A055129 -id:A055129 - cp's OEIS frontend

A107893 Triangle read by rows, related to A055129 (repunits in base k).

1, 2, 1, 3, 4, 2, 4, 11, 14, 6, 5, 26, 64, 66, 24, 6, 57, 244, 456, 384, 120, 7, 120, 846, 2556, 3744, 2640, 720, 8, 247, 2778, 12762, 28944, 34560, 20880, 5040, 9, 502, 8828, 59382, 195768, 352080, 353520, 186480, 40320, 10, 1013, 27488, 264012, 1216368, 3091320, 4587120, 3966480, 1854720, 362880

Offset: 1

Gary W. Adamson, May 26 2005

Second column of A107893 = Eulerian numbers (A000295) starting with 1: 1, 4, 11, 26, 57, ... Rightmost term in row n = (n-1)!.

Using the Jun 18 2009 formula of Johannes W. Meijer in A028246: Instead of a(n,1)=1 set a(n,1)=n. The result is A107893. - Werner Schulte, Dec 12 2016

			Binomial transform of Row 4 in the form: (4, 11, 14, 6, 0, 0, 0, ...) = Row 4 of A055129: 4, 15, 40, 85, ... which is generated from f(x) = x^3 + x^2 + x + 1; (x = 1,2,3, ...).
Triangle starts:
  1;
  2,   1;
  3,   4,   2;
  4,  11,  14,   6;
  5,  26,  64,  66,  24;
  6,  57, 244, 456, 384, 120;
  ...

Michael De Vlieger, Table of n, a(n) for n = 1..5050 (1 <= n <= 100)

Cf. A000295, A055129, A028246.

Table[Sum[Sum[(-1)^(k - j) Binomial[k, j] j^i, {j, 0, k}]/k, {i, n}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Dec 11 2016, after Jean-François Alcover at A028246 *)

n-th row = inverse binomial transform of n-th column of A055129, where the latter are generated from f(x) = x^(n-1) + x^(n-2) + ...+ x + 1; (x = 1, 2, 3, ...)
A(n,k) = Sum_{i=1..n} A028246(i,k) for 1 <= k <= n. - Werner Schulte, Dec 08 2016
The polynomials p(n,t) = Sum_{k=1..n} A(n,k)*t^k are given by p(1,t) = t and p(n+1,t) = t + t*(t+1)*(d/dt)p(n,t) for n >= 1. - Werner Schulte, Dec 12 2016

A053698 a(n) = n^3 + n^2 + n + 1.

Original entry on oeis.org

1, 4, 15, 40, 85, 156, 259, 400, 585, 820, 1111, 1464, 1885, 2380, 2955, 3616, 4369, 5220, 6175, 7240, 8421, 9724, 11155, 12720, 14425, 16276, 18279, 20440, 22765, 25260, 27931, 30784, 33825, 37060, 40495, 44136, 47989, 52060, 56355, 60880

Offset: 0

Henry Bottomley, Mar 23 2000

a(n) = 1111 in base n.

n^3 + n^2 + n + 1 = (n^2 + 1)*(n + 1), therefore a(n) is never prime. - Alonso del Arte, Apr 22 2014

			a(2) = 15 because 2^3 + 2^2 + 2 + 1 = 8 + 4 + 2 + 1 = 15.
a(3) = 40 because 3^3 + 3^2 + 3 + 1 = 27 + 9 + 3 + 1 = 40.
a(4) = 85 because 4^3 + 4^2 + 4 + 1 = 64 + 16 + 4 + 1 = 85.
From _Bruno Berselli_, Jan 02 2017: (Start)
The terms of the sequence are provided by the row sums of the following triangle (see the seventh formula in the previous section):
.   1;
.   3,   1;
.   9,   5,   1;
.  19,  13,   7,   1;
.  33,  25,  17,   9,   1;
.  51,  41,  31,  21,  11,   1;
.  73,  61,  49,  37,  25,  13,  1;
.  99,  85,  71,  57,  43,  29, 15,  1;
. 129, 113,  97,  81,  65,  49, 33, 17,  1;
. 163, 145, 127, 109,  91,  73, 55, 37, 19,  1;
. 201, 181, 161, 141, 121, 101, 81, 61, 41, 21, 1;
...
Columns from the first to the fifth, respectively: A058331, A001844, A056220 (after -1), A059993, A161532. Also, eighth column is A161549.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A237627 (subset of semiprimes).
Cf. A056106 (first differences).
Cf. A062158, A027444, A104878, A055129.

