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 1-10 of 18 results. Next

A019434 Fermat primes: primes of the form 2^(2^k) + 1, for some k >= 0.

Original entry on oeis.org

3, 5, 17, 257, 65537
Offset: 1

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Author

Keywords

Comments

It is conjectured that there are only 5 terms. Currently it has been shown that 2^(2^k) + 1 is composite for 5 <= k <= 32 (see Eric Weisstein's Fermat Primes link). - Dmitry Kamenetsky, Sep 28 2008
No Fermat prime is a Brazilian number. So Fermat primes belong to A220627. For proof see Proposition 3 page 36 in "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
This sequence and A001220 are disjoint (see "Other theorems about Fermat numbers" in Wikipedia link). - Felix Fröhlich, Sep 07 2014
Numbers n > 1 such that n * 2^(n-2) divides (n-1)! + 2^(n-1). - Thomas Ordowski, Jan 15 2015
From Jaroslav Krizek, Mar 17 2016: (Start)
Primes p such that phi(p) = 2*phi(p-1); primes from A171271.
Primes p such that sigma(p-1) = 2p - 3.
Primes p such that sigma(p-1) = 2*sigma(p) - 5.
For n > 1, a(n) = primes p such that p = 4 * phi((p-1) / 2) + 1.
Subsequence of A256444 and A256439.
Conjectures:
1) primes p such that phi(p) = 2*phi(p-2).
2) primes p such that phi(p) = 2*phi(p-1) = 2*phi(p-2).
3) primes p such that p = sigma(phi(p-2)) + 2.
4) primes p such that phi(p-1) + 1 divides p + 1.
5) numbers n such that sigma(n-1) = 2*sigma(n) - 5. (End)
Odd primes p such that ratio of the form (the number of nonnegative m < p such that m^q == m (mod p))/(the number of nonnegative m < p such that -m^q == m (mod p)) is a divisor of p for all nonnegative q. - Juri-Stepan Gerasimov, Oct 13 2020
Numbers n such that tau(n)*(number of distinct ratio (the number of nonnegative m < n such that m^q == m (mod n))/(the number of nonnegative m < n such that -m^q == m (mod n))) for nonnegative q is equal to 4. - Juri-Stepan Gerasimov, Oct 22 2020
The numbers of primitive roots for the five known terms are 1, 2, 8, 128, 32768. - Gary W. Adamson, Jan 13 2022
Prime numbers such that every residue is either a primitive root or a quadratic residue. - Keith Backman, Jul 11 2022
If there are only 5 Fermat primes, then there are only 31 odd order groups which have a 2-group automorphism group. See the Miles Englezou link for a proof. - Miles Englezou, Mar 10 2025

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 137-141, 197.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see Table 1, p. 458.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • Richard K. Guy, Unsolved Problems in Number Theory, A3.
  • Hardy and Wright, An Introduction to the Theory of Numbers, bottom of page 18 in the sixth edition, gives an heuristic argument that this sequence is finite.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 7, 70.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Subsequence of A147545 and of A334101. Cf. also A333788, A334092.
Cf. A045544.

Programs

Formula

a(n+1) = A180024(A049084(a(n))). - Reinhard Zumkeller, Aug 08 2010
a(n) = 1 + A001146(n-1), if 1 <= n <= 5. - Omar E. Pol, Jun 08 2018

A000215 Fermat numbers: a(n) = 2^(2^n) + 1.

