A125134 -id:A125134 - cp's OEIS frontend

A286094 Nonprime numbers n such that n^4 + n^3 + n^2 + n + 1 is prime.

1, 12, 22, 24, 28, 30, 40, 44, 50, 62, 63, 68, 74, 77, 85, 94, 99, 110, 117, 118, 120, 122, 129, 134, 143, 145, 154, 162, 164, 165, 172, 175

Offset: 1

Bernard Schott, May 02 2017

nonn

A065509 Union {this sequence} = A049409.

The corresponding prime numbers n^4 + n^3 + n^2 + n + 1 = 11111_n are in A193366; these Brazilian primes, except 5 which is not Brazilian, belong to A085104 and A285017.

			12 is in the sequence because 12^4 + 12^3 + 12^2 + 12 + 1 = 11111_12 = 22621, which is prime.

R. J. Cano, Table of n, a(n) for n = 1..10000
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38.

Cf. A049409, A065509, A085104, A088548, A125134, A190527, A193366, A285017.

Select[Range@ 414, And[! PrimeQ@ #, PrimeQ[Total[#^Range[0, 4]]]] &] (* Michael De Vlieger, May 03 2017 *)

isok(n)=if(n==1,5,if(ispseudoprime(n), 0, isprime(fromdigits([1, 1, 1, 1, 1], n))));
getfirstterms(n)={my(L:list, c:small); L=List(); c=0; forstep(k=1, +oo, 1, if(isok(k), listput(L, k); if(c++==n, break))); return(Vec(L))} \\ R. J. Cano, May 09 2017

A306845 Sophie Germain primes which are Brazilian.

Original entry on oeis.org

28792661, 78914411, 943280801, 7294932341, 30601685951, 919548423641, 2275869057821, 4172851565741, 4801096143881, 27947620155401, 29586967653101, 43573806645461, 119637719305001, 124484682222941, 148908227169101, 172723673300501

Offset: 1

Bernard Schott, Mar 13 2019

nonn

These terms point out that the conjecture proposed in Quadrature "No Sophie Germain prime is Brazilian (prime)" (see link) was false.
Giovanni Resta has found the first counterexample of Sophie Germain prime which is Brazilian. It's the 141385th Sophie Germain prime 28792661 = 1 + 73 + 73^2 + 73^3 + 73^4 = (11111)73. The other counterexamples have been found by _Michel Marcus.
These numbers are relatively rare: only 25 terms < 10^15.
The 47278 initial terms of this sequence are of the form (11111)_b. The successive bases b are 73, 94, 175, 292, 418, 979, 1228, 1429, ...
The first term which is not of this form has 32 digits, it is 14781835607449391161742645225951 = 1 + 1309 + ... + 1309^9 + 1309^10 = (11111111111)_1309 with a string of eleven 1's. In this case, the successive bases b are 1309, 1348, 2215, 2323, 2461, ...
If (b^q - 1)/(b - 1) is a term, necessarily q (prime) == 5 (mod 6) and b == 1 (mod 3). The smallest term for each pair (q,b) is: (5,73), (11,1309), (17,1945), (23,20413), (29,5023), (41,9565), (47,2764) (See link Jon Grantham, Hester Graves).
Other smallest pairs (q, b) are: (53, 139492), (59, 154501), (71, 7039), (83, 9325), (89, 78028), (101, 8869), (107, 86503), (113, 89986), (131, 429226), (137, 929620), (149, 1954), (167, 175), (173, 1368025). - David A. Corneth, Mar 13 2019

			78914411 is a term because 2 * 78914411 + 1 = 157828823 is prime, so 78914411 is Sophie Germain prime, then, 78914411 = 1 + 94 + 94^2 + 94^3 + 94^4 = (11111)_94 and 78914411 is also a Brazilian prime.

Jon Grantham, Hester Graves, Brazilian Primes Which Are Also Sophie Germain Primes, arXiv:1903.04577 [math.NT], 2019.
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

Intersection of A005384 and A085104.

Cf. A007528, A306849.

lista(lim)=my(v=List(), t, k); for(n=2, sqrt(lim), t=1+n; k=1; while((t+=n^k++)<=lim, if(isprime(t) && isprime(2*t+1), listput(v, t)))); v = vecsort(Vec(v), , 8); \\ Michel Marcus, Mar 13 2019

A333858 Numbers that are both Colombian and Brazilian.

