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 1-5 of 5 results.

A074981 Conjectured list of positive numbers which are not of the form r^i - s^j, where r,s,i,j are integers with r>0, s>0, i>1, j>1.

Original entry on oeis.org

6, 14, 34, 42, 50, 58, 62, 66, 70, 78, 82, 86, 90, 102, 110, 114, 130, 134, 158, 178, 182, 202, 206, 210, 226, 230, 238, 246, 254, 258, 266, 274, 278, 290, 302, 306, 310, 314, 322, 326, 330, 358, 374, 378, 390, 394, 398, 402, 410, 418, 422, 426
Offset: 1

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Author

Zak Seidov, Oct 07 2002

Keywords

Comments

This is a famous hard problem and the terms shown are only conjectured values.
The terms shown are not the difference of two powers below 10^19. - Don Reble
One can immediately represent all odd numbers and multiples of 4 as differences of two squares. - Don Reble
_Ed Pegg Jr_ remarks (Oct 07 2002) that the techniques of Preda Mihailescu (see MathWorld link) might make it possible to prove that 6, 14, ... are indeed members of this sequence.
Numbers n such that there is no solution to Pillai's equation. - T. D. Noe, Oct 12 2002
The terms shown are not the difference of two powers below 10^27. - Mauro Fiorentini, Jan 03 2020

Examples

			Examples showing that certain numbers are not in the sequence: 10 = 13^3 - 3^7, 22 = 7^2 - 3^3, 29 = 15^2 - 14^2, 31 = 2^5 - 1, 52 = 14^2 - 12^2, 54 = 3^4 - 3^3, 60 = 2^6 - 2^2, 68 = 10^2 - 2^5, 72 = 3^4 - 3^2, 76 = 5^3 - 7^2, 84 = 10^2 - 2^4, ... 342 = 7^3 - 1^2, ...
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, Sections D9 and B19.
  • P. Ribenboim, Catalan's Conjecture, Academic Press NY 1994.
  • T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge University Press, 1986.

Crossrefs

Subsequence of A016825 (see second comment of Don Reble).
n such that A076427(n) = 0. [Corrected by Jonathan Sondow, Apr 14 2014]
For a count of the representations of a number as the difference of two perfect powers, see A076427. The numbers that appear to have unique representations are listed in A076438.
For sequence with similar definition, but allowing negative powers, see A066510.

Extensions

Corrected by Don Reble and Jud McCranie, Oct 08 2002. Corrections were also sent in by Neil Fernandez, David W. Wilson, and Reinhard Zumkeller.

A076438 Numbers k which appear to have a unique representation as the difference of two perfect powers; that is, there is only one solution to Pillai's equation a^x - b^y = k, with a > 0, b > 0, x > 1, y > 1.

Original entry on oeis.org

1, 2, 10, 29, 30, 38, 43, 46, 52, 59, 122, 126, 138, 142, 146, 150, 154, 166, 170, 173, 181, 190, 194, 214, 222, 234, 263, 270, 282, 283, 298, 317, 318, 332, 338, 342, 347, 349, 354, 361, 370, 379, 382, 383, 386, 406, 419, 428, 436, 461, 467, 479, 484, 486
Offset: 1

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Author

T. D. Noe, Oct 12 2002

Keywords

Comments

This is the classic Diophantine equation of S. S. Pillai, who conjectured that there are only a finite number of solutions for each k. A generalization of Catalan's conjecture that a^x - b^y = 1 has only one solution. See A076427 for the number of solutions for each k. Interestingly, the unique solutions (k,a,x,b,y) fall into two groups: (A076439) those in which x and y are even numbers, so that k is the difference of squares, and (A076440) those requiring an odd power. This sequence was found by examining all perfect powers (A001597) less than 2^63-1. By examining a larger set of perfect powers, we may discover that some of these numbers do not have a unique representation.

References

  • R. K. Guy, Unsolved Problems in Number Theory, D9.
  • T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge University Press, 1986.

Crossrefs

A219551 Number of positive integer solutions to the equation |2^x - 3^y| = n.

Original entry on oeis.org

0, 3, 0, 0, 0, 3, 0, 2, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0
Offset: 0

Views

Author

Jonathan Sondow, Dec 09 2012

Keywords

Comments

Pillai (1931) proved that a(n) is finite for all n.
Hershfeld (1936) computed a(n) for n <= 10 and proved that a(n) <= 2 for all large n.
Stroeker and Tijdeman (1982) proved that a(n) <= 2 for all n > 13.
For additional comments, references, and links, see the crossrefs.
a(n) <= 1 except for n=1, 5, 7, 13, 23: see e,g, Bennett (2003). - Robert Israel, Mar 06 2017

Examples

			1 = 2^2 - 3 = 3 - 2 = 3^2 - 2^3.
5 = 2^3 - 3 = 2^5 - 3^3 = 3^2 - 2^2.
7 = 2^4 - 3^2 = 3^2 - 2.
23 = 2^5 - 3^2 = 3^3 - 2^2 and a(n) <= 2 for n > 13, so a(23) = 2.
		

