A086982 -id:A086982 - cp's OEIS frontend

A282344 The smallest square referenced in A086982 (Numbers n such that 10^n+1 is not squarefree).

121, 49, 121, 169, 121, 49, 121, 121, 49, 169, 121, 289, 121, 49, 121, 361, 121, 49, 169, 10201, 121, 49, 237169, 121, 49, 121, 5329, 121, 49, 121, 121, 169, 49, 121, 121, 49, 841, 121, 289, 121, 49, 121, 121, 49, 121, 169, 361, 121, 49, 121, 18769, 121, 49

Offset: 1

Robert Price, Feb 12 2017

nonn

Cf. A086982, A282345.

a(n) = A008833( 1+10^A086982(n) ). - R. J. Mathar, Feb 13 2017

A282345 Square root of the smallest square referenced in A086982 (Numbers n such that 10^n+1 is not squarefree).

Original entry on oeis.org

11, 7, 11, 13, 11, 7, 11, 11, 7, 13, 11, 17, 11, 7, 11, 19, 11, 7, 13, 101, 11, 7, 487, 11, 7, 11, 73, 11, 7, 11, 11, 13, 7, 11, 11, 7, 29, 11, 17, 11, 7, 11, 11, 7, 11, 13, 19, 11, 7, 11, 137, 11, 7

Offset: 1

Robert Price, Feb 12 2017

nonn

Cf. A086982, A282344.

a(n) = sqrt(A282344(n)). - R. J. Mathar, Feb 13 2017

A102567 Numbers k such that the concatenation of k with itself is a biperiod square.

Original entry on oeis.org

13223140496, 20661157025, 29752066116, 40495867769, 52892561984, 66942148761, 82644628100, 183673469387755102041, 326530612244897959184, 510204081632653061225, 734693877551020408164

Offset: 1

C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 15 2005

Also, numbers N associated with A106497.
Also, numbers k such that k concatenated with k-1 gives the product of two numbers which differ by 2. E.g., 13223140496//13223140495 = 36363636363 * 36363636365, where // denotes concatenation. - Giovanni Resta and Franklin T. Adams-Watters, Nov 13 2006
From Jianing Song, Nov 01 2024: (Start)
Numbers 10^(k-1) <= a <= 10^k - 1 such that a*(10^k + 1) is a square. Note that 10^k + 1 must be nonsquarefree, i.e., k is in A086982, otherwise a must be divisible by 10^k + 1, which is impossible.
Let v(p,m) be the p-adic valuation of m.
 - If p is not in A045616, then v(p,10^k+1) = r > 0 if and only if v(p,gcd(n,10^k+1)) = r-1.
 - If p is in A045616, let e be the multiplicative order of 10 modulo p, then v(p,10^k+1) > 0 if and only if e is even and k is an odd multiple of e/2, in which case v(p,10^k+1) = v(p,10^e-1) + v(p,k) = v(p,10^e-1) + v(p,gcd(k,10^k+1)).
This helps to find the terms. (End)

			13223140496 concatenated with 13223140496 is 1322314049613223140496 = 36363636364^2.
40495867769 is in the sequence because writing it twice gives the square number 4049586776940495867769 = 63636363637^2.

Andrew Bridy, Robert J. Lemke Oliver, Arlo Shallit, and Jeffrey Shallit, The Generalized Nagell-Ljunggren Problem: Powers with Repetitive Representations, Experimental Math, 28 (2019), 428-439.
R. Ondrejka, Problem 1130: Biperiod Squares, Journal of Recreational Mathematics, Vol. 14:4 (1981-82), 299. Solution by F. H. Kierstead, Jr., JRM, Vol. 15:4 (1982-83), 311-312.

David W. Wilson, Table of n, a(n) for n = 1..1098
Dr Barker, Can Numbers Like These Be Square?, YouTube video, 2023.
Andrew Bridy, Robert J. Lemke Oliver, Arlo Shallit, and Jeffrey Shallit, The Generalized Nagell-Ljunggren Problem: Powers with Repetitive Representations, preprint arXiv:1707.03894 [math.NT], July 14 2017.

