A147553 -id:A147553 - cp's OEIS frontend

A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

1, 3, 13, 21, 37, 39, 91, 1443, 3367, 9901, 157737, 333667, 999001, 3075403, 9226209, 14287143, 33336667, 99990001, 1171182883, 1224848037, 1286294191, 1397863441, 1428557143, 1469179621, 1535254357, 1568996211, 1753536967, 1792076241, 1839599913, 1891910811

Offset: 1

Hans Havermann, May 31 2014

Number of d-digit solutions for d = 1..100: 2, 5, 0, 3, 0, 3, 2, 3, 0, 39, 0, 2, 0, 106, 0, 3, 3, 2, 0, 441, 4, 14, 0, 5, 0, 15, 2, 283, 0, 23, 0, 61, 0, 24, 21, 4, 0, 22, 0, 240, 0, 34, 0, 96, 3, 30, 0, 6, 16, 281, 0, 216, 0, 22, 5, 3894, 2, 10, 0, 149, 2, 11, 0, 407, 0, 25, 0, 2136, 0, 53983, 0, 12, 1, 29, 11, 1872, 99, 20, 0, 6984, 0, 45, 0, 279, 32, 10, 5, 15928, 0, 213, 24, 791, 0, 20, 14, 44, 0, 713, 12, 89804.

Numbers n such that n divides 100^d+10^d+1, where 10^(d-1)<=n<10^d. - Robert Israel, Jan 11 2017

			21^2 divides 212121; 91^2 divides 919191; so both 21 and 91 are in the sequence.

Hans Havermann, Table of n, a(n) for n = 1..1366

Cf. A147553 (n^2 divides n.n), A147554 (primes in this sequence).

Contains A074992 and A168624.

Res:= {}:
for d from 1 to 15 do
  Res:= Res union select(t -> t >= 10^(d-1) and t < 10^d,
   numtheory:-divisors(100^d+10^d+1))
od:
sort(convert(Res,list)); # Robert Israel, Jan 11 2017

Do[d=Divisors[100^i+10^i+1];s=Select[d,Length[IntegerDigits[#]]==i&];If[Length[s]>0,Do[Print[s[[j]]],{j,Length[s]}]],{i,42}]

A147554 Primes p such that p^2 divides p.p.p where dot "." means concatenation.

Original entry on oeis.org

3, 13, 37, 9901, 333667, 99990001, 999999000001, 9999999900000001, 13168164561429877, 130654897808007778425046117

Offset: 1

Farideh Firoozbakht, Dec 26 2008

Primes p dividing 10^(2*d)+10^d+1 where d=ceiling(log(p)/log(10)) is the number of decimal digits in p. - Max Alekseyev
There is no prime p such that p^2 divides p.p.
All primes of the forms 10^(2m) - 10^m + 1 or (1/3)*(10^(2m) + 10^m + 1) are in the sequence.
Primes in A243162. - Hans Havermann, May 31 2014
a(11) > 10^158. - Max Alekseyev, Sep 11 2024

Cf. A147553, A243162.

A347541 Numbers q.r such that q*r divides q.r, when q and r have the same number of digits, "." means concatenation, and r may not begin with 0.

Original entry on oeis.org

11, 12, 15, 24, 36, 1352, 1734, 143143, 167334, 16673334, 1666733334, 166667333334, 16666673333334, 1666666733333334, 142857143142857143, 166666667333333334, 16666666673333333334, 1666666666733333333334, 166666666667333333333334, 16666666666673333333333334, 1666666666666733333333333334, 142857142857143142857142857143

Offset: 1

Bernard Schott, Oct 11 2021

Problem proposed on French site Diophante (see link).
We have to solve Diophantine equation q.r = q*10^m + r = k * q * r where m = length(q) = length(r). Some results:
k can only take values 2, 3, 6, 7, 11, and ratio r/q = 1, 2, 4 or 5.
There are exactly 3 subsequences of terms that are solutions, one finite and two infinites:
-> Finite subsequence: 11, 12, 15, 36, 1352.
-> Infinite subsequence with k = 7 and r = q = (10^(6h-3)+1)/7, h>=1 (A147553 \ {1}), hence terms are (10^(6h-3)+1)^2/7 for h>=1: {143143, 142857143142857143, ... }.
-> Infinite subsequence with k = 3 and r = 2q, with q = (10^h+2)/6, r = (10^h+2)/3 for h>= 1: {24, 1734, 167334, 16673334, ...} (A348589).
Consequence: integers q.q that are divisible by q*q are exactly integers such that q is a term of A147553. If q = A147553(1) = 1, then 11/(1*1) = 11, while for q = A147553(n), n>=2, then q.q / (q*q) = 7.
Note that the first five terms are the 2-digit Zuckerman numbers (A007602).

			One example for each possible value of k = q.r / (q*r).
a(1) = 11 and 11/(1*1) = 11.
a(2) = 12 and 12/(1*2) = 6.
a(5) = 36 and 36/(3*6) = 2.
a(7) = 1734 and 1734/(17*34) = 3.
a(8) = 143143 and 143143/(143*143) = 7.

Diophante, A1945 - Concaténations en tous genres (in French).
Giovanni Resta, Zuckerman numbers, Numbers Aplenty.

Cf. A007602, A147553, A348589.

A348589 a(n) = (10^n+2)^2 / 6.

Original entry on oeis.org

24, 1734, 167334, 16673334, 1666733334, 166667333334, 16666673333334, 1666666733333334, 166666667333333334, 16666666673333333334, 1666666666733333333334, 166666666667333333333334, 16666666666673333333333334, 1666666666666733333333333334

Offset: 1

Bernard Schott, Oct 24 2021

Numbers q.r such that q.r = 3*q*r, when q and r have the same number of digits, "." means concatenation, r = 2q and r may not begin with 0.
We must solve the Diophantine equation q.r = q*10^m+r = 3 * q*r where m = length(q) = length(r).
The number of solutions is infinite with (r, q) = ((10^n+2)/3, (10^n+2)/6) and n >= 1.
Note that 15 satisfies also q.r = 3*q*r, 15 = 3*1*5 with here r = 5*q.
For further information about the general equation q.r = k * q*r, see A347541.
Problem proposed on the French website Diophante (see link).

			a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.
a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.

Diophante, A1945 - Concaténations en tous genres (in French).
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A102807, A133384, A147553.

Subsequence of A347541.

Maple
```
seq((10^n+2)^2 / 6, n=1..14);
```

Table[(10^n + 2)^2/6, {n, 1, 14}] (* Amiram Eldar, Oct 24 2021 *)
LinearRecurrence[{111,-1110,1000},{24,1734,167334},20] (* Harvey P. Dale, Sep 05 2025 *)

def a(n): return (10**n+2)**2//6
print([a(n) for n in range(1, 15)]) # Michael S. Branicky, Oct 24 2021

a(n) = (10^n+2)^2 / 6.
a(n) = A133384(n-1)^2/6.
G.f.: 6*x*(4-155*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)). - Stefano Spezia, Oct 25 2021
a(n) = 3*A102807(n)/2. - Hugo Pfoertner, Oct 30 2021

cp's OEIS Frontend

A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A147554 Primes p such that p^2 divides p.p.p where dot "." means concatenation.

Views

Author

Keywords

Comments

Crossrefs

A347541 Numbers q.r such that q*r divides q.r, when q and r have the same number of digits, "." means concatenation, and r may not begin with 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A348589 a(n) = (10^n+2)^2 / 6.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula