A053696 -id:A053696 - cp's OEIS frontend

A208242 Perfect powers y^q with y > 1 and q > 1 which are Brazilian repunits with three or more digits in some base.

121, 343, 400

Offset: 1

Bernard Schott, Jan 11 2013

These three numbers are the only known solutions y^q of the Nagell-Ljunggren equation (b^m-1)/(b-1) = y^q with y > 1, q > 1, b > 1, m > 2. Yann Bugeaud and Maurice Mignotte propose two alternative conjectures:
A) The Nagell-Ljunggren equation has only these three solutions.
Considering the current state of our knowledge, this conjecture seems too ambitious, while the next one seems more reasonable.
B) The Nagell-Ljunggren equation has only a finite number of solutions.
This last conjecture is true if the abc conjecture is true (see article Bugeaud-Mignotte in link, p. 148).
Consequence: 121 is the only known square of prime which is Brazilian.
There are no other solutions for some base b < 10000.
Some theorems and results about this equation:
With the exception of the 3 known solutions,
1) for q = 2, there are no other solutions than 11^2 and 20^2,
2) there is no other solution if 3 divides m than 7^3,
3) there is no other solution if 4 divides m than 20^2. - Bernard Schott, Apr 29 2019
From David A. Corneth, Apr 29 2019: (Start)
Intersection of A001597 and A053696.
a(4) > 10^25 if it exists using constraints above.
In the Nagell-Ljunggren equation, we need b > 2. If b = 2, we get y^q = 2^m - 1 which by Catalan's conjecture has no solutions (see A001597). (End)

			121 = 11^2 =  (3^5 - 1)/ (3 - 1) = 11111_3.
343 =  7^3 = (18^3 - 1)/(18 - 1) =   111_18.
400 = 20^2 =  (7^4 - 1)/ (7 - 1) =  1111_7.

Y. Bugeaud and M. Mignotte, L'équation de Nagell-Ljunggren (x^n-1)/(x-1) = y^q, Enseign. Math. 48(2002), 147-168.

Cf. A001597, A053696, A220571 (Brazilian composites), A307745 (similar but with digits > 1).

is(n) = if(!ispower(n), return(0)); for(b=2, n-1, my(d=digits(n, b)); if(#d > 2 && vecmin(d)==1 && vecmax(d)==1, return(1))); 0 \\ Felix Fröhlich, Apr 29 2019

Small edits to the name by Bernard Schott, Apr 30 2019

A090503 Number of hyperplanes in a finite projective space (of some dimension d over some finite field of order q).

Original entry on oeis.org

7, 13, 15, 21, 31, 40, 57, 63, 73, 85, 91, 121, 127, 133, 156, 183, 255, 273, 307, 341, 364, 381, 400, 511, 553, 585, 651, 757, 781, 820, 871, 993, 1023, 1057, 1093, 1365, 1407, 1464, 1723, 1893, 2047, 2257, 2380, 2451, 2801, 2863, 3280, 3541, 3783, 3906, 4095, 4161, 4369, 4557, 4681, 5113, 5220, 5403, 5461, 6321, 6643, 6973

Offset: 1

Olivier Giraud (olivier.giraud(AT)bristol.ac.uk), Feb 01 2004

nonn

The number of tiles building the known pairs of Euclidean isospectral billiards are 7, 13, 15, 21, ... (see Refs Okada et al. and Buser et al.).

Subsequence of A053696. - Hans Havermann, Nov 21 2013

T. Tsuzuki, Finite groups and finite geometries, Cambridge University Press, 1982, p. 73.

Max Alekseyev, Table of n, a(n) for n = 1..1504 (contains all terms below 10^8)
P. Buser, J. H. Conway, P. Doyle and K.-D. Semmler, Isospectral domains
W. Cherowitzo, Finite projective spaces
Y. Okada and A. Shudo, Equivalence between isospectrality and isolength spectrality for a certain class of planar billiard domains, J. Phys. A: Math. Gen. 34 (2001), 5911-5922

Cf. A053696.

