OFFSET
1,1
COMMENTS
Michel Waldschmidt writes: Conjecture 1.3 (Pillai). Let k be a positive integer. The equation x^p - y^q = k where the unknowns x, y, p and q take integer values, all >= 2, has only finitely many solutions (x,y,p,q). This means that in the increasing sequence of perfect powers [A001597] the difference between two consecutive terms [the present sequence] tends to infinity. It is not even known whether for, say, k=2, Pillai's equation has only finitely many solutions. A related open question is whether the number 6 occurs as a difference between two perfect powers. See Sierpiński [1970], problem 238a, p. 116. - Jonathan Vos Post, Feb 18 2008
Are there are any adjacent equal terms? - Gus Wiseman, Oct 08 2024
REFERENCES
Wacław Sierpiński, 250 problems in elementary number theory, Modern Analytic and Computational Methods in Science and Mathematics, No. 26, American Elsevier, Warsaw, 1970, pp. 21, 115-116.
S. S. Pillai, On the equation 2^x - 3^y = 2^X - 3^Y, Bull, Calcutta Math. Soc. 37 (1945) 15-20.
LINKS
Daniel Forgues and T. D. Noe, Table of n, a(n) for n = 1..10000
Rafael Jakimczuk, Gaps between consecutive perfect powers, International Mathematical Forum, Vol. 11, No. 9 (2016), pp. 429-437.
Holly Krieger and Brady Haran, Catalan's Conjecture, Numberphile video (2018).
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
FORMULA
From Amiram Eldar, Jun 30 2023: (Start)
Formulas from Jakimczuk (2016):
Lim sup_{n->oo} a(n)/(2*n) = 1.
Lim inf_{n->oo} a(n)/(2*n)^(2/3 + eps) = 0. (End)
Can be obtained by inserting 0 between 3 and 6 in A375702 and then adding 1 to all terms. In particular, for n > 2, a(n+1) - 1 = A375702(n). - Gus Wiseman, Sep 14 2024
EXAMPLE
MATHEMATICA
Differences@ Select[Range@ 3200, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &] (* Michael De Vlieger, Jun 30 2016, after Ant King at A001597 *)
PROG
(Python)
from sympy import mobius, integer_nthroot
def A053289(n):
if n==1: return 3
def f(x): return int(n-2+x+sum(mobius(k)*(integer_nthroot(x, k)[0]-1) for k in range(2, x.bit_length())))
kmin, kmax = 1, 2
while f(kmax)+1 >= kmax:
kmax <<= 1
rmin, rmax = 1, kmax
while True:
kmid = kmax+kmin>>1
if f(kmid)+1 < kmid:
kmax = kmid
else:
kmin = kmid
if kmax-kmin <= 1:
break
while True:
rmid = rmax+rmin>>1
if f(rmid) < rmid:
rmax = rmid
else:
rmin = rmid
if rmax-rmin <= 1:
break
return kmax-rmax # Chai Wah Wu, Aug 13 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Mar 03 2000
STATUS
approved