Cf. A092118, A054214, A116163, A116136, A116279.

with(numtheory): Digits:=50:for d from 1 to 35 do tendp1:=10^d+1: tendp1fact:=ifactors(tendp1)[2]: n:=mul(piecewise(tendp1fact[i][2] mod 2=1,tendp1fact[i][1],1),i=1..nops(tendp1fact)):for i from ceil(sqrt((10^(d-1))/n)) to floor(sqrt((10^d-1)/n)) do printf("%d, ",n*i^2) od: od:

A102567L[n_] := Catenate@Table[Module[{fac = FactorInteger[10^k + 1], min}, If[Max@fac[[All, -1]] == 1, {}, min = Times @@ Cases[fac, {a_, A102567L%5B30%5D%20(*%20_JungHwan%20Min">?OddQ} :> a]; Table[min s^2, {s, Ceiling@Sqrt[10^(k - 1)/min], Floor@Sqrt[(10^k - 1)/min]}]]], {k, n}]; A102567L[30] (* _JungHwan Min, Dec 11 2016 *)
A102567Q = IntegerQ@Sqrt@FromDigits[Join[#, #] &@IntegerDigits[#]] & (* JungHwan Min, Dec 11 2016 *)

p = [3, 487, 56598313]; \\ A045616
b(n) = my(d = gcd(n, lift(Mod(10,n)^n)+1), s = 1); for(j=1, #p, my(e = znorder(Mod(10, p[j]))); if((e % 2 == 0) && (n % (e/2) == 0) && (n/(e/2) % 2 == 1), my(v = valuation(d, p[j])); d /= p[j]^v; s *= p[j]^((v+valuation(10^e-1, p[j]))\2))); my(f = factor(d)); for(i=1, #f~, s *= f[i,1]^((f[i,2]+1)\2)); s; \\ giving s such that 10^n + 1 = s^2*t where t is squarefree, considering only the three already-known terms of A045616
A102567_length_n(n) = my(t = (10^n+1)/b(n)^2, lowlim = 1+sqrtint(10^(n-1)\t), uplim = sqrtint((10^n-1)\t)); vector(uplim-lowlim+1, i, (lowlim-1+i)^2 * t) \\ terms of the form a^2*t such that 10^(n-1) <= a^2*t <= 10^n - 1
\\ Jianing Song, Nov 01 2024

from itertools import count, islice
from sympy import sqrt_mod
def A102567_gen(): # generator of terms
    for j in count(0):
        b = 10**j
        a = b*10+1
        for k in sorted(sqrt_mod(0,a,all_roots=True)):
            if a*b <= k**2 < a*(a-1):
                yield k**2//a
A102567_list = list(islice(A102567_gen(),10)) # Chai Wah Wu, Feb 19 2024

Entry revised by N. J. A. Sloane, Nov 14 2006 and also Nov 27 2006

Definition edited and reference added by William Rex Marshall, Nov 12 2010

A049096 Numbers k such that 2^k + 1 is divisible by a square > 1.

Original entry on oeis.org

3, 9, 10, 15, 21, 27, 30, 33, 39, 45, 50, 51, 55, 57, 63, 68, 69, 70, 75, 78, 81, 87, 90, 93, 99, 105, 110, 111, 117, 123, 129, 130, 135, 141, 147, 150, 153, 159, 165, 170, 171, 177, 182, 183, 189, 190, 195, 201, 204, 207, 210, 213, 219, 225, 230, 231, 234, 237, 243

Offset: 1

Labos Elemer

nonn

Conjecture: lim n -> infinity a(n)/n = C exists and 4 < C < 9/2. There seems to be a sequence of primes p such that p^2 never divides numbers of the form 2^x + 1: the first few are 2, 7, 23, 31. - Benoit Cloitre, Aug 20 2002
That sequence is A072936. - Robert Israel, Nov 20 2015
The first case where 2^n + 1 is divisible by a square that is coprime to n is n = 182 (where 2^182 + 1 is divisible by 1093^2). - Robert Israel, Jul 07 2014
From Robert Israel, Nov 20 2015: (Start)
Numbers n such that gcd(n, 2^n + 1) > 1 or n = k m where k is odd and 2 m is the order of 2 modulo a Wieferich prime.  See link "When p^2 divides 2^n + 1".
If n is in the sequence, then so is k*n for any odd k. (End)
The sequence consists of all odd multiples of { 3, 10, 55, 68, 78, 182, 301, 406, 666, ... }. - M. F. Hasler, Mar 06 2018

