A003020 -id:A003020 - cp's OEIS frontend

A095371 Number of distinct prime factors of record setting repunits (A328899).

0, 1, 2, 5, 7, 8, 10, 13, 14, 20, 21, 22, 26, 29, 32, 33, 34, 35, 40, 44, 55, 56, 63

Offset: 1

Labos Elemer, Jun 04 2004

Conjecture: a(24) = 73, a(25) = 94, a(26) = 99, a(27) >= 107, a(28) >= 127, a(29) >= 136, a(30) >= 140, a(31) >= 151, a(32) >= 159, a(33) >= 163, a(34) >= 178, a(35) >= 184, a(36) >= 213, a(37) >= 214. - Chai Wah Wu, Nov 01 2019

Cf. A067063, A003020, A001221, A002275, A095370, A004023.

r[n_] := (10^n-1)/9; L = {}; bst = -1; Do[v = PrimeNu[r[n]]; If[v > bst, bst = v; AppendTo[L, v]], {n, 60}]; L
(* or, based on the b-file of A095370: *)
w = Last /@ Cases[Import["https://oeis.org/A095370/b095370.txt", "Table"], {Integer, _Integer}]; L = {}; bst = -1; Do[ If[j > bst, AppendTo[L, bst = j]], {j, w}]; L (* _Giovanni Resta, Oct 30 2019 *)

Data corrected by Ray Chandler and N. J. A. Sloane, May 03 2017
Name edited by Giovanni Resta, Oct 30 2019
a(20)-a(21) from Chai Wah Wu, Oct 30 2019
a(22)-a(23) from Chai Wah Wu, Nov 01 2019

A081317 Primes p such that p divides 10^n-1, p is the largest prime producing decimal fraction period n and p is not the largest prime dividing 10^n-1.

Original entry on oeis.org

13, 52579, 8779, 2161, 69857, 909090909090909091, 459691, 549797184491917, 14175966169, 183411838171, 296557347313446299, 388847808493, 3404193829806058997303, 8985695684401, 297262705009139006771611927

Offset: 1

Hugo Pfoertner, Mar 18 2003

			a(1)=13 because the largest factor 37 in the factorization of 10^6-1=999999=3^3*7*11*13*37 already occurs in the factorization of 10^3-1=3^3*37 and produces only a decimal fraction period of 3. 1/37=0.027027027...., 1/13=0.0769230769230...

Max Alekseyev, Table of n, a(n) for n = 1..54 (first 50 terms from Ray Chandler)
Sam Wagstaff, The Cunningham Project, The Main Tables

Cf. A005422, A003020, A061075, A081318.

Numbers in A061075(n) such that A061075(n) is not equal to A005422(n). The corresponding values of n are given in A081318.

a(n) = A061075(A081318(n)). - Max Alekseyev, Apr 27 2022

More terms from Hans Havermann, May 31 2003

A147556 Largest prime factor of prime(n)-th repunit number.

Original entry on oeis.org

11, 37, 271, 4649, 513239, 265371653, 5363222357, 1111111111111111111, 11111111111111111111111, 77843839397, 57336415063790604359, 2212394296770203368013, 201763709900322803748657942361

Offset: 1

Farideh Firoozbakht, Dec 26 2008

nonn

The sequence of repunit primes is a subsequence of this sequence.

			Prime(15)=47 and (10^47-1)/9 = 35121409*316362908763458525001406154038726382279, so a(15)=316362908763458525001406154038726382279.

Table of n, a(n) for n = 1..70
Makoto Kamada, Factorizations of 11...11 (Repunit).

Cf. A000040, A002275, A003020, A004022, A006530, A019328, A147555.

Table[FactorInteger[FromDigits[PadRight[{},n,1]]][[-1,1]],{n,Prime[ Range[15]]}] (* Harvey P. Dale, Feb 23 2016 *)

a(n) = A003020(A000040(n)) = A006530(A002275(A000040(n))) = A006530(A019328(A000040(n))). - Ray Chandler, May 11 2017

Edited by Ray Chandler, Apr 06 2011
terms to a(66) in b-file from Ray Chandler, May 11 2017
a(67)-a(70) in b-file from Max Alekseyev, Apr 26 2022

A177928 Let n be the number whose square n^2 has the decimal expansion { d(1) d(2) ... d(D) }, and let q be the corresponding number whose decimal expansion is { d(2) d(3) ... d(D) d(1)}. Sequence lists numbers n dividing q.

Original entry on oeis.org

1, 2, 3, 9, 27, 33, 66, 99, 123, 246, 271, 333, 351, 407, 429, 462, 481, 518, 546, 567, 666, 693, 702, 715, 777, 814, 819, 924, 936, 999, 1434, 2151, 2868, 3333, 4521, 4818, 6666, 7227, 7373, 7535, 8631, 9042, 9999, 33333, 53658, 54546, 66666, 80487, 81819

Offset: 1

Michel Lagneau, May 15 2010

A178028 is a subsequence of this sequence.