[n^3+n^2+n+1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

A053698:=n->n^3 + n^2 + n + 1; seq(A053698(n), n=0..50); # Wesley Ivan Hurt, Apr 22 2014

Table[n^3 + n^2 + n + 1, {n, 0, 39}] (* Alonso del Arte, Apr 22 2014 *)
FromDigits["1111", Range[0, 50]] (* Paolo Xausa, May 11 2024 *)

Vec((1 + 5*x^2) / (1 - x)^4 + O(x^50)) \\ Colin Barker, Jan 02 2017

def a(n): return (n**3+n**2+n+1) # Torlach Rush, May 08 2024

For n >= 2, a(n) = (n^4-1)/(n-1) = A024002(n)/A024000(n) = A002522(n)*(n+1) = A002061(n+1) + A000578(n).
G.f.: (1+5*x^2) / (1-x)^4. - Colin Barker, Jan 06 2012
a(n) = -A062158(-n). - Bruno Berselli, Jan 26 2016
a(n) = Sum_{i=0..n} 2*n*(n-i)+1. - Bruno Berselli, Jan 02 2017
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Colin Barker, Jan 02 2017
a(n) = A104878(n+3,n) = A055129(4,n) for n > 0. - Mathew Englander, Jan 06 2021
E.g.f.: exp(x)*(x^3+4*x^2+3*x+1). - Nikolaos Pantelidis, Feb 06 2023

A053699 a(n) = n^4 + n^3 + n^2 + n + 1.

Original entry on oeis.org

1, 5, 31, 121, 341, 781, 1555, 2801, 4681, 7381, 11111, 16105, 22621, 30941, 41371, 54241, 69905, 88741, 111151, 137561, 168421, 204205, 245411, 292561, 346201, 406901, 475255, 551881, 637421, 732541, 837931, 954305, 1082401, 1222981, 1376831, 1544761, 1727605

Offset: 0

Henry Bottomley, Mar 23 2000

a(n) = 11111 in base n.

a(n) = Phi_5(n), where Phi_k is the k-th cyclotomic polynomial.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Michael Penn, What base makes me a perfect square??, YouTube video, 2022.
Index to values of cyclotomic polynomials of integer argument.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

5th row of the array A055129.

Cf. A104878.

[n^4+n^3+n^2+n+1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

A053699 := proc(n)
        numtheory[cyclotomic](5,n) ;
end proc:
seq(A053699(n),n=0..20) ; # R. J. Mathar, Feb 07 2014

f[n_]:=((1+n+n^2+n^3+n^4));Table[f[n],{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2010 *)
Join[{1},Table[Total[n^Range[0,4]],{n,40}]] (* Harvey P. Dale, Feb 02 2014 *)

A053699(n):=n^4 + n^3 + n^2 + n + 1$
makelist(A053699(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

a(n)=polcyclo(5,n) \\ Charles R Greathouse IV, Jul 19 2011

a(n) = n^4 + n^3 + n^2 + n + 1 = (n^5-1)/(n-1).
G.f.: (1 + 16*x^2 + 6*x^3 + x^4)/(1-x)^5. - Colin Barker, Jan 10 2012
E.g.f.: exp(x)*(1 + 4*x + 11*x^2 + 7*x^3 + x^4). - Stefano Spezia, Oct 03 2024
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, Aug 05 2025

A053716 a(n) = 1111111 in base n.

Original entry on oeis.org

7, 127, 1093, 5461, 19531, 55987, 137257, 299593, 597871, 1111111, 1948717, 3257437, 5229043, 8108731, 12204241, 17895697, 25646167, 36012943, 49659541, 67368421, 90054427, 118778947, 154764793, 199411801, 254313151, 321272407, 402321277, 499738093

Offset: 1

Henry Bottomley, Mar 23 2000

Evaluation of the seventh cyclotomic polynomial at n. - Joerg Arndt, Aug 27 2015

			a(3)=1093 because 1111111 base 3=729+243+81+27+9+3+1=121.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Index to values of cyclotomic polynomials of integer argument
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

7th row of the array A055129.

Cf. A104878.

[7] cat [(n^7-1)/(n-1): n in [2..35]]; // Vincenzo Librandi, Feb 08 2014

A053716 := proc(n)
    numtheory[cyclotomic](7,n) ;
end proc:
seq(A053716(n),n=1..20) ; # R. J. Mathar, Feb 07 2014

Table[FromDigits["1111111",n],{n,1,30}](*or*)Table[n^6+n^5+n^4+n^3+n^2+n+1,{n,1,60}] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
CoefficientList[Series[-(x^6 - 6 x^5 + 57 x^4 + 232 x^3 + 351 x^2 + 78 x + 7)/(x - 1)^7, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 08 2014 *)

a(n) = n^6+n^5+n^4+n^3+n^2+n+1 = (n^7-1)/(n-1).
G.f.: -x*(x^6-6*x^5+57*x^4+232*x^3+351*x^2+78*x+7)/(x-1)^7. - Colin Barker, Oct 29 2012
E.g.f.: exp(x)*(1 + 6*x + 57*x^2 + 122*x^3 + 76*x^4+ 16*x^5 + x^6) - 1. - Stefano Spezia, Oct 03 2024

A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

Original entry on oeis.org

0, 1, 4, 21, 156, 1555, 19608, 299593, 5380840, 111111111, 2593742460, 67546215517, 1941507093540, 61054982558011, 2085209001813616, 76861433640456465, 3041324492229179280, 128583032925805678351, 5784852794328402307380, 275941052631578947368421

Offset: 1

Henry Bottomley, Feb 21 2001

nonn

(n-1)-digit repunits in base n written in decimal.

			a(10)=111111111; i.e., just nine 1's (converted from base 10 to decimal).

