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 1-9 of 9 results.

A158236 The base for the numbers in A158235.

Original entry on oeis.org

3, 7, 22, 7, 18, 22, 30, 68, 68, 313, 313, 41, 313, 146, 653, 313, 499, 313, 710, 313, 653, 313, 423, 313, 292, 499, 313, 653, 313, 710, 439, 499, 653, 699, 99, 4366, 423, 653, 521, 581, 499, 710, 653, 653, 710, 4701, 653, 699, 4366, 710, 653, 4611, 787, 2272
Offset: 1

Views

Author

T. D. Noe, Mar 14 2009

Keywords

Comments

The prime 313 appears 10 times. Which numbers are allowed as bases?

Examples

			From _Michael De Vlieger_, Sep 08 2017: (Start)
Bases that appear, for numbers 2 <= m <= 101000 in A158235, i.e., for 1 <= n <= 100. f = frequency of b in a(n) for m in the range shown above:
Base b   f  Prime factors of b
------------------------------
     3   1  3
     7   2  7
    18   1  2, 3, 3
    22   2  2, 11
    30   1  2, 3, 5
    41   1  41
    68   2  2, 2, 17
    99   1  3, 3, 11
   146   1  2, 73
   239   2  239
   292   1  2, 2, 73
   313  10  313
   423   2  3, 3, 47
   439   1  439
   499   4  499
   521   1  521
   581   1  7, 83
   653   9  653
   699   2  3, 233
   710   6  2, 5, 71
   787   1  787
  1047   1  3, 349
  1353   1  3, 11, 41
  1425   1  3, 5, 5, 19
  1660   1  2, 2, 5, 83
  1714   1  2, 857
  2060   1  2, 2, 5, 103
  2174   1  2, 1087
  2198   1  2, 7, 157
  2272   5  2, 2, 2, 2, 2, 71
  2819   4  2819
  3019   1  3019
  4366  13  2, 37, 59
  4526   2  2, 31, 73
  4611   6  3, 29, 53
  4701   8  3, 1567
  6205   1  5, 17, 73
(End)

Links

Crossrefs

Cf. A158235, A158237.

Programs

A158237 The repdigit for the numbers in A158235.

Original entry on oeis.org

1, 1, 3, 4, 7, 12, 19, 13, 52, 3, 12, 21, 27, 127, 7, 48, 21, 75, 19, 108, 28, 147, 97, 192, 237, 84, 243, 63, 300, 76, 201, 189, 112, 109, 58, 3, 388, 175, 283, 247, 336, 171, 252, 343, 304, 7, 448, 436, 12, 475, 567, 13, 453, 57, 684
Offset: 1

Views

Author

T. D. Noe, Mar 14 2009

Keywords

Comments

Which "digits" are allowed? For instance, do we know that 2, 5, and 9 never appear?

Links

Crossrefs

Cf. A158235, A158236.

Programs

A158245 Primitive numbers in A158235.

Original entry on oeis.org

11, 20, 39, 49, 133, 247, 543, 1218, 1651, 1729, 2289, 3097, 4171, 4503, 6231, 7303, 7540, 7563, 8773, 9139, 12439, 16627, 16761, 17157, 20280, 22021, 22393, 37051, 37209, 37387, 45201, 47257, 51961, 65379, 66211, 69601, 100191, 103861, 105339
Offset: 1

Views

Author

T. D. Noe, Mar 14 2009

Keywords

Comments

The primitive terms in A158235 are the ones for which the repdigit is squarefree. Hence n^2 = d*(b^k-1)/(b-1) with d squarefree and d

Examples

			105339^2 = 333 in base 60817.

A167782 Numbers that are repdigits with length > 2 in some base.

Original entry on oeis.org

0, 7, 13, 15, 21, 26, 31, 40, 42, 43, 57, 62, 63, 73, 80, 85, 86, 91, 93, 111, 114, 121, 124, 127, 129, 133, 146, 156, 157, 170, 171, 172, 182, 183, 211, 215, 219, 222, 228, 241, 242, 255, 259, 266, 273, 285, 292, 307, 312, 314, 333, 341, 342, 343, 364, 365, 366
Offset: 1

Views

Author

Andrew Weimholt, Nov 12 2009

Keywords

Comments

Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
0 included at the suggestion of Franklin T. Adams-Watters (and others) as 0 = 000 in any base.

Examples

			26 is a term because 26_10 = 222_3.

Links

Crossrefs

Cf. A010785 (Repdigits (base 10)).
Cf. A167783 (Numbers that are repdigits with length > 2 in more than one base).
Cf. A053696 (Numbers which are repunits in some base).
Cf. A158235 (Numbers n whose square can be represented as a repdigit number in some base < n).