Cf. A000961, A108348.

a090503 n = a090503_list !! (n-1)
a090503_list = f [1..] where
   f (x:xs) = g $ tail a000961_list where
     g (q:pps) = h 0 $ map ((`div` (q - 1)) . subtract 1) $
                           iterate (* q) (q ^ 3) where
       h i (qy:ppys) | qy > x    = if i == 0 then f xs else g pps
                     | qy < x    = h 1 ppys
                     | otherwise = x : f xs
-- Reinhard Zumkeller, Nov 26 2013

isA090503[n_] := Module[{f = FactorInteger[n-1]}, For[i = 1, i <= Length[f], i++, For[j = 1, j <= f[[i, 2]], j++, q = f[[i, 1]]^j; If[q == n-1, Continue[]]; If[n*(q-1)+1 == q^IntegerExponent[n*(q-1)+1, q], Return[True]]]]; False]; Reap[For[n = 2, n <= 10^5, n++, If[isA090503[n], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Nov 21 2013, translated and adapted from Max Alekseyev's program *)

isA090503(n) = my(f,q); f=factor(n-1); for(i=1,matsize(f)[1], for(j=1,f[i,2], q=f[i,1]^j; if(q==n-1,next); if( n*(q-1)+1 == q^valuation(n*(q-1)+1,q), return(q)); )); 0 /* Max Alekseyev, Nov 20 2013 */

Numbers of the form (q^(d+1)-1)/(q-1), d>=2, q=p^m with m>=1 and p prime.

Missing terms provided by Jean-François Alcover and Wouter Meeussen; edited by M. F. Hasler, Nov 20 2013

PARI program and further terms in a b-file added by Max Alekseyev, Nov 20 2013

A125598 a(n) = ((n+1)^(n-1) - 1)/n.

Original entry on oeis.org

0, 1, 5, 31, 259, 2801, 37449, 597871, 11111111, 235794769, 5628851293, 149346699503, 4361070182715, 139013933454241, 4803839602528529, 178901440719363487, 7143501829211426575, 304465936543600121441

Offset: 1

Alexander Adamchuk, Nov 26 2006

nonn

Odd prime p divides a(p-2).
a(n) is prime for n = {3,4,6,74, ...}; prime terms are {5, 31, 2801, ...}.
a(n) is the (n-1)-th generalized repunit in base (n+1). For example, a(5) = 259 which is 1111 in base 6. - Mathew Englander, Oct 20 2020

G. C. Greubel, Table of n, a(n) for n = 1..350

Cf. A000272 (n^(n-2)), A125599.

Cf. other sequences of generalized repunits, such as A125118, A053696, A055129, A060072, A031973, A173468, A023037, A119598, A085104, and A162861.

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

Mathematica
```
Table[((n+1)^(n-1)-1)/n, {n,25}]
```

[gaussian_binomial(n,1,n+2) for n in range(0,18)] # Zerinvary Lajos, May 31 2009

a(n) = ((n+1)^(n-1) - 1)/n.
a(n) = (A000272(n+1) - 1)/n.
a(2k-1)/(2k+1) = A125599(k) for k>0.
From Mathew Englander, Dec 17 2020: (Start)
a(n) = (A060072(n+1) - A083069(n-1))/2.
For n > 1, a(n) = Sum_{k=0..n-2} (n+1)^k.
For n > 1, a(n) = Sum_{j=0..n-2} n^j*C(n-1,j+1). (End)

A325658 Brazilian composites of the form 1 + b + b^2 + b^3 + ... + b^k, b > 1, k > 1.

Original entry on oeis.org

15, 21, 40, 57, 63, 85, 91, 111, 121, 133, 156, 183, 255, 259, 273, 341, 343, 364, 381, 400, 507, 511, 553, 585, 651, 703, 781, 813, 820, 871, 931, 993, 1023, 1057, 1111, 1191, 1261, 1333, 1365, 1407, 1464, 1555, 1561, 1641, 1807, 1885, 1893, 1981, 2047, 2071, 2163, 2257, 2353

Offset: 1

Bernard Schott, May 12 2019

Composites that are repunits in base b >= 2 with three or more digits. If the Goormaghtigh conjecture is true, there are no composite numbers which can be represented as a string of three or more 1's in a base >= 2 in more than one way (A119598).
Only three known perfect powers belong to this sequence: 121, 343 and 400 (A208242).
Except for 121, each term of this sequence have also at least one Brazilian representation with only 2 digits.