References

  • S. Pillai, On the inequality 0 < a^x - b^y <= n, Journal Indian Math. Soc., 19 (1931), 1-11.
  • R. J. Stroeker and R. Tijdeman, Diophantine equations, Computational methods in number theory, Part 2, Math. Cent. Tracts, 155 (1982), 321-369.

Crossrefs

Programs

  • Mathematica
    Clear[seq]; seq[m_] := seq[m] = (Clear[a]; a[A219551%20=%20seq%5Bm%5D%20(*%20_Jean-Fran%C3%A7ois%20Alcover">] = 0; Do[n = Abs[2^x - 3^y]; a[n] = a[n] + 1, {x, 1, m}, {y, 1, m}]; Table[a[n], {n, 0, 10}]); seq[m = 1]; While[seq[m] != seq[m - 1], m = 2*m]; A219551 = seq[m] (* _Jean-François Alcover, Dec 13 2012 *)

Formula

a(2n) = a(3n) = 0.
a(n) <= 2 for n > 13.

Extensions

a(11) - a(30) from Robert Israel, Mar 06 2017

A262958 Numbers whose base-b expansions, for both b=3 and b=4, include no digits other than 1 and b-1.

Original entry on oeis.org

1, 5, 7, 13, 23, 53, 125, 215, 373, 1367, 1373, 1375, 3551, 4093, 5471, 5495, 5503, 30581, 30589, 32765, 32767, 56821, 56831, 89557, 96119, 96215, 96223, 97655, 98135, 98141, 98143, 98167, 98293, 98303, 351743, 352093, 521599, 521693, 521717, 521719, 524119, 524149, 875893, 875903, 884725, 884735
Offset: 1

Views

Author

Robin Powell, Oct 05 2015

Keywords

Comments

1, 7 and 32767 also share this property in base 2; their binary expansions consist only of a sequence of 1s.

Examples

			53 is 1222 in base 3 and 311 in base 4; it only uses the digit 1 or the largest digit in the two bases and is therefore a term.
Similarly 215 is 21222 in base 3 and 3113 in base 4 so it is also a term.
		

Crossrefs

Programs

  • Mathematica
    Select[Range@ 1000000, Last@ DigitCount[#, 3] == 0 && Total@ Rest@ Drop[DigitCount[#, 4], {3}] == 0 &] (* Michael De Vlieger, Oct 05 2015 *)
    Join[{1,5},Flatten[Table[Select[FromDigits[#,3]&/@Tuples[{1,2},n], Union[ IntegerDigits[ #,4]] =={1,3}&],{n,20}]]] (* Harvey P. Dale, Jun 14 2016 *)
  • PARI
    is(n)=!setsearch(Set(digits(n,3)),0) && #setintersect(Set(digits(n,4)),[0,2])==0 \\ Charles R Greathouse IV, Oct 12 2015
    
  • Python
    from gmpy2 import digits
    def f1(n):
        s = digits(n,3)
        m = len(s)
        for i in range(m):
            if s[i] == '0':
                return(int(s[:i]+'1'*(m-i),3))
        return n
    def f2(n):
        s = digits(n,4)
        m = len(s)
        for i in range(m):
            if s[i] == '0':
                return(int(s[:i]+'1'*(m-i),4))
            if s[i] == '2':
                return(int(s[:i]+'3'+'1'*(m-i-1),4))
        return n
    A262958_list = []
    n = 1
    for i in range(10**4):
        m = f2(f1(n))
        while m != n:
            n, m = m, f2(f1(m))
        A262958_list.append(m)
        n += 1 # Chai Wah Wu, Oct 30 2015

A236211 Numbers c > 0 for which there exist integers a > 1 and b > 1 such that the equation a^x - b^y = c has two solutions in positive integers x, y.

Original entry on oeis.org

1, 3, 4, 5, 9, 10, 13, 89, 275, 1215, 4900
Offset: 1

Views

Author

Jonathan Sondow, Jan 23 2014

Keywords

Comments

Bennett proved that if a, b, c are nonzero integers with a > 1 and b > 1, then the equation a^x - b^y = c has at most two solutions in positive integers x and y.
Bennett conjectured that if a, b, c are positive integers with a > 1 and b > 1, then the equation a^x - b^y = c has at most one solution in positive integers x and y, except for the triples (a,b,c) = (3,2,1), (2,5,3), (6,2,4), (2,3,5), (15,6,9), (13,3,10), (2,3,13), (91,2,89), (280,5,275), (6,3,1215), (4930,30,4900). If this is true, then the present sequence is complete.

Examples

			3 - 2 = 3^2 - 2^3 = 1.
2^3 - 5 = 2^7 - 5^3 = 3.
6 - 2 = 6^2 - 2^5 = 4.
2^3 - 3 = 2^5 - 3^3 = 5.
15 - 6 = 15^2 - 6^3 = 9.
13 - 3 = 13^3 - 3^7 = 10.
2^4 - 3 = 2^8 - 3^5 = 13.
91 - 2 = 91^2 - 2^13 = 89.
280 - 5 = 280^2 - 5^7 = 275.
6^4 - 3^4 = 6^5 - 3^8 = 1215.
4930 - 30 = 4930^2 - 30^5 = 4900.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, D9.
  • T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge University Press, 1986.

Crossrefs

Cf. A207079 and the OEIS link.
Showing 1-5 of 5 results.