Original entry on oeis.org

3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937
Offset: 0

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Author

Keywords

Comments

It is conjectured that just the first 5 numbers in this sequence are primes.
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
For n>0, Fermat numbers F(n) have digital roots 5 or 8 depending on whether n is even or odd (Koshy). - Lekraj Beedassy, Mar 17 2005
This is the special case k=2 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058. - Seppo Mustonen, Sep 04 2005
For n>1 final two digits of a(n) are periodically repeated with period 4: {17, 57, 37, 97}. - Alexander Adamchuk, Apr 07 2007
For 1 < k <= 2^n, a(A007814(k-1)) divides a(n) + 2^k. More generally, for any number k, let r = k mod 2^n and suppose r != 1, then a(A007814(r-1)) divides a(n) + 2^k. - T. D. Noe, Jul 12 2007
From Daniel Forgues, Jun 20 2011: (Start)
The Fermat numbers F_n are F_n(a,b) = a^(2^n) + b^(2^n) with a = 2 and b = 1.
For n >= 2, all factors of F_n = 2^(2^n) + 1 are of the form k*(2^(n+2)) + 1 (k >= 1).
The products of distinct Fermat numbers (in their binary representation, see A080176) give rows of Sierpiński's triangle (A006943). (End)
Let F(n) be a Fermat number. For n > 2, F(n) is prime if and only if 5^((F(n)-1)/4) == sqrt(F(n)-1) (mod F(n)). - Arkadiusz Wesolowski, Jul 16 2011
Conjecture: let the smallest prime factor of Fermat number F(n) be P(F(n)). If F(n) is composite, then P(F(n)) < 3*2^(2^n/2 - n - 2). - Arkadiusz Wesolowski, Aug 10 2012
The Fermat primes are not Brazilian numbers, so they belong to A220627, but the Fermat composites are Brazilian numbers so they belong to A220571. For a proof, see Proposition 3 page 36 on "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
It appears that this sequence is generated by starting with a(0)=3 and following the rule "Write in binary and read in base 4". For an example of "Write in binary and read in ternary", see A014118. - John W. Layman, Jul 30 2013
Conjecture: the numbers > 5 in this sequence, i.e., 2^2^k + 1 for k>1, are exactly the numbers n such that (n-1)^4-1 divides 2^(n-1)-1. - M. F. Hasler, Jul 24 2015

Examples

			a(0) = 1*2^1 + 1 = 3 = 1*(2*1) + 1.
a(1) = 1*2^2 + 1 = 5 = 1*(2*2) + 1.
a(2) = 1*2^4 + 1 = 17 = 2*(2*4) + 1.
a(3) = 1*2^8 + 1 = 257 = 16*(2*8) + 1.
a(4) = 1*2^16 + 1 = 65537 = 2048*(2*16) + 1.
a(5) = 1*2^32 + 1 = 4294967297 = 641*6700417 = (10*(2*32) + 1)*(104694*(2*32) + 1).
a(6) = 1*2^64 + 1 = 18446744073709551617 = 274177*67280421310721 = (2142*(2*64) + 1)*(525628291490*(2*64) + 1).
		

References

  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 2nd. ed., 2001; see p. 3.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 7.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 87.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • R. K. Guy, Unsolved Problems in Number Theory, A3.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 14.
  • E. Hille, Analytic Function Theory, Vol. I, Chelsea, N.Y., see p. 64.
  • T. Koshy, "The Digital Root Of A Fermat Number", Journal of Recreational Mathematics Vol. 32 No. 2 2002-3 Baywood NY.
  • M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001.
  • C. S. Ogilvy and J. T. Anderson, Excursions in Number Theory, Oxford University Press, NY, 1966, p. 36.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see pp. 18, 59.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 202.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 6-7, 70-75.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 148-149.

Crossrefs

a(n) = A001146(n) + 1 = A051179(n) + 2.
See A004249 for a similar sequence.
Cf. A080176 for binary representation of Fermat numbers.