Original entry on oeis.org

7, 20, 31, 42, 64, 75, 86, 108, 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 211, 222, 244, 255, 266, 288, 299, 310, 312, 323, 334, 345, 356, 378, 400, 411, 413, 424, 435, 446, 468, 490, 501, 512, 514, 525, 536, 558, 580, 591, 602, 615, 626, 637, 648, 670, 681, 692

Offset: 1

Bernard Schott, Apr 08 2020

121 is the only square of prime in this sequence.

			20 is a term because it is not of the form m + sum of digits of m for any m < 20, so 20 is Colombian and 20 = (22)_9, so 20 is also Brazilian.

Wikipédia, Nombre brésilien (in French).
Index entries for sequences related to Brazilian Numbers.
Index entries for sequences related to Colombian Numbers.

Intersection of A003052 and A125134.

brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; n = 700; Select[Complement[Range[n], Union @ Table[Plus @@ IntegerDigits[k] + k, {k, 1, n}]], brazQ] (* Amiram Eldar, Apr 08 2020 after T. D. Noe at A125134 *)

A340796 a(n) is the smallest number with exactly n divisors that are Brazilian.

Original entry on oeis.org

1, 7, 14, 24, 40, 48, 60, 84, 140, 144, 120, 168, 252, 700, 240, 336, 560, 360, 420, 672, 1120, 2304, 960, 720, 1008, 1080, 840, 2184, 1800, 1260, 2016, 5376, 8960, 2160, 1680, 2880, 4032, 3600, 7056, 19600, 3960, 2520, 3360, 6480, 9072, 9900, 6300, 11520, 16128

Offset: 0

Bernard Schott, Jan 21 2021

Primes can be partitioned into Brazilian primes and non-Brazilian primes. If two distinct primes each larger than 11 are in the same category then the larger one has a multiplicity that is smaller than or equal to that of the smaller prime. - David A. Corneth, Jan 24 2021

			Of the eight divisors of 24, three are Brazilian numbers: 8, 12 and 24, and there is no smaller number with three Brazilian divisors, hence a(3) = 24.

David A. Corneth, Table of n, a(n) for n = 0..427
David A. Corneth, More terms
Wikipédia, Nombre brésilien (in French).
Index entries for sequences related to Brazilian numbers.

Cf. A085104, A125134, A190300, A220570, A308851, A340795, A340797.

Similar with: A087997 (palindromes), A333456 (Niven), A335038 (Zuckerman).

brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; d[n_] := DivisorSum[n, 1 &, brazQ[#] &]; m = 30; s = Table[0, {m}]; c = 0; n = 1; While[c < m, i = d[n]; If[i < m && s[[i + 1]] == 0, c++; s[[i + 1]] = n]; n++]; s (* Amiram Eldar, Jan 21 2021 *)

isokb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1))); \\ A125134
isok(k, n) = sumdiv(k, d, isokb(d)) == n;
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Jan 23 2021

More terms from Amiram Eldar, Jan 21 2021

A340797 Integers whose number of divisors that are Brazilian sets a new record.

Original entry on oeis.org

1, 7, 14, 24, 40, 48, 60, 84, 120, 168, 240, 336, 360, 420, 672, 720, 840, 1260, 1680, 2520, 3360, 5040, 7560, 10080, 15120, 20160, 25200, 27720, 30240, 43680, 45360, 50400, 55440, 65520, 83160, 98280, 110880, 131040, 166320, 196560, 221760, 262080, 277200, 327600

Offset: 1

Bernard Schott, Jan 24 2021

Corresponding number of Brazilian divisors: 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 14, 15, 17, 18, 19, 26, ...

Observation: the 58 consecutive highly composite numbers from A002182(12) = 240 to A002182(69) = 2095133040 (maybe more, according to conjectured terms) are also terms of this sequence.

			40 has 8 divisors: {1, 2, 4, 5, 8, 10, 20, 40} of which 4 are Brazilian: {8, 10, 20, 40}. No positive integer smaller than 40 has as many as four Brazilian divisors; hence 40 is a term.

David A. Corneth, Table of n, a(n) for n = 1..85
David A. Corneth, Conjectured terms
Wikipédia, Nombre brésilien (in French).
Index entries for sequences related to Brazilian numbers.

Cf. A002182, A125134, A220570, A308851, A340795, A340796.