			121 = (11111)_3, 133 = (111)_11 = (77)_18.

Robert Israel, Table of n, a(n) for n = 1..10000
Yann Bugeaud and T. N. Shorey, On the Diophantine Equation (x^m - 1)/(x-1) = (y^n - 1)/(y-1), Pacific Journal of Mathematics, Vol. 207, No 1, November 2002.
Sean A. Irvine, Java program (github)
Michel Waldschmidt, Lecture on the abc conjecture and some of its consequences, 6th World Conference on 21st Century Mathematics 2013, Lahore, p. 14 (Goormaghtigh conjecture).
Wikipedia, Goormatigh conjecture.

Complement of A085104 with respect to A053696.
Intersection of A053696 and A220571.
Cf. A119598, A208242, A325659.

N:= 3000:
Res:= NULL:
for m from 2 while 1+m+m^2 <= N do
  for k from 2 do
    v:= (m^(k+1)-1)/(m-1);
    if v > N then break fi;
    if not isprime(v) then  Res:= Res, v fi
od od:
sort(convert({Res},list)); # Robert Israel, May 13 2019

lista(nn) = {forcomposite(n=1, nn, for(b=2, sqrtint(n), my(d=digits(n, b), sd=Set(d)); if ((#d >= 3) && (#sd == 1) && (sd[1] == 1), print1(n, ", "); break);););} \\ Michel Marcus, May 18 2019

A376298 Numbers which are the sum of at least three successive terms of a geometric progression.

Original entry on oeis.org

7, 13, 14, 15, 21, 26, 28, 30, 31, 35, 39, 40, 42, 43, 45, 49, 52, 56, 57, 60, 62, 63, 65, 70, 73, 75, 77, 78, 80, 84, 85, 86, 90, 91, 93, 98, 104, 105, 111, 112, 114, 117, 119, 120, 121, 124, 126, 127, 129, 130, 133, 135, 140, 143, 146, 147, 150, 154, 155, 156, 157

Offset: 1

Andrew Howroyd, Sep 19 2024

nonn

Multiples of terms in A053696.

Numbers of the form m*(b^n-1)/(b-1) for n > 2 and b > 1, m > 0.

			7 is a term because 7 = 1 + 2 + 4.
13 is a term because 13 = 1 + 3 + 9.
14 is a term because 14 = 2 + 4 + 8.
15 is a term because 15 = 1 + 2 + 4 + 8.

Andrew Howroyd, Table of n, a(n) for n = 1..10000
Wikipedia, Geometric progression.

Cf. A053696, A084733, A085104.

B(k,lim)={vector(logint(lim*(k-1)+1,k)-2, i, (k^(i+2) - 1)/(k-1))}
upto(lim=200)={my(v=concat(vector(sqrtint(lim)-1, k, B(k+1,lim)))); Set(concat(vector(#v, i, my(t=v[i]); t*[1..lim\t])))}

A275766 a(n) = (5^(2*(n + 1)) - 1)/4.

Original entry on oeis.org

156, 3906, 97656, 2441406, 61035156, 1525878906, 38146972656, 953674316406, 23841857910156, 596046447753906, 14901161193847656, 372529029846191406, 9313225746154785156, 232830643653869628906, 5820766091346740722656, 145519152283668518066406

Offset: 1

Gionata Neri, Aug 07 2016

It seems that these terms are the only numbers n such that n and n + 1 are in A053696.

			3906 written in base 5 is 111111 and 3907 written in base 62 is 111.

Colin Barker, Table of n, a(n) for n = 1..700
Index entries for linear recurrences with constant coefficients, signature (26,-25).

Cf. A003463, A053696, A125831.