			9 is here because 2^9 + 1 = 513 is divisible by 9.
99 is here because 2^99 + 1 = 3^3*19*67*683*5347*20857*242099935645987 is divisible by 9, i.e. is not squarefree.

Robert Israel, Table of n, a(n) for n = 1..10000
Robert Israel, When p^2 divides 2^n + 1

Cf. A001220, A049093, A049094, A049095, A072936, A282269, A282270.

Cf. A086982, which is just the same with base b = 10 instead of b = 2.

[n: n in [3..220] | not IsSquarefree(2^n+1)]; // Vincenzo Librandi, Mar 08 2018

remove(n -> numtheory:-issqrfree(2^n+1), [$1..250]); # Robert Israel, Jul 07 2014

Select[Range[243], !SquareFreeQ[2^# + 1] &] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011*)

is(n)=!issquarefree(2^n+1) \\ Altug Alkan, Nov 20 2015

For any a(n+1) - a(n) <= 6 since numbers of form 3^a*(2k+1) a > 0, k >= 0, are in the sequence (2^(3*(2k+1) + 1 is divisible by 9). So are numbers of the form 20k + 10 since 2^(20k+10) + 1 is divisible by 25, 110k + 55 since 2^(110k+55) + 1 is divisible by 11^2, 78 + 156k since 2^(156k+78) + 1 is divisible by 13^2 ... - Benoit Cloitre, Aug 20 2002

More terms from James Sellers, Dec 16 1999
More terms from Vladeta Jovovic, Apr 12 2002
Missing term 182 added by Rainer Rosenthal, Nov 01 2005

A086981 a(n) = smallest k where (10^k+1)=0 mod prime(n)^2, or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 21, 11, 39, 136, 171, 253, 406, 0, 0, 0, 0, 1081, 0, 1711, 1830, 0, 0, 292, 0, 0, 1958, 4656, 202, 1751, 0, 5886, 6328, 2667, 8515, 548, 3197, 11026, 0, 6123, 0, 13861, 0, 15931, 16290, 0, 18528, 9653, 0, 3165, 24753, 0, 26106, 27028, 0, 3615, 6275

Offset: 1

Ray Chandler, Jul 27 2003

nonn

For a given a(n)>0, all the values of k such that (10^k+1)=0 mod prime(n)^2 are given by the sequence a(n)*A005408, i.e. odd multiples of a(n). For example, for n=5, prime(5)=11, a(n)=11, the set of values of k for which (10^k+1)=0 mod 11^2 is 11*A005408=11,33,55,77,99,... All the terms of the sequence a(n) are integer multiples of prime(n) for primes <1000 except for a(93)=243 where prime(93)=487.

			a(4)=21 since 21 is least value of k for which (10^k+1)=0 mod 7^2.

Charles R. Greathouse IV, Sep 10, 2008, Table of n, a(n) for n = 1..4000

Cf. A000040, A086982.

A061051 Smallest square of the form [n digits][same n digits][further digits].

Original entry on oeis.org

0, 225, 40401, 2042041, 317231721, 16198161984, 3921203921209, 400000040000001, 23391004233910041, 1100298301100298304, 141162631214116263129, 1322314049613223140496, 3171326702963171326702969, 107786983188610778698318864, 29726516052320297265160523209, 1003781781031081003781781031081

Offset: 0

Erich Friedman, May 26 2001

a(n) <= (2*10^n+1)^2. This bound is tight for n = 2, 7. Are there other values of n for which this bound is tight? For n = 11, there are no [further digits] block, i.e. the smallest square has 2n digits. This is true for all n in A086982. For instance, a(21) = 183673469387755102041183673469387755102041, a(33) = 132231404958677685950413223140496132231404958677685950413223140496. - Chai Wah Wu, Mar 25 2020

