A088323 -id:A088323 - cp's OEIS frontend

A085104 Primes of the form 1 + n + n^2 + n^3 + ... + n^k, n > 1, k > 1.

7, 13, 31, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801, 2971, 3307, 3541, 3907, 4423, 4831, 5113, 5701, 6007, 6163, 6481, 8011, 8191, 9901, 10303, 11131, 12211, 12433, 13807, 14281, 17293, 19183, 19531, 20023

Offset: 1

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jul 03 2003

Primes that are base-b repunits with three or more digits for at least one b >= 2: Primes in A053696. Subsequence of A000668 U A076481 U A086122 U A165210 U A102170 U A004022 U ... (for each possible b). - Rick L. Shepherd, Sep 07 2009
From Bernard Schott, Dec 18 2012: (Start)
Also known as Brazilian primes. The primes that are not Brazilian primes are in A220627.
The number of terms k+1 is always an odd prime, but this is not enough to guarantee a prime, for example 111 = 1 + 10 + 100 = 3*37.
The inverses of the Brazilian primes form a convergent series; the sum is slightly larger than 0.33 (see Theorem 4 of Quadrature article in the Links). (End)
It is not known whether there are infinitely many Brazilian primes. See A002383. - Bernard Schott, Jan 11 2013
Primes of the form (n^p - 1)/(n - 1), where p is odd prime and n > 1. - Thomas Ordowski, Apr 25 2013
Number of terms less than 10^n: 1, 5, 14, 34, 83, 205, 542, 1445, 3880, 10831, 30699, 88285, ..., . - Robert G. Wilson v, Mar 31 2014
From Bernard Schott, Apr 08 2017: (Start)
Brazilian primes fall into two classes:
1) when n is prime, we get sequence A023195 except 3 which is not Brazilian,
2) when n is composite, we get sequence A285017. (End)
The conjecture proposed in Quadrature "No Sophie Germain prime is Brazilian (prime)" (see link Bernard Schott, Quadrature, Conjecture 1, page 36) is false. Thanks to Giovanni Resta, who found that a(856) = 28792661 = 1 + 73 + 73^2 + 73^3 + 73^4 = (11111)73 is the 141385th Sophie Germain prime. - _Bernard Schott, Mar 08 2019

			13 is a term since it is prime and 13 = 1 + 3 + 3^2 = 111_3.
31 is a term since it is prime and 31 = 1 + 2 + 2^2 + 2^3 + 2^4 = 11111_2.
From _Hartmut F. W. Hoft_, May 08 2017: (Start)
The sequence represented as a sparse matrix with the k-th column indexed by A006093(k+1), primes minus 1, and row n by A000027(n+1). Traversing the matrix by counterdiagonals produces a non-monotone ordering.
    2    4      6        10             12          16
2  7    31     127      -              8191        131071
3  13   -      1093     -              797161      -
4  -    -      -        -              -           -
5  31   -      19531    12207031       305175781   -
6  43   -      55987    -              -           -
7  -    2801   -        -              16148168401 -
8  73   -      -        -              -           -
9  -    -      -        -              -           -
10  -    -      -        -              -           -
11  -    -      -        -              -           50544702849929377
12  157  22621  -        -              -           -
13  -    30941  5229043  -              -           -
14  211  -      8108731  -              -           -
15  241  -      -        -              -           -
16 -    -      -        -              -           -
17  307  88741  25646167 2141993519227  -           -
18  -    -      -        -              -           -
19  -    -      -        -              -           -
20  421  -      -        10778947368421 -           689852631578947368421
21  463  -      -        17513875027111 -           1502097124754084594737
22  -    245411 -        -              -           -
23  -    292561 -        -              -           -
24  601  346201 -        -              -           -
Except for the initial values in the respective sequences the rows and columns as labeled in the matrix are:
column  2:  A002383            row 2:  A000668
column  4:  A088548            row 3:  A076481
column  6:  A088550            row 4:  -
column 10:  A162861            row 5:  A086122.
(End)

Daniel Lignon, Dictionnaire de (presque) tous les nombres entiers, Ellipses, Paris, 2012, page 174.

