%I #79 Oct 11 2024 12:44:08
%S 3,4,1,7,9,2,5,4,13,15,17,19,21,4,3,16,25,27,20,9,18,13,33,35,19,18,
%T 39,41,43,28,17,47,49,51,53,55,57,59,61,39,24,65,67,69,71,35,38,75,77,
%U 79,81,47,36,85,87,89,23,68,71,10,12,95,97,99,101,103,40,65,107,109,100
%N First differences of consecutive perfect powers (A001597).
%C _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
%C Are there are any adjacent equal terms? - _Gus Wiseman_, Oct 08 2024
%D 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.
%D S. S. Pillai, On the equation 2^x - 3^y = 2^X - 3^Y, Bull, Calcutta Math. Soc. 37 (1945) 15-20.
%H Daniel Forgues and T. D. Noe, <a href="/A053289/b053289.txt">Table of n, a(n) for n = 1..10000</a>
%H Rafael Jakimczuk, <a href="http://dx.doi.org/10.12988/imf.2016.6232">Gaps between consecutive perfect powers</a>, International Mathematical Forum, Vol. 11, No. 9 (2016), pp. 429-437.
%H Holly Krieger and Brady Haran, <a href="https://www.youtube.com/watch?v=Us-__MukH9I&t=0s">Catalan's Conjecture</a>, Numberphile video (2018).
%H Michel Waldschmidt, <a href="https://arxiv.org/abs/math/0312440">Open Diophantine problems</a>, arXiv:math/0312440 [math.NT], 2003-2004.
%F a(n) = A001597(n+1) - A001597(n). - _Jonathan Vos Post_, Feb 18 2008
%F From _Amiram Eldar_, Jun 30 2023: (Start)
%F Formulas from Jakimczuk (2016):
%F Lim sup_{n->oo} a(n)/(2*n) = 1.
%F Lim inf_{n->oo} a(n)/(2*n)^(2/3 + eps) = 0. (End)
%F 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
%e Consecutive perfect powers are A001597(14) = 121, A001597(13) = 100, so a(13) = 121 - 100 = 21.
%t Differences@ Select[Range@ 3200, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &] (* _Michael De Vlieger_, Jun 30 2016, after _Ant King_ at A001597 *)
%o (Python)
%o from sympy import mobius, integer_nthroot
%o def A053289(n):
%o if n==1: return 3
%o 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())))
%o kmin, kmax = 1,2
%o while f(kmax)+1 >= kmax:
%o kmax <<= 1
%o rmin, rmax = 1, kmax
%o while True:
%o kmid = kmax+kmin>>1
%o if f(kmid)+1 < kmid:
%o kmax = kmid
%o else:
%o kmin = kmid
%o if kmax-kmin <= 1:
%o break
%o while True:
%o rmid = rmax+rmin>>1
%o if f(rmid) < rmid:
%o rmax = rmid
%o else:
%o rmin = rmid
%o if rmax-rmin <= 1:
%o break
%o return kmax-rmax # _Chai Wah Wu_, Aug 13 2024
%Y For non-perfect-powers (A007916) we have A375706.
%Y The union is A023055.
%Y For prime-powers (A000961 or A246655) we have A057820.
%Y Sorted positions of first appearances are A376268, complement A376519.
%Y For second differences we have A376559.
%Y Ascending and descending points are A376560 and A376561.
%Y A001597 lists perfect-powers.
%Y A112344 counts integer partitions into perfect-powers, factorizations A294068.
%Y A333254 gives run-lengths of differences between consecutive primes.
%Y Cf. A025475, A069623, A219551.
%Y Cf. A007921, A036263, A045542, A052410, A053707, A174965, A336416, A375735, A375736, A375740, A376562.
%K nonn
%O 1,1
%A _Labos Elemer_, Mar 03 2000