cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A341058 Numbers that have only one divisor that is Brazilian.

Original entry on oeis.org

7, 8, 10, 12, 13, 15, 18, 22, 27, 31, 33, 34, 38, 43, 46, 49, 51, 55, 57, 58, 69, 73, 74, 82, 85, 87, 94, 95, 106, 111, 115, 118, 121, 122, 123, 125, 127, 134, 141, 142, 145, 157, 158, 159, 166, 169, 177, 178, 183, 185, 187, 194, 201, 202, 205, 206, 209, 211, 213, 214, 218
Offset: 1

Views

Author

Bernard Schott, Feb 15 2021

Keywords

Comments

m is a term iff m is a Brazilian prime (A085104), or m is the square of a Brazilian prime, or m = 121, the only square of prime that is Brazilian, or m = p*q >= 10 with p>q are non-Brazilian primes, or m is the cube of a Brazilian prime, or m = 12 or 18 (see corresponding examples).

Examples

			One example for each type of terms that has k divisors:
-> k=2: 7 is a Brazilian prime, hence 7 = 111_2  is a term.
-> k=3: 169 has three divisors {1, 13, 169} and 13 = 111_3 is the only divisor of 169 that is Brazilian, hence 169 is a term.
-> k=3: 121 has three divisors {1, 11, 121} and 121 = 11111_3, hence, 121 that is the only square of prime that is Brazilian, is a term.
-> k=4: 34 has four divisors {1, 2, 17, 34} and 34 = 22_16  is the only divisor of 34 that is Brazilian, hence 34 is a term.
-> k=4: 27 has four divisors {1, 3, 9, 27} and 27 = 33_8 is the only divisor of 27 that is Brazilian, hence 27 is a term.
-> k=6: only two cases: 12 and 18, these integers have each 6 divisors and only 12 = 22_5 and 18 = 33_5 are Brazilian.

Crossrefs

Cf. A125134, A220627, A308851, A340795, A341057.
Subsequence: A085104.

Programs

  • Mathematica
    brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length @ Union[IntegerDigits[n, b]] > 1, b++]; b < n - 1]; Select[Range[200], DivisorSum[#, 1 &, brazQ[#1] &] == 1 &] (* Amiram Eldar, Feb 16 2021 *)

Formula

A340795(a(n)) = 1.

A343774 Primes of the form (c^k+1)/(c+1) not having a representation in the form (b^q-1)/(b-1), where b, c > 1 and k, q > 2.

Original entry on oeis.org

3, 11, 61, 521, 547, 683, 2731, 9091, 13421, 19141, 43691, 61681, 152381, 174763, 185641, 224071, 398581, 909091, 1151041, 1623931, 1824841, 2031671, 2796203, 3341101, 4778021, 5200081, 7027567, 8987221, 10678711, 15790321, 22796593, 25058741, 31224301, 32222107
Offset: 1

Views

Author

Bernard Schott, Apr 29 2021

Keywords

Comments

The exponents k, q are necessarily primes.
Equivalently: primes of the form (c^k+1)/(c+1) that are not Brazilian: intersection of A059055 and A220627.
Except for 3 where k = 3, all the terms of this sequence are of the form (c^k+1)/(c+1) with k prime >= 5.
The only known prime of this form with k prime >= 5 that is not present is 43 = (2^7+1)/(2+1) because also 43 = (7^3+1)/(7+1) = (6^3-1)/(6-1) = 111_6, so 43 belongs to A002383.

Examples

			3 = (2^3+1)/(2+1) is not Brazilian, hence 3 is a term.
11 = (2^5+1)/(2+1) is not Brazilian, hence 11 is a term.
547 = (3^7+1)/(3+1) is not Brazilian, hence 547 is a term.
9091 = (10^5+1)/(10+1) is not Brazilian, hence 9091 is a term.

Links

Crossrefs

Primes of the form (b^k-1)/(b-1) = A085104 (Brazilian primes).
Primes of the form (c^q+1)/(c+1) = A059055.
Primes of the form (b^k-1)/(b-1) and (c^q+1)/(c+1): A002383 \ {3} is a subsequence, but, maybe the intersection (conjecture).
Primes of the form (b^k-1)/(b-1) but not (c^q+1)/(c+1) = A225148.
Primes of the form (c^q+1)/(c+1) but not (b^k-1)/(b-1) = this sequence.
Primes neither of the form (c^q+1)/(c+1) nor (b^k-1)/(b-1) = A343775.