Programs

  • Haskell
    a000215 = (+ 1) . (2 ^) . (2 ^)  -- Reinhard Zumkeller, Feb 13 2015
    
  • Maple
    A000215 := n->2^(2^n)+1;
  • Mathematica
    Table[2^(2^n) + 1, {n, 0, 8}] (* Alonso del Arte, Jun 07 2011 *)
  • Maxima
    A000215(n):=2^(2^n)+1$ makelist(A000215(n),n,0,10); /* Martin Ettl, Dec 10 2012 */
    
  • PARI
    a(n)=if(n<1,3*(n==0),(a(n-1)-1)^2+1)
    
  • Python
    def a(n): return 2**(2**n) + 1
    print([a(n) for n in range(9)]) # Michael S. Branicky, Apr 19 2021

Formula

a(0) = 3; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get the empty product, i.e., 1, plus 2, giving 3 = a(0). - Benoit Cloitre, Sep 15 2002 [edited by Daniel Forgues, Jun 20 2011]
The above formula implies that the Fermat numbers (being all odd) are coprime.
Conjecture: F is a Fermat prime if and only if phi(F-2) = (F-1)/2. - Benoit Cloitre, Sep 15 2002
A000120(a(n)) = 2. - Reinhard Zumkeller, Aug 07 2010
If a(n) is composite, then a(n) = A242619(n)^2 + A242620(n)^2 = A257916(n)^2 - A257917(n)^2. - Arkadiusz Wesolowski, May 13 2015
Sum_{n>=0} 1/a(n) = A051158. - Amiram Eldar, Oct 27 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = A249119.
Product_{n>=0} (1 - 1/a(n)) = 1/2. (End)
a(n) = 2*A077585(n) + 3. - César Aguilera, Jul 26 2023
a(n) = 2*2^A000225(n) + 1. - César Aguilera, Jul 11 2024

A190300 Composite numbers that are not Brazilian.

Original entry on oeis.org

4, 6, 9, 25, 49, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721, 4489, 5041, 5329, 6241, 6889, 7921, 9409, 10201, 10609, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 22201, 22801, 24649, 26569, 27889, 29929, 32041, 32761, 36481, 37249, 38809, 39601, 44521, 49729
Offset: 1

Views

Author

N. J. A. Sloane, May 14 2011

Keywords

Comments

Other than the term 6 and the missing term 121, is this sequence the same as A001248? - Nathaniel Johnston, May 24 2011
From Bernard Schott, Dec 04 2012: (Start)
Yes, because
1) 4 is not a Brazilian number [4 = 100_2].
2) 6 is not a Brazilian number [6 = 110_2 = 20_3 = 12_4].
3) Theorem 1, page 32 of Quadrature article mentioned in links: If n > 7 is not Brazilian, then n is a prime or the square of a prime.
4) Theorem 5, page 37 of Quadrature article mentioned in links: The only square of prime number which is Brazilian is 121 = 11^2 = 11111_3.
(End)
There is an infinity of composite numbers that are not Brazilian: Corollary 2, page 37 of Quadrature article in links (consider the sequence of squares of prime numbers for p >= 13). - Bernard Schott, Dec 17 2012
Also semiprimes that are not Brazilian. - Bernard Schott, Apr 11 2019

Examples

			a(10) = p_10^2 = 29^2 = 841.
		

Crossrefs

Intersection of A002808 and A220570.
Intersection of A001358 and A220570.

Programs

  • Maple
    4, 6, 9, 25, 49,seq(ithprime(i)^2, i=6..100); # Robert Israel, Apr 17 2019
  • Mathematica
    brazBases[n_] := Select[Range[2, n - 2], Length[Union[IntegerDigits[n, #]]] == 1 &]; Select[Range[2, 10000], ! PrimeQ[#] && brazBases[#] == {} &] (* T. D. Noe, Dec 26 2012 *)
    f[n_] := Block[{b = 2}, While[ Length@ Union@ IntegerDigits[n, b] != 1, b++]; b]; k = 4; lst = {}; While[k < 50001, If[ !PrimeQ@ k && 1 + f@ k == k, AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Mar 30 2014 *)
  • PARI
    isnotb(n) = my(c=0, d); for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), c=n; break); c++); (c==max(n-3, 0)); \\ A220570
    lista(nn) = forcomposite(n=1, nn, if (isnotb(n), print1(n, ", "))); \\ Michel Marcus, Apr 14 2019