Jon E. Schoenfield, Table of n, a(n) for n = 1..10831 (terms up to 10^10; terms 1..3880 from T. D. Noe)
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

Cf. A189891 (complement), A125134 (Brazilian numbers), A220627 (Primes that are non-Brazilian).
Cf. A003424 (n restricted to prime powers).
Cf. A053696, A086930, A059055.
Equals A023195 \3 Union A285017 with empty intersection.
Primes of the form (b^k-1)/(b-1) for b=2: A000668, b=3: A076481, b=5: A086122, b=6: A165210, b=7: A102170, b=10: A004022.
Primes of the form (b^k-1)/(b-1) for k=3: A002383, k=5: A088548, k=7: A088550, k=11: A162861.

a085104 n = a085104_list !! (n-1)
a085104_list = filter ((> 1) . a088323) a000040_list
-- Reinhard Zumkeller, Jan 22 2014

max = 140; maxdata = (1 - max^3)/(1 - max); a = {}; Do[i = 1; While[i = i + 2; cc = (1 - m^i)/(1 - m); cc <= maxdata, If[PrimeQ[cc], a = Append[a, cc]]], {m, 2, max}]; Union[a] (* Lei Zhou, Feb 08 2012 *)
f[n_] := Block[{i = 1, d, p = Prime@ n}, d = Rest@ Divisors[p - 1]; While[ id = IntegerDigits[p, d[[i]]]; id != Reverse@ id || Union@ id != {1}, i++]; d[[i]]]; Select[ Range[2, 60], 1 + f@# != Prime@# &] (* Robert G. Wilson v, Mar 31 2014 *)

list(lim)=my(v=List(),t,k);for(n=2,sqrt(lim), t=1+n;k=1; while((t+=n^k++)<=lim,if(isprime(t), listput(v,t))));vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Jan 08 2013

A085104_vec(N,L=List())=forprime(K=3,logint(N+1,2),for(n=2,sqrtnint(N-1,K-1),isprime((n^K-1)\(n-1))&&listput(L,(n^K-1)\(n-1))));Set(L) \\ M. F. Hasler, Jun 26 2018

A010051(a(n)) * A088323(a(n)) > 1. - Reinhard Zumkeller, Jan 22 2014

More terms from David Wasserman, Jan 26 2005

A053696 Numbers that can be represented as a string of three or more 1's in a base >= 2.

Original entry on oeis.org

7, 13, 15, 21, 31, 40, 43, 57, 63, 73, 85, 91, 111, 121, 127, 133, 156, 157, 183, 211, 241, 255, 259, 273, 307, 341, 343, 364, 381, 400, 421, 463, 507, 511, 553, 585, 601, 651, 703, 757, 781, 813, 820, 871, 931, 993, 1023, 1057, 1093, 1111, 1123, 1191

Offset: 1

Henry Bottomley, Mar 23 2000

Numbers of the form (b^n-1)/(b-1) for n > 2 and b > 1. - T. D. Noe, Jun 07 2006
Numbers m that are nontrivial repunits for any base b >= 2. For k = 2 (I use k for the exponent since n is used as the index in a(n)) we get (b^k-1)/(b-1) = (b^2-1)/(b-1) = b+1, so every integer m >= 3 is a 2-digit repunit in base b = m-1. And for n = 1 (the 1-digit degenerate repunit) we get (b-1)/(b-1) = 1 for any base b >= 2. If we considered all k >= 1 we would get the sequence of all positive integers except 2 since it is the smallest uniform base used in positional representation (2 might be seen as the "repunit" in a nonpositional base representation such as the Roman numerals where 2 is expressed as II). - Daniel Forgues, Mar 01 2009
These repunits numbers belong to Brazilian numbers (A125134) (see Links: "Les nombres brésiliens" - section IV, p. 32). - Bernard Schott, Dec 18 2012
The Brazilian primes (A085104) belong to this sequence. - Bernard Schott, Dec 18 2012

			a(5) = 31 because 31 can be written as 111 base 5 (or indeed 11111 base 2).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1172 terms from T. D. Noe)
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

Cf. A090503 (a subsequence), A119598 (numbers that are repunits in four or more bases), A125134, A085104.

