login
Primes of the form (p^x - 1)/(p^y - 1), where p is prime, y > 1, and y is the largest proper divisor of x.
2

%I #68 Mar 29 2023 08:59:26

%S 5,17,73,257,757,65537,262657,1772893,4432676798593,48551233240513,

%T 378890487846991,3156404483062657,17390284913300671,

%U 280343912759041771,319913861581383373,487014306953858713,5559917315850179173,7824668707707203971,8443914727229480773,32564717507686012813

%N Primes of the form (p^x - 1)/(p^y - 1), where p is prime, y > 1, and y is the largest proper divisor of x.

%C Complement of A023195 relative to A003424.

%C Only eight primes of this form don't exceed 1.275*10^10 (see Bateman and Stemmler):

%C (1) three of the form (p^9 - 1)/(p^3 - 1): 73 (p=2), 757 (p=3), 1772893 (p=11);

%C (2) four of the form (2^x - 1)/(2^y - 1) with x = 2y: 5 (x=4), 17 (x=8), 257 (x=16), 65537 (x=32); and

%C (3) the prime 262657 = (2^27 - 1)/(2^9 - 1).

%C Some of these prime numbers are not Brazilian, these are Fermat primes > 3: 5, 17, 257, 65537, so they are in A220627.

%C The other primes are Brazilian so they are in A085104, example: (p^9 - 1)/(p^3 - 1) = 111_{p^3} with 73 = 111_8, 757 = 111_27, 1772893 = 111_1331, also 262657 = 111_512 [See section V.4 of Quadrature article in Links] (comment improved in Mar 03 2023).

%C Comments from _Don Reble_, Jul 28 2022 (Start)

%C This is an easy sequence that looks hard.

%C Note that x must be a power of a prime; otherwise (p^x-1)/(p^y-1) has too many cyclotomic factors.

%C Almost all values are (p^9-1)/(p^3-1). The exceptions below 10^45

%C are the Fermat primes 5, 17, 257, 65537 and also

%C 262657, 4432676798593, 5559917315850179173,

%C 227376585863531112677002031251,

%C 467056170954468301850494793701001,

%C 36241275390490156321975496980895092369525753,

%C 284661951906193731091845096405947222295673201 (see examples).

%C (End)

%H Don Reble, <a href="/A187823/b187823.txt">Table of n, a(n) for n = 1..50000</a>

%H Paul T. Bateman and Rosemarie M. Stemmler, <a href="https://doi.org/10.1215/ijm/1255631815">Waring's problem for algebraic number fields and primes of the form (p^r-1)/(p^d-1)</a>, Illinois J. Math. 6 (1962), pp. 142-156.

%H Bernard Schott, <a href="/A125134/a125134.pdf">Les nombres brésiliens</a>, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

%H <a href="/index/Br#Brazilian_numbers">Index entries for sequences related to Brazilian numbers</a>.

%e 5 = (2^4 - 1)/(2^2 - 1)= 11_{2^2} = 11_4.

%e 17 = (2^8 - 1)/(2^4 - 1) = 11_{2^4} = 11_16.

%e 257 = (2^16 - 1)/(2^8 - 1) = 11_{2^8} = 11_256.

%e 757 = (3^9 - 1)/(3^3 - 1) = 111_{3^3} = 111_27.

%e 262657 = (2^27 - 1)/(2^9 - 1) = 111_{2^9} = 111_512.

%e 655357 = (2^32 - 1)/(2^16 - 1) = 11_{2^16} = 11_655356.

%e 4432676798593 = (2^49 - 1)/(2^7 - 1) = 1111111_{2^7} = 1111111_128.

%e 5559917315850179173 = (11^27 - 1)/(11^9 - 1) = 111_{11^3} = 111_1331.

%e 227376585863531112677002031251 = (5^49 - 1)/(5^7 - 1) = 1111111_{5^7}.

%e 467056170954468301850494793701001 = (43^25 - 1)/(43^5 - 1) = 11111_{43^5}.

%e 36241275390490156321975496980895092369525753 = (263^27 - 1)/(263^9 - 1).

%e 284661951906193731091845096405947222295673201 = (167^25 - 1)/(167^5 - 1).

%Y Equals A003424 \ A023195.

%Y Cf. A085104, A220627.

%K nonn

%O 1,1

%A _Bernard Schott_, Dec 27 2012

%E a(9)-a(16) from _Don Reble_, Jul 28 2022

%E a(17)-a(20) from _Don Reble_, Mar 21 2023