Programs

  • PARI
    /* In PARI versions < 2.6, define: digits(n,b) = if(n=b^2+b+1,d=digits(n,b);if(is_repdigit(d),print(n," = ",d," base ",b));b++)) \\ Michael B. Porter

A167783 Numbers that are repdigits with length > 2 in more than one base.

Original entry on oeis.org

31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 3906, 4095, 4557, 6560, 7566, 7812, 8191, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698
Offset: 1

Views

Author

Andrew Weimholt, Nov 12 2009

Keywords

Comments

Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
From Daniel Forgues, Nov 13 2009: (Start)
0 = 00 = 000 = 0000 = 00000 = 000000 = 0000000 = 00000000 = ... in any positional number representation (includes fixed base radix b > 1, mixed base radix with each b_i > 1, i >= 0, such as factorial and primorial based radix...)
The sequence definition should be read as:
Nonnegative integers that are repdigits with length > 2 in more than one fixed base radix b > 1.
Considering all fixed and mixed base radix would include many more nonnegative integers (but not the integers 1 to 6) which are repdigits with length > 2 in more than one radix. (End)
From Bernard Schott, Aug 08 2017: (Start)
In this sequence data, the first number which is repdigit, with length > 2, in more than two bases is the twelfth Mersenne number 4095 with four Brazilian representations: M_12 = 4095 = 111111111111_2 = 333333_4 = 7777_8 = (15 15 15)_16.
The Mersenne number M_15 is the first number which is repdigit in exactly three bases with M_15 = 32767 = 111111111111111_2 = 77777_8 = (31 31 31)_32.
Only two numbers are repunits in more than one base: the Mersenne primes 31 and 8191 (Examples and A119598).
Some numbers are once repunit and once multiple of a Brazilian prime such that Mersenne number M_9 = 511 = 7 * 73 = 111111111_2 = 7 * 111_8 = 777_8.
Some numbers are once repunit and once multiple of a composite repunit such that Mersenne number M_6 = 63 = 3 * 21 = 111111_2 = 3 * 111_4 = 333_4.
Some numbers are repdigits in two different bases: 546 = 666_9 = 222_16. (End)

Examples

			31 is in the list because 31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90;
10507 = {19 19 19}_23 = 111_102.

Links

Crossrefs

Cf. A167782 (numbers that are repdigits with length > 2 in some base).
Cf. A010785 (repdigits (base 10)).
Cf. A053696 (numbers which are repunits in some base).
Cf. A158235 (numbers n whose square is a repdigit in some base < n).
Cf. A290869 (Numbers that are repdigits with length > 2 in more than two bases).

Programs

Extensions

a(41)-a(44) from Bernard Schott, Aug 08 2017

A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.

Original entry on oeis.org

1521, 1600, 2401, 2744, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361
Offset: 1

Views

Author

Bernard Schott, Apr 26 2019

Keywords

Comments

The terms of this sequence are solutions y^m of the Diophantine equation a * (b^q - 1)/(b-1) = y^m with 1 < a < b, y >= 2, q >= 3, m >= 2. This equation has been studied by Kustaa A. Inkeri in Acta Arithmetica; some terms of this sequence come from his article where the author limits the study of this equation to bases b <= 100.
The case a = 1 is clarified in A208242; it corresponds to the Nagell-Ljunggren equation.
The sequence has infinitely many terms because the Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions. - Giovanni Resta, Apr 26 2019
The corresponding solutions (x, y) of this Diophantine equation are (A028231, A341671).
The integers y such that y^2 (m = 2) satisfies this equation are in A158235, except 11 and 20 corresponding to a = 1. - Bernard Schott, Apr 27 2019

Examples

			3 * (22^3-1)/(22-1) = 39^2 and (333)_22 = 39^2 = 1521.
58 * (99^4-1)/(99-1) = 7540^2 and (AAAA)_99 = 7540^2 = 56851600 where A is the symbol for 58 in base 99.

Links

Crossrefs

Subsequence of A001597 and of A125134.
Cf. A158235, A208242 (a=1, that is, with repunits).
Cf. A028231, A341671.

Programs

  • Mathematica
    rupQ[n_, mx_] := Block[{t, x, p}, p = x^2 + x + 1; While[(t = p /. x -> mx) <= n && Reduce[p == n && x >= mx, x, Integers] === False, p = x*p + 1]; t <= n]; repdQ[n_] := AnyTrue[ Rest@ Most@ Divisors@ n, rupQ[n/#, #+1] &]; ex = 2; up = 10^7; L = {}; While[2^ex <= up, L = Union[L, Parallelize@ Select[ Range[2, Floor[ up^(1/ex)] ]^ex, repdQ]]; ex = NextPrime@ ex]; L (* Giovanni Resta, Apr 27 2019 *)
  • PARI
    isokb(n) = for(b=2, n-2, d=digits(n, b); if((#d > 2) && (vecmin(d)==vecmax(d)) && (d[1] > 1), return (1))); 0;
isok(n) = ispower(n) && isokb(n); \\ Michel Marcus, Apr 28 2019

Extensions

More terms from Giovanni Resta, Apr 26 2019

A158912 Primitive numbers n whose square can be represented as a four "digit" repdigit number in some base < n.