Similar with: A093036 (palindromes), A340548 (repdigits), A340549 (repunits), A340637 (Niven), A340638 (Zuckerman).

brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[ IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; d[n_] := DivisorSum[n, 1 &, brazQ[#] &]; dm = -1; s = {}; Do[d1 = d[n]; If[d1 > dm, dm = d1; AppendTo[s, n]], {n, 1, 1000}]; s (* Amiram Eldar, Jan 24 2021 *)

isb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1))); \\ A125134
nbd(n) = sumdiv(n, d, isb(d)); \\ A340795
lista(nn) = {my(m=-1); for (n=1, nn, my(x = nbd(n)); if (x > m, print1(n, ", "); m = x););} \\ Michel Marcus, Jan 24 2021

a(20)-a(36) from Michel Marcus, Jan 24 2021

a(37)-a(44) from Amiram Eldar, Jan 24 2021

A187823 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.

Original entry on oeis.org

5, 17, 73, 257, 757, 65537, 262657, 1772893, 4432676798593, 48551233240513, 378890487846991, 3156404483062657, 17390284913300671, 280343912759041771, 319913861581383373, 487014306953858713, 5559917315850179173, 7824668707707203971, 8443914727229480773, 32564717507686012813

Offset: 1

Bernard Schott, Dec 27 2012

nonn

Complement of A023195 relative to A003424.
Only eight primes of this form don't exceed 1.275*10^10 (see Bateman and Stemmler):
(1) three of the form (p^9 - 1)/(p^3 - 1): 73 (p=2), 757 (p=3), 1772893 (p=11);
(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
(3) the prime 262657 = (2^27 - 1)/(2^9 - 1).
Some of these prime numbers are not Brazilian, these are Fermat primes > 3: 5, 17, 257, 65537, so they are in A220627.
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).
Comments from Don Reble, Jul 28 2022 (Start)
This is an easy sequence that looks hard.
Note that x must be a power of a prime; otherwise (p^x-1)/(p^y-1) has too many cyclotomic factors.
Almost all values are (p^9-1)/(p^3-1). The exceptions below 10^45
    are the Fermat primes 5, 17, 257, 65537 and also
    262657, 4432676798593, 5559917315850179173,
    227376585863531112677002031251,
    467056170954468301850494793701001,
    36241275390490156321975496980895092369525753,
    284661951906193731091845096405947222295673201 (see examples).
(End)

			5 = (2^4 - 1)/(2^2 - 1)= 11_{2^2} = 11_4.
17 = (2^8 - 1)/(2^4 - 1) = 11_{2^4} = 11_16.
257 = (2^16 - 1)/(2^8 - 1) = 11_{2^8} = 11_256.
757 = (3^9 - 1)/(3^3 - 1) = 111_{3^3} = 111_27.
262657 = (2^27 - 1)/(2^9 - 1) = 111_{2^9} = 111_512.
655357 = (2^32 - 1)/(2^16 - 1) = 11_{2^16} = 11_655356.
4432676798593 = (2^49 - 1)/(2^7 - 1) = 1111111_{2^7} = 1111111_128.
5559917315850179173 = (11^27 - 1)/(11^9 - 1) = 111_{11^3} = 111_1331.
227376585863531112677002031251 = (5^49 - 1)/(5^7 - 1) = 1111111_{5^7}.
467056170954468301850494793701001 = (43^25 - 1)/(43^5 - 1) = 11111_{43^5}.
36241275390490156321975496980895092369525753 = (263^27 - 1)/(263^9 - 1).
284661951906193731091845096405947222295673201 = (167^25 - 1)/(167^5 - 1).

Don Reble, Table of n, a(n) for n = 1..50000
Paul T. Bateman and Rosemarie M. Stemmler, Waring's problem for algebraic number fields and primes of the form (p^r-1)/(p^d-1), Illinois J. Math. 6 (1962), pp. 142-156.
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.
Index entries for sequences related to Brazilian numbers.

Equals A003424 \ A023195.

Cf. A085104, A220627.

a(9)-a(16) from Don Reble, Jul 28 2022

a(17)-a(20) from Don Reble, Mar 21 2023

A189891 Complement of A085104.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 79, 80

Offset: 1

N. J. A. Sloane, May 14 2011

nonn

Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

A253260 Brazilian squares.