When n divides q, n divides d(D)*(10^D - 1) because q = 10*n^2 - d(D)*(10^D - 1). If n is prime, n divides (10^D - 1); for example, the prime term 271 divides 10^5 - 1 = 99999 = 271*369.

			429 is in the sequence because 429^2 = 184041 and 840411/429 = 1959.

Eric Weisstein's World of Mathematics, Repunits

Cf. A002275, A003020, A005422, A067063, A102380, A178028.

for n from 1 to 10^6 do: d:=convert(n^2, base, 10):n1:=nops(d):s:=sum('d[i]*10^i','i'=1..n1-1)+d[n1]:if irem(s,n)=0 then printf(`%d, `,n):else fi:od:

Select[Range[100000], Mod[FromDigits[RotateLeft[IntegerDigits[#^2]]], #] == 0 &] (* T. D. Noe, Jul 27 2012 *)

A366165 a(n) is the least k > 0 such that 10^(2n-1) - k can be written as a product jm, where j and m have an equal number of decimal digits.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 3, 1, 5, 3, 1, 6, 1, 7, 1, 2, 2, 1, 4, 7, 5, 1, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 10, 4, 3, 3, 10, 1, 2, 3, 1, 1, 1, 7, 1, 1

Offset: 1

Hugo Pfoertner, Oct 04 2023

a(n) <= 10 since 10^(2n-1)-10 = (10^(n-1)+1)(10^n-10). A consequence is that j and m in the product both have n decimal digits. - Chai Wah Wu, Oct 05 2023

			n a(n) 10^(2n-1)-a(n)       j       m
1  1   9                    1       9
2  1   999                 27      37
3  1   99999              123     813
4  1   9999999           2151    4649
5 10   999999990        10001   99990
6  1   99999999999     194841  513239
7  3   9999999999997  2769823 3610339
More than one pair (j,m) may exist, e.g., 9 = 1*9 = 3*3.

Cf. A002275, A003020, A004022, A057951, A327435.

A067272 are the solutions for even exponents of 10, corresponding to (j,m) = (9,9), (99,99), (999,999), ... .

a366165(n)={my (p10=10^(2*n-1)); for (dd=1, p10, my (d=p10-dd); fordiv (d, x, fordiv (d, y, if (x*y==d && #digits(x)==#digits(y), return(dd)))))};

from itertools import count, takewhile
from sympy import divisors
def A366165(n):
    a, l1, l2 = 10**((n<<1)-1), 10**(n-1), 10**n
    for k in count(1):
        b = a-k
        if any(l1<=db for d in takewhile(lambda m:m*m<=b, divisors(b))):
            return k # Chai Wah Wu, Oct 05 2023

a(33)-a(35) from Chai Wah Wu, Oct 05 2023

a(36)-a(46) from Chai Wah Wu, Oct 07 2023

A366921 a(n) is the least prime factor > 3 of 10^n - 1.

Original entry on oeis.org

11, 37, 11, 41, 7, 239, 11, 37, 11, 21649, 7, 53, 11, 31, 11, 2071723, 7, 1111111111111111111, 11, 37, 11, 11111111111111111111111, 7, 41, 11, 37, 11, 3191, 7, 2791, 11, 37, 11, 41, 7, 2028119, 11, 37, 11, 83, 7, 173, 11, 31, 11, 35121409, 7, 239, 11, 37, 11, 107

Offset: 2

Hugo Pfoertner, Oct 28 2023

nonn

			a(2) = 11 because 99 = 3^2 * 11 = 3^A366922(2) * A003020(2).
a(9) = 37 because 10^9 - 1 = 3^4 * 37 * 333667 = 3^A366922(9) * 37 * A003020(9).

Cf. A002283, A003020, A102347, A366922.

PARI
```
a366921(n) = factor(10^n-1)[2,1]
```

A366922 a(n) is the exponent of 3 in the prime factorization of 10^n - 1.

Original entry on oeis.org

2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 5, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 5, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 6, 2, 2, 3, 2, 2, 3, 2, 2, 4

Offset: 1

Hugo Pfoertner, Oct 28 2023

1

Cf. A002283, A003020, A007949, A051064, A057951, A102347, A366921.

a[n_]:=IntegerExponent[10^n-1,3]; Array[a,90] (* Stefano Spezia, Oct 28 2023 *)

PARI
```
a366922(n) = valuation(10^n-1,3)
```

def A366922(n):
    c = 0
    a, b = divmod(10**n-1, 3)
    while b == 0:
        a, b = divmod(a, 3)
        c += 1
    return c # Chai Wah Wu, Oct 29 2023

a(n) = A007949(10^n - 1).

a(n) = A007949(n) + 2 = A051064(n) + 1.

A269503 Largest prime factor of A138148(n).

