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.

Showing 1-6 of 6 results.

A340795 a(n) is the number of divisors of n that are Brazilian.

Original entry on oeis.org

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

Views

Author

Bernard Schott, Jan 21 2021

Keywords

Comments

The cases a(n) = 0 and a(n) = 1 are respectively detailed in A341057 and A341058.

Examples

			For n = 16, the divisors are 1, 2, 4, 8 and 16. Only 8 = 22_3 and 16 = 22_7 are Brazilian numbers, so a(16) = 2.
For n = 30, the divisors are 1, 2, 3, 5, 6, 10, 15 and 30. Only 10 = 22_4, 15 = 33_4 and 30 = 33_9 are Brazilian numbers, so a(30) = 3.
For n = 49, the divisors are 1, 7 and 49. Only 7 = 111_2 is Brazilian, so a(49) = 1 although 49 that is square of prime <> 121 is not Brazilian.

Links

Crossrefs

Cf. A085104, A125134, A220627, A308851, A340796, A340797, A341057, A341058.

Programs

  • Mathematica
    brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; a[n_] := DivisorSum[n, 1 &, brazQ[#] &]; Array[a, 100] (* Amiram Eldar, Jan 21 2021 *)
  • PARI
    isb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1))); \\ A125134
a(n) = sumdiv(n, d, isb(d)); \\ Michel Marcus, Jan 24 2021

A337941 Numbers whose divisors are all Zuckerman numbers (A007602).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 24, 1111111111111111111, 11111111111111111111111
Offset: 1

Views

Author

Bernard Schott, Oct 01 2020

Keywords

Comments

Inspired by A337741.
Zuckerman numbers are numbers that are divisible by the product of their digits (see link).
The next term is the repunit prime R_317 which is too large to include in the data.
Primes in this sequence are 2, 3, 5, 7 and all the repunit primes (see A004023).
This sequence is infinite if and only if there are infinitely many repunit primes.

Examples

			6 is a term since all the divisors of 6, i.e., 1, 2, 3 and 6, are Zuckerman numbers.

Links

Crossrefs

Subsequence of A007602.
Similar sequences: A062687, A190217, A308851, A329419, A337741.
Cf. A004023, A335037, A335038.
Cf. A004022 (subsequence of prime repunits).

Programs

  • Mathematica
    zuckQ[n_] := (prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod]; Select[Range[24], AllTrue[Divisors[#], zuckQ] &] (* Amiram Eldar, Oct 01 2020 *)
  • PARI
    isok(m) = {fordiv(m, d, my(p=vecprod(digits(d))); if (!p || (d % p), return (0))); return (1);} \\ Michel Marcus, Oct 05 2020

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

Views

Author

Bernard Schott, Jan 21 2021

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

Cf. A085104, A125134, A190300, A220570, A308851, A340795, A340797.
Similar with: A087997 (palindromes), A333456 (Niven), A335038 (Zuckerman).

Programs

  • Mathematica
    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 *)
  • PARI
    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

Extensions

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

Views

Author

Bernard Schott, Jan 24 2021

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A002182, A125134, A220570, A308851, A340795, A340796.
Similar with: A093036 (palindromes), A340548 (repdigits), A340549 (repunits), A340637 (Niven), A340638 (Zuckerman).

Programs

  • Mathematica
    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 *)
  • PARI
    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

Extensions

a(20)-a(36) from Michel Marcus, Jan 24 2021
a(37)-a(44) from Amiram Eldar, Jan 24 2021

A341057 Numbers without Brazilian divisors.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 11, 17, 19, 23, 25, 29, 37, 41, 47, 53, 59, 61, 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, 289, 293, 311
Offset: 1

Views

Author

Bernard Schott, Feb 04 2021

Keywords

Comments

The first 16 terms are the first 16 terms of A220570 (non-Brazilian numbers), then a(17) = 53 while A220570(17) = 49.
m is a term iff m = 1, or m = 6, or m is a non-Brazilian prime (A220627) or m is the square of a non-Brazilian prime, except for 121 that is Brazilian (see examples).

Examples

			One example for each type of terms that has k divisors:
-> k=1: 1 is the smallest number not Brazilian, hence 1 is the first term.
-> k=2: 17 is a prime non-Brazilian, hence 17 is a term.
-> k=3: 25 has three divisors {1, 5, 25} that are all not Brazilian, hence 25 is another term.
-> k=4: 6 has four divisors {1, 2, 3, 6} that are all not Brazilian, hence 6 is the term that has the largest number of divisors.

Crossrefs

Cf. A125134, A340795, A308851, A341058 (with 1 Brazilian divisor).
Subsequence of A220570 (non-Brazilian numbers).
Supersequence of A220627 (non-Brazilian primes).

Programs

  • Mathematica
    brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length @ Union @ IntegerDigits[n, b] > 1, b++]; b < n - 1]; q[n_] := AllTrue[Divisors[n], ! brazQ[#] &]; Select[Range[300], q] (* Amiram Eldar, Feb 04 2021 *)
  • PARI
    isb(n) = for(b=2, n-2, my(d=digits(n, b)); if(vecmin(d)==vecmax(d), return(1))); \\ A125134
isok(n) = fordiv(n, d, if (isb(d), return(0))); return(1); \\ Michel Marcus, Feb 07 2021

Formula

A340795(a(n)) = 0.

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.
Showing 1-6 of 6 results.

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