Formula

a(1) = 2^2 = p_1^2, a(2) = 2*3 = p_1*p_2, a(3) = 3^2 = p_2^2, a(4) = 5^2 = p_3^2, a(5) = 7^2 = p_4^2, a(6) = 13^2 = p_6^2, ..., for n >= 6, a(n) = p_n^2, where p_k is the k-th prime number. - Bernard Schott, Dec 04 2012

Extensions

a(6)-a(24) from Nathaniel Johnston, May 24 2011
a(25) onward from Robert G. Wilson v, Mar 30 2014

A288783 Brazilian numbers which have only one Brazilian representation.

Original entry on oeis.org

7, 8, 10, 12, 13, 14, 16, 20, 22, 27, 33, 34, 35, 38, 39, 43, 46, 51, 55, 58, 65, 69, 73, 74, 77, 81, 82, 87, 94, 95, 106, 115, 118, 119, 121, 122, 123, 125, 127, 134, 141, 142, 143, 145
Offset: 1

Views

Author

Bernard Schott, Jun 15 2017

Keywords

Comments

These numbers could be called 1-Brazilian numbers.
The smallest number of this sequence is 7 = 111_2 which is also the smallest Brazilian number (A125134) and the smallest Brazilian prime (A085104), and as such belongs to A329383.
a(2) = 8 is the smallest composite Brazilian number and so the smallest even composite Brazilian with 8 = 22_3 (A220571).
a(10) = 27 is the smallest odd composite Brazilian in this sequence because 27 = 33_8 but 15 is the smallest odd composite Brazilian with 15 = 1111_2 = 33_4 so with two representations.
121 is the only square of prime which is Brazilian with 121 = 11111_3.
In this sequence, there are:
1) The Brazilian primes (except for 31 and 8191) and the only square of prime 121 which are all repunits in a base >= 2 with a string of at least three 1's.
2) The composite numbers which are such that n = a * b = (aa)_(b-1) with 1 < a < b-1 with only one such product a * b.

Examples

			13 = 111_3; 127 = 1111111_2.
20 = 2 * 10 = 22_9; 55 = 5 * 11 = 55_10; 69 = 3 * 23 = 33_22.
31 = 11111_2 = 111_5 so 31 is not a term.
		

References

  • D. Lignon, Dictionnaire de (presque) tous les nombres entiers, Ellipses, 2012, page 420.

Crossrefs

Programs

  • Mathematica
    Select[Range@ 145, Function[n, Count[Range[2, n - 2], b_ /; SameQ @@ IntegerDigits[n, b]] == 1]] (* Michael De Vlieger, Jun 16 2017 *)

A290017 Brazilian numbers which have exactly four Brazilian representations.

Original entry on oeis.org

40, 48, 63, 72, 90, 112, 114, 132, 162, 170, 176, 208, 222, 266, 285, 304, 306, 366, 368, 380, 399, 405, 438, 455, 464, 496, 512, 518, 555, 567, 592, 650, 651, 656, 665, 682, 686, 688, 752, 762, 812, 848, 891, 915, 931, 942, 944, 976, 992, 999, 1024, 1029, 1053, 1072, 1106, 1136, 1168
Offset: 1

Views

Author

Bernard Schott, Jul 28 2017

Keywords

Comments

These numbers could be called 4-Brazilian numbers.
All these numbers are composite with six to twelve divisors.
The smallest number of this sequence is 40 with 40 = 1111_3 = 55_7 = 44_9 = 22_19. The number 40 is a highly Brazilian number in A329383.

Examples

			48 = 6 * 8 = 66_7 = 4 * 12 = 44_11 = 3 * 16 = 33_15 = 2 * 24 = 22_23.
63 = 111111_2 = 3 * 21 = 33_20 = 333_4 = 7 * 9 = 77_8.
		