Cf. A108348.

a053696 n = a053696_list !! (n-1)
a053696_list = filter ((> 1) . a088323) [2..]
-- Reinhard Zumkeller, Jan 22 2014, Nov 26 2013

N:= 10^4: # to get all terms <= N
V:= Vector(N):
for b from 2 while (b^3-1)/(b-1) <= N do
  inds:= [seq((b^k-1)/(b-1), k=3..ilog[b](N*(b-1)+1))];
  V[inds]:= 1;
od:
select(t -> V[t] = 1, [$1..N]); # Robert Israel, Dec 10 2015

fQ[n_] := Block[{d = Rest@ Divisors[n - 1]}, Length@ d > 2 && Length@ Select[ IntegerDigits[n, d], Union@# == {1} &] > 1]; Select[ Range@ 1200, fQ]
lim=1000; Union[Reap[Do[n=3; While[a=(b^n-1)/(b-1); a<=lim, Sow[a]; n++], {b, 2, Floor[Sqrt[lim]]}]][[2, 1]]]
Take[Union[Flatten[With[{l=Table[PadLeft[{},n,1],{n,3,100}]}, Table[ FromDigits[#,n]&/@l,{n,2,100}]]]],80] (* Harvey P. Dale, Oct 06 2011 *)

list(lim)=my(v=List(),e,t);for(b=2,sqrt(lim),e=3;while((t=(b^e-1)/(b-1))<=lim,listput(v,t);e++));vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Oct 06 2011

list(lim)=my(v=List(),e,t);for(b=2,lim^(1/3),e=4;while((t=(b^e-1)/(b-1))<=lim,listput(v,t);e++));vecsort(concat(Vec(v), vector((sqrtint (lim\1*4-3)-3)\2,i,i^2+3*i+3)),,8) \\ Charles R Greathouse IV, May 30 2013

a(n) ~ n^2 since as n grows the density of repunits of degree 2 among all the repunits tends to 1. - Daniel Forgues, Dec 09 2008

A088323(a(n)) > 1. - Reinhard Zumkeller, Jan 22 2014

A119598 Numbers that are repunits in four or more bases.

Original entry on oeis.org

1, 31, 8191

Offset: 1

Sergio Pimentel, Jun 01 2006

Except for first term, numbers which can be represented as a string of three or more 1's in a base >=2 in more than one way; subset of A053696.
No more terms less than 2^44 = 17592186044416. - Ray Chandler, Jun 08 2006
Let the 4-tuple (a,b,m,n) be a solution to the exponential Diophantine equation (a^m-1)/(a-1)=(b^n-1)/(b-1) with a>1, b>a, m>2 and n>2. Then (a^m-1)/(a-1) is in this sequence. The terms 31 and 8191 correspond to the solutions (2,5,5,3) and (2,90,13,3), respectively. No other solutions with n=3 and b<10^5. The Mathematica code finds repunits in increasing order and prints solutions. - T. D. Noe, Jun 07 2006
Following the Goormaghtigh conjecture (Links), 31 and 8191 which are both Mersenne numbers, are the only primes which are Brazilian in two different bases. - Bernard Schott, Jun 25 2013

			a(1)=1 is a repunit in every base. a(2)=31 is a repunit in bases 1, 2, 5 and 30. a(3)=8191 is a repunit in bases 1, 2, 90 and 8190.
31 and 8191 are Brazilian numbers in two different bases:
31 = 11111_2 = 111_5,
8191 = 1111111111111_2 = 111_90.