Original entry on oeis.org

16, 36, 64, 81, 100, 121, 144, 196, 225, 256, 324, 400, 441, 484, 576, 625, 676, 729, 784, 900, 1024, 1089, 1156, 1225, 1296, 1444, 1521, 1600, 1764, 1936, 2025, 2116, 2304, 2401, 2500, 2601, 2704, 2916, 3025, 3136, 3249, 3364, 3600, 3844, 3969, 4096, 4225, 4356, 4624, 4761, 4900, 5184

Offset: 1

Derek Orr, Apr 30 2015

Trivially, all even squares > 4 will be in this sequence.
The only square of a prime which is Brazilian is 121. - Bernard Schott, May 01 2017
Intersection of A000290 and A125134. - Felix Fröhlich, May 01 2017
Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1); then Product_{n>=1} r(n) = (15/17) * (35/37) * (63/65) * (40/41) * (99/101) * (60/61) * (143/145) * (195/197) * ... = (150 * Pi) / (61 * sinh(Pi)) = 0.668923905.... - Dimitris Valianatos, Feb 27 2019

			From _Bernard Schott_, May 01 2017: (Start)
a(1) = 16 = 4^2 = 22_7.
a(6) = 121 = 11^2 = 11111_3. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..405
Bernard Schott, Les nombres brésiliens Quadrature, no. 76, avril-juin 2010, théorème 5, page 37.

Cf. A000290, A125134.

fQ[n_]:=Module[{b=2, found=False}, While[b1, b++]; bVincenzo Librandi, May 02 2017 *)

for(n=4, 10^4, for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d)&&issquare(n), print1(n, ", "); break)))

A253261 Odd Brazilian squares.

Original entry on oeis.org

81, 121, 225, 441, 625, 729, 1089, 1225, 1521, 2025, 2401, 2601, 3025, 3249, 3969, 4225, 4761, 5625, 5929, 6561, 7225, 7569, 8281, 8649, 9025, 9801, 11025, 12321, 13225, 13689, 14161, 14641, 15129, 15625, 16641, 17689, 18225, 19881, 20449, 21025, 21609, 23409, 24025, 25281, 25921, 27225, 28561

Offset: 1

Derek Orr, Apr 30 2015

121 is believed to be the only number of the form p^2 for prime p.

The previous comment conjectures the 1 and the 121 are the only difference with respect to A062532. - R. J. Mathar, Jul 25 2015

Cf. A125134, A257521.

for(n=4, 10^5, for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d)&&(n+1)%2==0&&issquare(n), print1(n, ", "); break)))

A258165 Odd non-Brazilian numbers > 1.

Original entry on oeis.org

3, 5, 9, 11, 17, 19, 23, 25, 29, 37, 41, 47, 49, 53, 59, 61, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 131, 137, 139, 149, 151, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 223, 227, 229, 233, 239, 251, 257, 263, 269, 271, 277, 281, 283, 289, 293, 311, 313, 317, 331, 337

Offset: 1

Daniel Lignon and Robert G. Wilson v, May 22 2015

nonn

Complement of A257521 in A144396 (odd numbers > 1).
The terms are only odd primes or squares of odd primes.
Most odd primes are present except those in A085104.
All terms which are not primes are squares of odd primes except 121 = 11^2.

			11 is present because there is no base b < 11 - 1 = 10 such that the representation of 11 in base b is a repdigit (all digits are equal). In fact, we have: 11 = 1011_2 = 102_3 = 23_4 = 21_5 = 15_6 = 14_7 = 13_8 = 12_9, and none of these representations are repdigits. - _Bernard Schott_, Jun 21 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..1150

Cf. A085104, A125134, A190300, A220570, A220571, A220627, A242397, A253260, A253261, A257521.

fQ[n_] := Block[{b = 2}, While[b < n - 1 && Length@ Union@ IntegerDigits[n, b] > 1, b++]; b+1 == n]; Select[1 + 2 Range@ 170, fQ]

forstep(n=3, 300, 2, c=1; for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), c=0;  break));if(c,print1(n,", "))) \\ Derek Orr, May 27 2015

from sympy.ntheory.factor_ import digits
l=[]
for n in range(3, 301, 2):
    c=1
    for b in range(2, n - 1):
        d=digits(n, b)[1:]
        if max(d)==min(d):
            c=0
            break
    if c: l.append(n)
print(l) # Indranil Ghosh, Jun 22 2017, after PARI program

cp's OEIS Frontend

A286094 Nonprime numbers n such that n^4 + n^3 + n^2 + n + 1 is prime.

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A306845 Sophie Germain primes which are Brazilian.

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A333858 Numbers that are both Colombian and Brazilian.

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A340796 a(n) is the smallest number with exactly n divisors that are Brazilian.

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A340797 Integers whose number of divisors that are Brazilian sets a new record.

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A187823 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.

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A189891 Complement of A085104.

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A253260 Brazilian squares.

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