Crossrefs

Programs

A290018 Numbers with exactly five Brazilian representations: bases 1 < b_1 < b_2 < b_3 < b_4 < b_5 < n-1 such that n is a repdigit in base b_i.

Original entry on oeis.org

60, 80, 84, 96, 108, 126, 140, 150, 156, 160, 198, 200, 204, 220, 224, 234, 255, 260, 273, 276, 294, 308, 315, 340, 342, 348, 350, 352, 372, 392, 414, 416, 460, 476, 486, 490, 492, 495, 500, 516, 522, 525, 544, 550, 558, 564, 572, 580, 608, 620, 636, 644, 675, 693, 708, 726, 735, 736
Offset: 1

Views

Author

Bernard Schott, Aug 07 2017

Keywords

Comments

These numbers could be called 5-Brazilian numbers.
All these numbers are composite with 8 to 13 divisors.
The smallest term is 60 and as such is a highly Brazilian number that belongs to A329383.

Examples

			60 = 66_9 = 55_11 = 44_14 = 33_19 = 22_29 and tau(60) = 12.
80 = 2222_3 = 22_39 = 44_19 = 55_15 = 88_9 and tau(80) = 10.
255 = 11111111_2 = 3333_4 = 33_84 = 55_50 = (15 15)_16 and tau(255) = 8.
4096 = (32 32)_127 = (16 16)_255 = 88_511 = 44_1023 = 22_2047 and tau(4096) = 13.
		

Crossrefs

k-Brazilian numbers: A220570 (0), A288783 (1), A290015 (2), A290016 (3), A290017 (4), this sequence (5).

Programs

A290015 Brazilian numbers which have exactly two Brazilian representations.

Original entry on oeis.org

15, 18, 21, 26, 28, 30, 31, 32, 44, 45, 50, 52, 56, 57, 62, 64, 68, 75, 76, 85, 86, 91, 92, 93, 98, 99, 110, 111, 116, 117, 129, 133, 146, 147, 148, 153, 164, 175, 183, 188, 207, 212, 215, 219, 236, 243, 244, 245, 259, 261, 268, 275, 279, 284, 314, 316, 325, 332, 338, 341, 343, 356, 363, 365, 369, 381, 387, 388
Offset: 1

Views

Author

Bernard Schott, Jul 17 2017

Keywords

Comments

These numbers could be called 2-Brazilian numbers.
The smallest number of this sequence is 15 which is also the smallest odd composite Brazilian in A257521 with 15 = 11111_2 = 33_4. The number 15 is highly Brazilian in A329383.
Following the Goormaghtigh conjecture, only two primes, 31 and 8191, which are both Mersenne numbers, are Brazilian in two different bases (A119598).

Examples

			18 = 2 * 9 = 22_8 = 3 * 6 = 33_5.
26 = 2 * 13 = 2 * 111_3 = 222_3 = 22_12.
31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90.
		

Crossrefs

Programs

  • Maple
    bresilienbaseb:=proc(n,b)
    local r,q,coupleq:
    if n0 then
    return [couple[1]+1,r]
    else
    return [0,0]
    end if
    end if
    end proc;
    bresil:=proc(n)
    local b,L,k,t:
    k:=0:
    for b from 2 to (n-2) do
    t:=bresilienbase(n,b):
    if t[1]>0 then
    k:=k+1
    L[k]:=[b,t[1],t[2]]:
    end if:
    end do:
    seq(L[i],i=1..k);
    end proc;
    nbbresil:=n->nops([bresil(n)]);
    #Numbers 2 times Brazilian
    for n from 1 to 100 do if nbbresil(n)=2 then print(n,bresil(n)) else fi; od:
  • Mathematica
    Flatten@ Position[#, 2] &@ Table[Count[Range[2, n - 2], ?(And[Length@ # != 1, Length@ Union@ # == 1] &@ IntegerDigits[n, #] &)], {n, 400}] (* _Michael De Vlieger, Jul 18 2017 *)

A290016 Brazilian numbers which have exactly three Brazilian representations.