Harry J. Smith, Table of n, a(n) for n=1..200

Cf. A000169, A037205, A055869, A060073, A125118, A228275.

Cf. other sequences of generalized repunits, such as A053696, A055129, A031973, A125598, A173468, A023037, A119598, A085104, and A162861.

[0] cat [ (n^(n-1) -1)/(n-1) : n in [2..25]]; // G. C. Greubel, Aug 15 2022

Join[{0},Array[(#^(#-1)-1)/(#-1)&,20,2]] (* Harvey P. Dale, Jun 04 2013 *)

a(n) = if (n==1, 0, (n^(n - 1) - 1)/(n - 1)); \\ Harry J. Smith, Jul 01 2009

[0]+[(n^(n-1) -1)/(n-1) for n in (2..25)] # G. C. Greubel, Aug 15 2022

a(n+1) = Sum_{k=1..n} n^(k-1)*C(n, k). - Olivier Gérard, Jun 26 2001 [Corrected by Mathew Englander, Dec 15 2020]
a(n) = Sum_{j=2..n} n^(n-j). - Zerinvary Lajos, Sep 11 2006
a(n+1) = A125118(n,n). - Reinhard Zumkeller, Nov 21 2006
a(n) = Integral_{x=1/n..1} 1/x^n dx. - Francesco Daddi, Aug 01 2011
a(n) = A037205(n-1)/(n-1) = A060073(n)*(n-1) = A023037(n) - A000169(n).
a(n) = [x^n] x^2/((1 - x)*(1 - n*x)). - Ilya Gutkovskiy, Oct 04 2017
a(n) = 1 + A228275(n, n-2) for n >= 2. - Mathew Englander, Dec 14 2020

Name edited by Michel Marcus, Dec 14 2020

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

Original entry on oeis.org

1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111

Offset: 1

Reinhard Zumkeller, Nov 21 2006

			First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Repunit

This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Columns (adjusting offset as necessary): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
Cf. A023037, A031973, A125119, A125120 (row sums).

[((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022

Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)

def A125118(n,k): return ((k+1)^n -1)/k
flatten([[A125118(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022

T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020

A053700 a(n) = 111111 in base n.

Original entry on oeis.org

6, 63, 364, 1365, 3906, 9331, 19608, 37449, 66430, 111111, 177156, 271453, 402234, 579195, 813616, 1118481, 1508598, 2000719, 2613660, 3368421, 4288306, 5399043, 6728904, 8308825, 10172526, 12356631, 14900788, 17847789, 21243690, 25137931

Offset: 1

Henry Bottomley, Mar 23 2000

			a(3)=364 because 111111 base 3 = 243 + 81 + 27 + 9 + 3 + 1 = 121.

Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

6th row of the array A055129.

Cf. A104878.

Table[(n^5+n^4+n^3+n^2+n+1),{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)

a(n)=(n^6-1)/(n-1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = n^5 + n^4 + n^3 + n^2 + n + 1 = (n^6-1)/(n-1).
G.f.: x*(6 + 27*x + 76*x^2 + 6*x^3 + 6*x^4 - x^5)/(1-x)^6. - Colin Barker, May 08 2012
E.g.f.: exp(x)*(1 + 5*x + 26*x^2 + 32*x^3 + 11*x^4+ x^5) - 1. - Stefano Spezia, Oct 03 2024

A066180 a(n) = smallest base b so that repunit (b^prime(n) - 1) / (b - 1) is prime, where prime(n) = n-th prime; or 0 if no such base exists.