Y. Bugeaud and T. N. Shorey, On the diophantine equation (x^m - 1)/(x-1) = (y^n - 1)/(y-1), Pacific Journal of Mathematics 207:1 (2002), pp. 61-75.
Jon Grantham, No new Goormaghtigh primes up to 10^700, arXiv:2410.03677 [math.NT], 2024.
Eric Weisstein's World of Mathematics, Repunit
Wikipedia, Goormaghtigh conjecture

Cf. A002275, A053696, A055129, A088323.
Cf. A053696 (numbers of the form (b^k-1)/(b-1)).
Cf. A145461: bases 5 and 90 are 2 exceptions (Goormaghtigh's conjecture).
Cf. A085104 (Brazilian primes).

fQ[n_] := Block[{d = Rest@Divisors[n - 1]}, Length@d > 2 && Length@Select[IntegerDigits[n, d], Union@# == {1} &] > 2]; Do[ If[ fQ@n, Print@n], {n, 10^8/3}] (* Robert G. Wilson v *)
nn=1000; pow=Table[3, {nn}]; t=Table[If[n==1, Infinity, (n^3-1)/(n-1)], {n,nn}]; While[pos=Flatten[Position[t,Min[t]]]; !MemberQ[pos,nn], If[Length[pos]>1, Print[{pos,pow[[pos]],t[[pos[[1]]]]}]]; Do[n=pos[[i]]; pow[[n]]++; t[[n]]=(n^pow[[n]]-1)/(n-1), {i,Length[pos]}]] (* T. D. Noe, Jun 07 2006 *)

def isrep(n, b):
  while n >= b:
    n, r = divmod(n, b)
    if r != 1: return False
  return n == 1
def agen():
  yield 1
  n = 2
  while True:
    reps = 2 # n is a repunit in bases 1 and n-1
    for b in range(2, n-1):
      if isrep(n, b): reps += 1
      if reps == 4: yield n; break
    n += 1
for m in agen(): print(m) # Michael S. Branicky, Jan 31 2021

Edited by Ray Chandler, Jun 08 2006

A341434 a(n) is the number of bases 1 < b < n in which n is divisible by its product of digits.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 3, 2, 2, 1, 5, 2, 3, 4, 6, 1, 5, 1, 5, 4, 4, 1, 9, 2, 2, 4, 5, 1, 7, 3, 9, 4, 2, 3, 12, 1, 2, 3, 10, 1, 7, 2, 7, 7, 2, 1, 15, 2, 5, 3, 6, 1, 10, 3, 10, 4, 3, 1, 14, 1, 2, 7, 14, 3, 8, 1, 6, 3, 6, 1, 20, 2, 3, 8, 7, 3, 7, 1, 16, 7, 2, 1, 14

Offset: 1

Amiram Eldar, Feb 11 2021

			a(3) = 1 since 3 is divisible by its product of digits only in base 2: 3 = 11_2 and 1*1 | 3.
a(6) = 2 since 6 is divisible by its product of digits in 2 bases: in base 4, 6 = 12_4 and 1*2 | 6, and in base 5, 6 = 11_5 and 1*1 | 6.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A007602, A068953, A080221, A085104, A088323, A119598.

q[n_, b_] := (p = Times @@ IntegerDigits[n, b]) > 0 && Divisible[n, p]; a[n_] := Count[Range[2, n], _?(q[n, #] &)]; Array[a, 100]

a(n) = sum(b=2, n-1, my(x=vecprod(digits(n, b))); x && !(n%x)); \\ Michel Marcus, Feb 12 2021

a(n) > 0 for all numbers n > 2 since n in base b = n-1 is 11.
a(n) > 1 for all even numbers > 4 since n in base b = n-2 is 12. Similarly, a(n) > 1 for all composite numbers > 4 since if n = k*m, then n is divisible by its product of digits in bases n-m and n-k.
a(p) > 1 for primes p in A085104.
a(p) > 2 for primes p in A119598 (i.e., 31, 8191, ...).
a(n) >= A088323(n), with equality if n = 4 or if n is a prime.

cp's OEIS Frontend

A085104 Primes of the form 1 + n + n^2 + n^3 + ... + n^k, n > 1, k > 1.

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A053696 Numbers that can be represented as a string of three or more 1's in a base >= 2.

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A119598 Numbers that are repunits in four or more bases.

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A341434 a(n) is the number of bases 1 < b < n in which n is divisible by its product of digits.

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Mathematica

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