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

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

Author

Bernard Schott, May 12 2019

Keywords

• nonn
• base

Comments

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.

Examples

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

Links

• 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.

Crossrefs

Complement of A085104 with respect to A053696.

Intersection of A053696 and A220571.

Cf. A119598, A208242, A325659.

Programs

Maple

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

PARI

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