Original entry on oeis.org

101, 13, 137, 9091, 9901, 909091, 5882353, 52579, 333667, 9091, 99990001, 1058313049, 265371653, 909091, 2906161, 21993833369, 999999000001, 909090909090909091, 1111111111111111111, 909091, 1056689261, 549797184491917, 11111111111111111111111

Offset: 1

Altug Alkan, May 11 2016

nonn

Largest prime factor of (10^(n+1)+1)*(10^n-1)/9.

			a(4) = 9091 because largest prime factor of 111101111 is 9091.

Max Alekseyev, Table of n, a(n) for n = 1..344

Cf. A003020, A003021, A006530, A138148.

seq(max(max(numtheory:-factorset((10^n-1)/9)),
max(numtheory:-factorset(10^(n+1)+1))), n=1..30); # Robert Israel, May 11 2016

a006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
for(n=1, 25, print1(a006530((10^(2*n+1)-1)/9-10^n), ", "));

a(n) = max(A003020(n),A003021(n+1)) for n >= 2. - Robert Israel, May 11 2016

A385579 Smallest prime factor that the repunit(n) = (10^n-1)/9 shares with at least one other binary vector of the same length in base 10, or 1 if they are coprime.

Original entry on oeis.org

1, 1, 1, 1, 11, 1, 3, 1, 11, 3, 11, 1, 3, 53, 11, 3, 11, 1, 3, 1, 11, 3, 11, 1, 3, 41, 11, 3, 11, 3191, 3, 2791, 11, 3, 11, 41, 3, 2028119, 11, 3, 11, 83, 3, 173, 11, 3, 11, 35121409, 3, 239, 11, 3, 11, 107, 3, 41, 11, 3, 11

Offset: 0

Dmytro Inosov, Jul 03 2025

a(n) is the smallest prime factor that divides both the decimal repunit (10^n-1)/9 and at least one other smaller decimal number consisting of only 0's and 1's.

a(n)=1 iff n is a term in A385537 (indices of repunits coprime with all other binary vectors of the same length).

			a(3) = 1 because 111 = 3*37 is coprime with all other nonzero binary vectors of length 3, which are 001, 010, 011, 100, 101, 110. None of them is divisible by 3 or 37.
a(4) = 11 because 11 is the smallest prime factor of 1111 which it shares, for example, with the binary vector 0011.

Cf. A067063, A003020, A378761, A385537, A385539.

F[d_] := Min[Select[Table[Min[Transpose[FactorInteger[GCD[FromDigits[IntegerDigits[i,2]],(10^d-1)/9]]][[1]]], {i, 1, 2^d-2}], # > 1 &]];
Table[If[# < \[Infinity], #, 1] &[F[n]], {n, 0, 25}]

a(n) = my(x=(10^n-1)/9, m=oo, b=0, z); for (i=1, 2^n-2, my(y=fromdigits(binary(i))); if ((z=gcd(y, x)) != 1, b=1; m = min(m, vecmin(factor(z)[,1])); ); ); if (b, m, 1); \\ Michel Marcus, Jul 03 2025

If a(n) > 1, A067063(n) <= a(n) <= A003020(n).

A119761 Values n such that the largest prime factor of repunit R_n=(10^n -1)/9 is a unique prime,i.e.,whose reciprocal has unique decimal period length.

Original entry on oeis.org

2, 3, 4, 6, 9, 10, 12, 14, 18, 19, 23, 24, 36, 38, 39, 46, 48, 62, 78, 93, 96, 106, 120, 134, 150, 186, 196, 240, 268, 294, 300, 317, 320

Offset: 1

Lekraj Beedassy, Jun 18 2006

			18 is in the sequence because R_18=3^2*7*11*13*19*37*52579*333667 and 333667 is the only prime with period 9: 1/333667 =0.000002997000002997...

M. Kamada, Factorization of 11...11(Repunits)

Cf. A002275, A003020, A007498, A040017.

Corrected and extended by Hans Havermann, Jun 22 2006

a(25)-a(33) from Ray Chandler, May 09 2017

cp's OEIS Frontend

A095371 Number of distinct prime factors of record setting repunits (A328899).

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A081317 Primes p such that p divides 10^n-1, p is the largest prime producing decimal fraction period n and p is not the largest prime dividing 10^n-1.

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A147556 Largest prime factor of prime(n)-th repunit number.

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A177928 Let n be the number whose square n^2 has the decimal expansion { d(1) d(2) ... d(D) }, and let q be the corresponding number whose decimal expansion is { d(2) d(3) ... d(D) d(1)}. Sequence lists numbers n dividing q.

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A366165 a(n) is the least k > 0 such that 10^(2*n-1) - k can be written as a product j*m, where j and m have an equal number of decimal digits.

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A366921 a(n) is the least prime factor > 3 of 10^n - 1.

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A366922 a(n) is the exponent of 3 in the prime factorization of 10^n - 1.

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A269503 Largest prime factor of A138148(n).

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A366165 a(n) is the least k > 0 such that 10^(2n-1) - k can be written as a product jm, where j and m have an equal number of decimal digits.