Programs

  • PARI
    isc(p) = for (b=2, p, my(k=3); while ((x=(b^k+1)/(b+1)) <= p, if (x == p, return (1)); k = nextprime(k+1); ); );
isnotb(p) = for (b=2, p-1, my(d=digits(p, b), md=vecmin(d)); if ((#d > 2) && (md == 1) && (vecmax(d) == 1), return (0)); ); return (1);
isok(p) = isprime(p) && isc(p) && isnotb(p); \\ Michel Marcus, May 01 2021

Extensions

More terms from Michel Marcus, Apr 30 2021

A343775 Primes that are neither of the form (c^q+1)/(c+1) and nor of the form (b^k-1)/(b-1) for any b, c > 1 and k, q primes > 2.

Original entry on oeis.org

2, 5, 17, 19, 23, 29, 37, 41, 47, 53, 59, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 131, 137, 139, 149, 151, 163, 167, 173, 179, 181, 191, 193, 197, 199, 223, 227, 229, 233, 239, 251, 257, 263, 269, 271, 277, 281, 283, 293, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367
Offset: 1

Views

Author

Bernard Schott, Apr 29 2021

Keywords

Comments

Equivalently, non-Brazilian primes that are not of the form (c^q+1)/(c+1) for some c > 1, q prime > 2.
Equals A220627 \ A059055.

Crossrefs

Primes of the form (b^k-1)/(b-1) = A085104 (Brazilian primes).
Primes of the form (c^q+1)/(c+1) = A059055.
Primes of the form (b^k-1)/(b-1) and also (c^q+1)/(c+1): A002383 \ {3} is a subsequence, but, maybe the intersection (conjecture).
Primes of the form (b^k-1)/(b-1) but not (c^q+1)/(c+1) = A225148.
Primes of the form (c^q+1)/(c+1) but not (b^k-1)/(b-1) = A343774.
Primes neither of the form (c^q+1)/(c+1) nor (b^k-1)/(b-1) = this sequence.

Programs

  • PARI
    isc(p) = for (b=2, p, my(k=3); while ((x=(b^k+1)/(b+1)) <= p, if (x == p, return (1)); k = nextprime(k+1); ); );
isnotb(p) = for (b=2, p-1, my(d=digits(p, b), md=vecmin(d)); if ((#d > 2) && (md == 1) && (vecmax(d) == 1), return (0)); ); return (1);
isok(p) = isprime(p) && !isc(p) && isnotb(p); \\ Michel Marcus, May 01 2021

A361914 Primes that are repunits with three or more digits for exactly one base b >= 2.

Original entry on oeis.org

7, 13, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801, 2971, 3307, 3541, 3907, 4423, 4831, 5113, 5701, 6007, 6163, 6481, 8011, 9901, 10303, 11131, 12211, 12433, 13807, 14281, 17293, 19183, 19531, 20023, 20593, 21757, 22621, 22651, 23563
Offset: 1

Views

Author

Bernard Schott, Mar 29 2023

Keywords

Comments

Brazilian primes that have exactly one Brazilian representation as a repunit.
As these primes p satisfy beta(p) = tau(p) / 2 (= 1), where beta = A220136 and tau = A000005, this sequence is a subsequence of A326380.
Equals A085104 \ {31, 8191}, since according to the Goormaghtigh conjecture (link), 31 and 8191 which are both Mersenne numbers, are the only primes which are Brazilian in two different bases.
The three following sequences realize a partition of the set of primes: A220627 (primes not Brazilian), this sequence (primes 1-Brazilian) and {31,8191} (primes 2-Brazilian).

Examples

			7 = 111_2 is a term.
13 = 111_3 is a term.
19 = 11_18 is not a term.
31 = 11111_5 = 111_5 is not a term.
127 = 1111111_2 is a term.
8191 = 1111111111111_2 = 111_90 is not a term.

Links

Crossrefs

Cf. A000005, A085104, A220136, A220627, A326380.
Equals A326380 \ {A326385 Union A326387}.
Subsequence of A288783.

Programs

Previous Showing 11-14 of 14 results.

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