Table[(5^(2 (n + 1)) - 1)/4, {n, 16}] (* or *)
Rest@ CoefficientList[Series[6 x (26 - 25 x)/((1 - x) (1 - 25 x)), {x, 0, 16}], x] (* Michael De Vlieger, Aug 28 2016 *)

Vec(6*x*(26-25*x)/((1-x)*(1-25*x)) + O(x^20)) \\ Colin Barker, Aug 24 2016

a(n) = 5^(2*n+2)\4 \\ Charles R Greathouse IV, Aug 28 2016

a(n) = ((A125831(n+1))^3 - 1)/(A125831(n+1) - 1) - 1.
a(n) = A003463(2*(n+1)).
a(n) = 26*a(n-1) - 25*a(n-2), a(1) = 156, a(2) = 3906.
G.f.: 6*x*(26-25*x) / ((1-x)*(1-25*x)). - Colin Barker, Aug 24 2016

A282090 Totient numbers (A002202) of the form 1 + k + k^2 + k^3 +...+ k^i (i > 1, k > 1).

Original entry on oeis.org

40, 156, 400, 820, 1464, 2380, 3280, 3616, 3906, 5220, 7240, 9724, 12720, 19608, 20440, 25260, 30784, 37060, 60880, 66430, 70644, 81400, 93196, 97656, 106080, 120100, 135304, 151740, 169456, 177156, 188500, 254080, 265720, 278916, 333340, 363024, 394420, 427576, 462540, 499360

Offset: 1

Altug Alkan, Feb 06 2017

nonn

Totient numbers of the form (k^(i+1) - 1)/(k - 1) where k and i are both odd numbers that are greater than 1.

			40 is a term because 1 + 3 + 9 + 27 = 40 is a totient number.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Intersection of A002202 and A053696.

Cf. A281962.

list(lim)=my(v=List(), e, t); for(b=2, sqrt(lim), e=3; while((t=(b^e-1)/(b-1))<=lim, if(istotient(t),listput(v, t)); e++)); vecsort(Vec(v), , 8) \\ Ray Chandler, Feb 08 2017

Terms confirmed by Ray Chandler, Feb 08 2017

A290969 The least positive integer that is a repdigit with length > 2 in exactly n bases.

Original entry on oeis.org

1, 7, 31, 32767, 4095, 435356467, 16777215, 68719476735, 281474976710655

Offset: 0

Bernard Schott, Aug 16 2017

a(10) <= 1152921504606846975 = 2^60 - 1.
a(7) and following terms > 3*10^9. - Giovanni Resta, Aug 16 2017
a(7) <= 2^36-1 and a(8) <= 2^48-1. - Michel Marcus, Aug 17 2017
In fact, we have equality in both cases. - Rémy Sigrist, Aug 21 2017
Except for a(5) = (6^12 - 1) / 5, all the numbers in the data through a(8) are Mersenne numbers A000225. - Bernard Schott, Aug 27 2017

			a(1) = 7 = 111_2.
a(2) = 31 = 11111_2 = 111_5.
a(3) = 32767 = (R_15)_2 = 77777_8 = (31,31,31)_32.

Cf. A000225, A053696, A125134, A167782, A167783, A290869.

a(7) from Rémy Sigrist, Aug 19 2017

a(8) from Rémy Sigrist, Aug 21 2017

A325659 Smallest Brazilian composite in base n >= 2 which can be represented as a string of three or more 1's in this base.

Original entry on oeis.org

15, 40, 21, 156, 259, 57, 585, 91, 111, 133, 1885, 183, 2955, 3616, 273, 5220, 343, 381, 8421, 9724, 507, 553, 14425, 651, 703, 20440, 813, 871, 931, 993, 1057, 37060, 1191, 1261, 1333, 1407, 56355, 1561, 1641, 70644, 1807, 1893, 1981, 2071, 2163, 2257, 2353, 2451, 127551

Offset: 2

Bernard Schott, May 12 2019

Also the smallest Brazilian composite of the form (n^k - 1)/(n - 1) with k > 2.
For Mersenne numbers = (11...11)_2 = 2^k-1 in A000225, there is a smaller integer which is Brazilian prime: 7, so 7 is the first term of A285642 and another one is the smaller composite 15, so 15 is the first term of this sequence.
For numbers (11...11)_3 = (3^k-1)/2 in A003462,  there is also a smaller integer which is Brazilian prime:13, so 13 is the second term of A285642 and another one is the smaller Brazilian composite: 40, so 40 is the second term of this sequence.
For numbers like (11...11)_4 = (4^k-1)/3, the terms are respectively 0 in A285642 because there is no Brazilian prime of this type and 21 for composite numbers of this sequence, and so on.