Original entry on oeis.org

24, 36, 42, 54, 66, 70, 78, 88, 100, 102, 104, 105, 124, 128, 130, 135, 136, 138, 152, 154, 165, 171, 172, 174, 182, 184, 186, 189, 190, 195, 196, 225, 230, 231, 232, 238, 242, 246, 248, 250, 256, 258, 272, 282, 286, 290, 292, 296, 297, 310, 318, 322, 328, 333, 344, 345
Offset: 1

Author

Bernard Schott, Jul 27 2017

Keywords

Comments

These numbers could be called 3-Brazilian numbers.
All these numbers are composite with six to ten different divisors.
The smallest number of this sequence is 24 with 24 = 44_5 = 33_7 = 22_11. The number 24 is highly Brazilian in A329383.

Examples

			36 = 4 * 9 = 44_8 = 3 * 12 = 33_11 = 2 * 18 = 22_19.
42 = 2 * 21 = 22_20 = 222_4 = 3 * 14 = 33_13.
124 = 4 * 31 = 44_30 = 444_5 = 2 * 62 = 22_61.
272 = 8 * 34 = 88_33 = 4 * 68 = 44_67 = 2 * 136 = 22_135.
		

Programs

A059711 Smallest base in which n is a repdigit.

Original entry on oeis.org

2, 2, 3, 2, 3, 4, 5, 2, 3, 8, 4, 10, 5, 3, 6, 2, 7, 16, 5, 18, 9, 4, 10, 22, 5, 24, 3, 8, 6, 28, 9, 2, 7, 10, 16, 6, 8, 36, 18, 12, 3, 40, 4, 6, 10, 8, 22, 46, 7, 48, 9, 16, 12, 52, 8, 10, 13, 7, 28, 58, 9, 60, 5, 2, 15, 12, 10, 66, 16, 22, 9, 70, 11, 8, 36, 14, 18, 10, 12, 78, 3, 26, 40, 82, 11, 4
Offset: 0

Author

Erich Friedman, Feb 19 2001

Keywords

Comments

Numbers n such that a(n) < n - 1 correspond to Brazilian numbers (A125134); conversely, positive numbers n such that a(n) >= n - 1 correspond to A220570. - Rémy Sigrist, Apr 04 2018

Examples

			a(13) = 3 since 13 in base 3 is 111.
		

Programs

  • PARI
    a(n) = for (b=2, oo, if (#Set(digits(n, b))<=1, return (b))) \\ Rémy Sigrist, Apr 04 2018

Formula

From Rémy Sigrist, Apr 04 2018: (Start)
a(n) <= n - 1 for any n >= 3.
a(2^n-1) = 2 for any n >= 0.
a(A048328(n)) <= 3 for any n >= 0.
a(A048329(n)) <= 4 for any n >= 0.
a(A048330(n)) <= 5 for any n >= 0.
a(A048331(n)) <= 6 for any n >= 0.
a(A048332(n)) <= 7 for any n >= 0.
a(A048333(n)) <= 8 for any n >= 0.
a(A048334(n)) <= 9 for any n >= 0.
a(A010785(n)) <= 10 for any n >= 0.
a(A048335(n)) <= 11 for any n >= 0.
a(A048336(n)) <= 12 for any n >= 0.
a(A048337(n)) <= 13 for any n >= 0.
a(A048338(n)) <= 14 for any n >= 0.
a(A048339(n)) <= 15 for any n >= 0.
a(A048340(n)) <= 16 for any n >= 0.
(End)

Extensions

Example clarified by Harvey P. Dale, Oct 11 2015
Terms a(0) = 2, a(1) = 2 and a(2) = 3 prepended by Rémy Sigrist, Apr 04 2018

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

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.
		

Crossrefs

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
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