 1, these terms M are such that 21 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1. Actually 21 is A329914(1) and a(1) = A329915(1) = 91, and the terms >=91 form the set {M_21}; for example, 21 * 909091 = 1(909091)1. - Bernard Schott, Dec 01 2019

			Digit-pattern P=[ab..z] repeating integers equal formally with P*(-1+10^(Ln))/(-1+10^L), where L is the length of pattern;
a(9) divides A002275(38) repunit. See A095371.

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 (101,-100).

Cf. A002275, A095371.
Cf. A001562, A015585, A054416, A097209, A100047, A152577.
Cf. A329914, A329915.

Mathematica
```
Table[1+90*(100^n-1)/99, {n, 0, 20}]
```

a(n) = 1 + 90*(-1 + 100^n)/99 = (10^(2*n+1) + 1)/11. - Rick L. Shepherd, Aug 01 2004
From Colin Barker, Jul 03 2013: (Start)
a(n) = 101*a(n-1) - 100*a(n-2).
G.f.: -(10*x-1)/((x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 10*(exp(99*x) - 1)/11). - Elmo R. Oliveira, Mar 15 2025

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

A329914 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.

Original entry on oeis.org

21, 23, 27, 29, 33, 39, 57, 59, 69, 71, 83, 87, 99, 101, 107

Offset: 1

Bernard Schott, Nov 24 2019

The idea of this sequence comes from the 21-digit integer 112359550561797732809 in Penguin dictionary (see reference) with this property: "The smallest number which, when 1 is placed at both ends, the number is multiplied by 99". The terms of this sequence are the other numbers k that have the same property than 99 and the corresponding smallest numbers M in each set {M_k} are in A329915 (see link).
The Diophantine equation to solve is 1M1 = k * M  with M that has q digits, this is equivalent to 10^(q+1) + 1 = (k-10) * M, with number of zeros in 10^(q+1) + 1 = q also.
Some results coming from this Diophantine equation:
q >= 2 and 21 <= k <= 110, so this sequence is finite. The integers (k-10) end with 1, 3, 7 or 9, hence k also.
Integer (k-10) must be a divisor of 10^(q+1)+1 = A000533(q+1).
For k = 21, there is 21 * 91 = 1[91]1 but also 21 * 9091 = 1[9091]1; hence, 91 and 9091 are terms of M_21.
Since 10^(q+1)+1 mod (k-10) is periodic and the period length cannot exceed k-10, it is easy to check that the sequence is indeed full. - Giovanni Resta, Nov 26 2019

			23 * 77 = 1[77]1, so k = 23 is a term and 13 * 77 = 1001; remark: number M = 77 has 2 digits and 10^3+1 has 2 zeros also.
29 * 52631579 = 1[52631579]1, so 29 is a term et 19 * 52631579 = 10^9 + 1 = 1000000001.

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

Bernard Schott, Array with values of (q,k,M)

Cf. A000533, A329915 (corresponding numbers M).

Select[Range[21, 110], GCD[10, # - 10] == 1 && MemberQ[Mod[10^Range[#] + 1, # - 10], 0] &] (* Giovanni Resta, Nov 26 2019 *)

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.

A351320 a(n) is the unique integer k such that k * A116436(n) = 1.A116436(n).1 where "." stands for concatenation.

Original entry on oeis.org

111, 101, 87, 23, 21, 83, 21, 21, 27, 101, 87, 29, 23, 21, 33, 21, 83, 21, 39, 101, 87, 23, 21, 21, 21, 83, 101, 87, 59, 23, 21, 99, 57, 21, 27, 21, 101, 87, 29, 23, 21, 83, 69, 21, 71, 21, 101, 87, 33, 23, 21, 21, 83, 21, 101, 87, 23, 21, 27, 21, 39, 21, 83, 101, 87, 29, 23, 21, 21, 107, 21, 101

Offset: 1

Bernard Schott, Feb 07 2022

Except for a(1) = 111, which is unique, all terms appear infinitely many times and belong to this set of fifteen integers: {21, 23, 27, 29, 33, 39, 57, 59, 69, 71, 83, 87, 99, 101, 107}; see A329914.

The corresponding indices where these integers appear the first time are respectively: 5, 4, 9, 12, 15, 19, 33, 29, 43, 45, 6, 3, 32, 2, 70.

			A116436(1) = 1 and 111 * 1 = 1.1.1, hence a(1) = 111.
A116436(2) = 11 and 101 * 11 = 1.11.1, hence a(2) = 101.
A116436(32) = 112359550561797752809 and 99 * 112359550561797752809 = 1.112359550561797752809.1 hence a(32) = 99 (see Penguin reference).

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

Cf. A116436, A329914, A329915.

M such that k*M=1M1 for: A095372 \ {1} (k=21), A331630 (k=23), A351237 (k=83), A351238 (k=87), A351239 (k=101).

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
a(n) = {my(v6=[], i=1); while (#v6 < n, v6 = concat(v6, A116436(i)); i++); my(x= v6[n]); my(k=1); while (eval(Str(1, x, 1)) % x, k++); eval(Str(1, x, 1))/x;} \\ Michel Marcus, Feb 10 2022

a(n) * A116436(n) = 1.A116436(n).1 where "." stands for concatenation.

cp's OEIS Frontend

A095372 1+integers repeating "90" decimal digit pattern.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A329914 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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views