			40401 is the first square to have the first two digits the same as the next two digits

Chai Wah Wu, Table of n, a(n) for n = 0..37

Cf. A086982.

from sympy import integer_nthroot
def A061051(n):
    if n == 0:
        return 0
    nstart = 10**(n-1)
    nend = 10*nstart
    for i in range(nstart,nend):
        k = int(str(i)*2)
        if integer_nthroot(k,2)[1]:
            return k
    for i in range(nstart,nend):
        si = str(i)*2
        for sj in '014569':
            k = int(si+sj)
            if integer_nthroot(k,2)[1]:
                return k # Chai Wah Wu, Mar 25 2020

One more term from Vladeta Jovovic, Jun 02 2001

a(7)-a(15) from Chai Wah Wu, Mar 25 2020

A074109 Smallest n-digit squarefree number.

Original entry on oeis.org

2, 10, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000002, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001, 100000000000000001

Offset: 1

Amarnath Murthy, Aug 26 2002

For n>=3, a(n) = 10^(n-1)+1 unless n-1 is in A086982. - Robert Israel, Mar 23 2020

Cf. A074110, A086982.

Join[{2}, Table[x = 10^n; k = 0; While[! SquareFreeQ[y = x + k], k++]; y, {n, 19}]] (* Jayanta Basu, Aug 01 2013 *)

More terms from Matthew Conroy, Aug 27 2002

A077712 a(1) = 1, a(n) = the smallest squarefree number > a(n-1) which contains all the digits of a(n-1).

Original entry on oeis.org

1, 10, 101, 110, 1001, 1010, 10001, 10010, 100001, 100010, 1000001, 1000010, 10000001, 10000010, 100000001, 100000010, 1000000001, 1000000010, 10000000001, 10000000010, 100000000010, 1000000000001, 10000000000001

Offset: 1

Amarnath Murthy, Nov 19 2002

Conjecture: Terms contain only two types of digits, i.e., 0 and 1.
Beginning with a(3), sequence follows a regular pattern: 10^2+1, 10^2+10, 10^3+1, 10^3+10, etc. until at a(21) the pattern is disrupted by 10^11+1, which is not squarefree (see A086982). 10^12+10 is also absent from the sequence since it is also not squarefree. The pattern resumes after this disruption until the next occurrence of 10^k+1 which is not squarefree, k=21, 33, 39, 55, ... The conjecture that the sequence is composed of terms containing only the digits 0 and 1 is certainly true up to 10^406+1 where both it and 10^407+1 are not squarefree. Indeed beginning with a(3) the terms contain exactly two 1 digits and the rest 0's up to this point. The term following 10^406+10 will introduce a third nonzero digit, perhaps a 1, but the pattern of the sequence changes dramatically at this point. - Ray Chandler, Aug 02 2003
Term following a(739)=10^406+10 is a(740)=10^407+11 so the conjecture is still in play. - Ray Chandler, Aug 05 2003

Cf. A086981, A086982.

Subsequence of A005117.

More terms from Ray Chandler, Aug 02 2003

Offset corrected by Mohammed Yaseen, Aug 16 2023

cp's OEIS Frontend

A282344 The smallest square referenced in A086982 (Numbers n such that 10^n+1 is not squarefree).

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A282345 Square root of the smallest square referenced in A086982 (Numbers n such that 10^n+1 is not squarefree).

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A102567 Numbers k such that the concatenation of k with itself is a biperiod square.

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A049096 Numbers k such that 2^k + 1 is divisible by a square > 1.

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A086981 a(n) = smallest k where (10^k+1)=0 mod prime(n)^2, or 0 if no such k exists.

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A061051 Smallest square of the form [n digits][same n digits][further digits].

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A074109 Smallest n-digit squarefree number.

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A077712 a(1) = 1, a(n) = the smallest squarefree number > a(n-1) which contains all the digits of a(n-1).

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