			15 = (1111)_2, 40 = (1111)_3, 21 = (111)_4, 156 = (1111)_5.

Wikipedia, Nombre brésilien

Subsequence of A053696, A220571, A325658.

Cf. A285642 (same with Brazilian primes).

a(n) = {my(k=4, x); while (isprime(x=(n^(k-1)-1)/(n-1)), k++); x;} \\ Michel Marcus, May 17 2019

More terms from Michel Marcus, May 17 2019

A326775 For any n >= 0, let b >= 2 be the smallest base where n has all digits equal, say to d; a(n) = d.

Original entry on oeis.org

0, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 5, 4, 1, 2, 3, 1, 1, 2, 1, 4, 5, 2, 1, 6, 1, 5, 3, 4, 1, 6, 5, 4, 1, 2, 1, 6, 1, 2, 1, 4, 5, 6, 1, 4, 3, 7, 1, 6, 1, 2, 5, 4, 7, 6, 1, 2, 3, 2, 1, 7, 1, 2

Offset: 0

Rémy Sigrist, Jul 28 2019

A059711 gives base b.
From Bernard Schott, Aug 17 2019: (Start)
a(n) = 1 iff n is A220570, then n = 11_(n-1) or, n is in A053696, then n = 11..11_b for some base b.
a(n) = 2 if n = 2 * p, p prime >= 5.
a(n) = 3 if n = 3 * p, p prime >= 11.
There are k = 2 equal digits in the representation of n in the corresponding base b, except when n is a term of A167782, in which case the number k of equal digits is >= 3. (End)
n = (b^k - 1)/(b - 1) * a(n) so a(n) | n for n > 0. Furthermore a(n) <= sqrt(n). - David A. Corneth, Aug 21 2019
If b is the smallest base such that n=d*b^k+...+d*b^0 (A059711) (d=a(n) is the repdigit) then n mod b = (d*b^k+...+d*b^0) mod b = (d*b^k+...+d*b^1) mod b + (d*b^0) mod b = 0 + (d*1) mod b. Since d is less than the base we end up with the formula n mod b = d. - Jon Maiga, May 31 2021

			For n = 45:
- we have:
     b  45 in base b  Repdigit ?
     -  ------------  ----------
     2  101101        no
     3  1200          no
     4  231           no
     5  140           no
     6  113           no
     7  63            no
     8  55            yes, with d = 5
- hence a(45) = 5.

David A. Corneth, Table of n, a(n) for n = 0..9999

Cf. A059711, A053696, A060775, A220570.

a(n) = for (b=2, oo, if (#Set(digits(n,b))<=1, return (n%b)))

# with library / without (faster for large n)
from sympy.ntheory import digits
def is_repdigit(n, b): return len(set(digits(n, b)[1:])) == 1
def is_repdigit(n, b):
  if n < b: return True
  n, r = divmod(n, b)
  onlyd = r
  while n > b:
    n, r = divmod(n, b)
    if r != onlyd: return False
  return n == onlyd
def a(n):
  for b in range(2, n+3):
    if is_repdigit(n, b): return n%b
print([a(n) for n in range(87)]) # Michael S. Branicky, May 31 2021

n is a multiple of a(n).

a(n) = n mod A059711(n). - Jon Maiga, May 31 2021

cp's OEIS Frontend

A208242 Perfect powers y^q with y > 1 and q > 1 which are Brazilian repunits with three or more digits in some base.

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PARI

Extensions

A090503 Number of hyperplanes in a finite projective space (of some dimension d over some finite field of order q).

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Haskell

Mathematica

PARI

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A125598 a(n) = ((n+1)^(n-1) - 1)/n.

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Magma

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Sage

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A325658 Brazilian composites of the form 1 + b + b^2 + b^3 + ... + b^k, b > 1, k > 1.

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Maple

PARI

A376298 Numbers which are the sum of at least three successive terms of a geometric progression.

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PARI

A275766 a(n) = (5^(2*(n + 1)) - 1)/4.

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Mathematica

PARI

PARI

Formula

A282090 Totient numbers (A002202) of the form 1 + k + k^2 + k^3 +...+ k^i (i > 1, k > 1).

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