Original entry on oeis.org

20, 1218, 7540, 20280, 1373090, 4903600, 23308460, 316540365, 527813814, 4091661910, 16718044954, 1034412600, 142903074740, 1827252574658, 31036383127940
Offset: 1

Views

Author

T. D. Noe, Mar 30 2009

Keywords

Examples

			20^2 = 1111 in base 7
1218^2 = (21)(21)(21)(21) in base 41
7540^2 = (58)(58)(58)(58) in base 99

Crossrefs

A158235, A158245

A326710 Squares m such that beta(m) = (tau(m) - 1)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Original entry on oeis.org

1, 121, 400, 1521, 1600, 2401, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361, 47155689
Offset: 1

Views

Author

Bernard Schott, Sep 14 2019

Keywords

Comments

As tau(m) = 2 * beta(m) + 1 is odd, the terms of this sequence are squares.
There are 3 classes of terms in this sequence (see examples):
1) The singleton {1} with 1^2 = 1.
2) The singleton {121}. Indeed, 121 is the only known square of prime that is Brazilian because 121 is a solution y^q of the Nagell-Ljunggren equation y^q = (b^m-1)/(b-1) with y = 11, q =2, b = 3, m = 5 (see A208242).
3) Squares of composites which have one Brazilian representation with three digits or more. These integers form A326711. We don't know if there exist squares of composites which have two or more Brazilian representations with three digits or more, consequently, there is no sequence with beta(m) = (tau(m) + k)/2, with k odd >= 1.

Examples

			One example for each type:
1) 1 is not Brazilian, tau(1) = 1 and beta(1) = (tau(1) - 1)/2 = 0.
2) 121 = 11^2 = 11111_3, tau(121) = 3 and beta(121) = (tau(121) - 1)/2 = 1.
3) 1521 = 39^2 = 333_22 = (13,13)_116 = 99_168 = 33_506. The divisors of 1521 are {1, 3, 9, 13, 39, 117, 169, 507, 1521} so tau(1521) = 9 and beta(1521) = (tau(1521) - 1)/2 = 4.

Links

Crossrefs

Cf. A326707 (tau(m)-3)/2, this sequence (tau(m)-1)/2.
Subsequence of A000290.

Programs

  • Mathematica
    brazQ[n_, b_] := Length@Union@IntegerDigits[n, b] == 1; beta[n_] := Sum[Boole @ brazQ[n, b], {b, 2, n - 2}]; aQ[n_] := beta[n] == (DivisorSigma[0, n] - 1)/2; Select[Range[6867]^2, aQ] (* Amiram Eldar, Sep 14 2019 *)

Formula

a(n+1) = (A158235(n))^2 for n >= 1.

A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

Original entry on oeis.org

3, 39, 543, 7563, 105339, 1467183, 20435223, 284625939, 3964327923, 55215964983, 769059181839, 10711612580763, 149193516948843, 2077997624703039, 28942773228893703, 403120827579808803, 5614748812888429539, 78203362552858204743, 1089232326927126436863, 15171049214426911911339
Offset: 1

Views

Author

Bernard Schott, Feb 17 2021

Keywords

Comments

Corresponding x are in A028231.
This equation belongs to the family of equations studied by Kustaa A. Inkeri, y^m = a * (x^q-1)/(x-1) with here: m=2, a=3, q=3. This equation is exhibed in A307745 by Giovanni Resta to prove that this sequence has infinitely many terms.
This Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions because the Pell-Fermat equation u^2 - 3*v^2 = -2 also has infinitely many solutions. The corresponding (u,v) are in (A001834, A001835) and for each pair (u,v), the corresponding solutions of 3*(x^2+x+1) = y^2 are x = (3*u*v-1)/2 and y = 3*(u^2+1)/2.
Note that if y = 3*z, this equation becomes 3*z^2 = x^2+x+1 with solutions (x, z) = (A028231, A001570).

Examples

			The first few values for (x,y) are (1,3), (22,39), (313,543), (4366,7563), (60817,105339), ...

Links

Crossrefs

Cf. A001834, A001835, A001570, A028231, A307745.
Subsequence of A158235, for a(n)>3.

Programs

  • Mathematica
    f[x_] := Sqrt[3*(x^2 + x + 1)]; f /@ LinearRecurrence[{15, -15, 1}, {1, 22, 313}, 20] (* Amiram Eldar, Feb 17 2021 *)

Formula

a(n) = 3*A001570(n). - Hugo Pfoertner, Feb 17 2021
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

Extensions

More terms from Amiram Eldar, Feb 17 2021
Showing 1-9 of 9 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.