Original entry on oeis.org

2, 2, 2, 2, 5, 2, 2, 2, 10, 6, 2, 61, 14, 15, 5, 24, 19, 2, 46, 3, 11, 22, 41, 2, 12, 22, 3, 2, 12, 86, 2, 7, 13, 11, 5, 29, 56, 30, 44, 60, 304, 5, 74, 118, 33, 156, 46, 183, 72, 606, 602, 223, 115, 37, 52, 104, 41, 6, 338, 217, 13, 136, 220, 162, 35, 10, 218, 19, 26, 39

Offset: 1

Frank Ellermann, Dec 15 2001

nonn

Is a(n) = 0 possible?
Let p be the n-th prime; Cp(x) be the p-th cyclotomic polynomial (x^p - 1)/(x - 1); a(n) is the least k > 1 such that Cp(k) is prime.
The values associated with a(5) and a(8) through a(70) have been certified prime with Primo. (a(1) through a(4), a(6) and a(7) give prime(2), prime(4), prime(11), prime(31), prime(1028) and prime(12251), respectively.)

			a(5) = 5 because 11 is the 5th prime; (b^5 - 1)/(b - 1) is composite for b = 2,3,4 and prime ((5^11 - 1)/4 = 12207031) for b = 5.
b = 61 for prime(12) = 37 because (61^37 - 1)/60 is prime and 61 is the least base b that makes (b^37 - 1)/(b - 1) a prime.

Paulo Ribenboim, "The New Book of Prime Numbers Records", Springer, 1996, p. 353.

Robert G. Wilson v, Table of n, a(n) for n = 1..300 (terms 1..200 from Charles R Greathouse IV).
H. Dubner, Generalized repunit primes, Math. Comp., 61 (1993), 927-930.
Andy Steward, Titanic Prime Generalized Repunits.
Eric Weisstein's World of Mathematics, Repunit.
H. C. Williams and E. Seah, Some primes of the form: (a^n - 1)/(a - 1), Mathematics of Computation 23, 1979.

Cf. A004023 (prime repunits in base 10), A000043 (prime repunits in base 2, Mersenne primes), A055129 (table of repunits).

Cf. A084732, A085398.

Table[p = Prime[n]; b = 1; While[b++; ! PrimeQ[(b^p - 1)/(b - 1)]]; b, {n, 1, 70}] (* Lei Zhou, Oct 07 2011 *)

/* This program assumes (probable) primes exist for each n. */
/* All 70 (probable) primes found by this program have been proved prime. */
gen_repunit(b,n) = (b^prime(n)-1)/(b-1);
for(n=1,70, b=1; until(isprime(p), b++; p=gen_repunit(b,n)); print1(b,","));

a(n) = A085398(prime(n)).

Sequence extended to 16 terms by Don Reble, Dec 18 2001
More terms from Rick L. Shepherd, Sep 14 2002
Entry revised by N. J. A. Sloane, Jul 23 2006

A104878 A sum-of-powers number triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 15, 13, 5, 1, 1, 6, 31, 40, 21, 6, 1, 1, 7, 63, 121, 85, 31, 7, 1, 1, 8, 127, 364, 341, 156, 43, 8, 1, 1, 9, 255, 1093, 1365, 781, 259, 57, 9, 1, 1, 10, 511, 3280, 5461, 3906, 1555, 400, 73, 10, 1, 1, 11, 1023, 9841, 21845

Offset: 0

Paul Barry, Mar 28 2005

Columns are partial sums of the columns of A004248. Row sums are A104879. Diagonal sums are A104880.

The rows of this triangle (apart from the initial "1" in each row) are the antidiagonals of rectangle A055129. The diagonals of this triangle (apart from the initial "1") are the rows of rectangle A055129. The columns of this triangle (apart from the leftmost column) are the same as the columns of rectangle A055129 but shifted downward. - Mathew Englander, Dec 21 2020

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 15, 13,  5,  1;
  1,  6, 31, 40, 21,  6,  1;
  ...

Cf. A004248 (first differences by column), A104879 (row sums), A104880 (antidiagonal sums), A125118 (version of this triangle with fewer terms).
This triangle (ignoring the leftmost column) is a rotation of rectangle A055129.
Columns (adjusting offset as necessary): A000012, A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
T(2n,n) gives A031973.

A104878 :=proc(n,k): if k = 0 then 1 elif k=1 then n elif k>=2 then (k^(n-k+1)-1)/(k-1) fi: end: for n from 0 to 7 do seq(A104878(n,k), k=0..n) od; seq(seq(A104878(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Aug 21 2011

T(n, k) = if(k=1, n, if(k<=n, (k^(n-k+1)-1)/(k-1), 0));
G.f. of column k: x^k/((1-x)(1-k*x)). [corrected by Werner Schulte, Jun 05 2019]
T(n, k) = A069777(n+1,k)/A069777(n,k). [Johannes W. Meijer, Aug 21 2011]
T(n, k) = A055129(n+1-k, k) for n >= k > 0. - Mathew Englander, Dec 19 2020

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

cp's OEIS Frontend

A107893 Triangle read by rows, related to A055129 (repunits in base k).

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A053698 a(n) = n^3 + n^2 + n + 1.

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A053699 a(n) = n^4 + n^3 + n^2 + n + 1.

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A053716 a(n) = 1111111 in base n.

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A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

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A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

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