A053289 First differences of consecutive perfect powers (A001597).

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, 39, 41, 43, 28, 17, 47, 49, 51, 53, 55, 57, 59, 61, 39, 24, 65, 67, 69, 71, 35, 38, 75, 77, 79, 81, 47, 36, 85, 87, 89, 23, 68, 71, 10, 12, 95, 97, 99, 101, 103, 40, 65, 107, 109, 100

Offset: 1

Author

Labos Elemer, Mar 03 2000

Keywords

• nonn

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

Examples

			Consecutive perfect powers are A001597(14) = 121, A001597(13) = 100, so a(13) = 121 - 100 = 21.
		

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.

Crossrefs

For non-perfect-powers (A007916) we have A375706.

The union is A023055.

For prime-powers (A000961 or A246655) we have A057820.

Sorted positions of first appearances are A376268, complement A376519.

For second differences we have A376559.

Ascending and descending points are A376560 and A376561.

A001597 lists perfect-powers.

A112344 counts integer partitions into perfect-powers, factorizations A294068.

A333254 gives run-lengths of differences between consecutive primes.

Cf. A025475, A069623, A219551.

Cf. A007921, A036263, A045542, A052410, A053707, A174965, A336416, A375735, A375736, A375740, A376562.

Programs

Mathematica

Differences@ Select[Range@ 3200, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &] (* Michael De Vlieger, Jun 30 2016, after Ant King at A001597 *)

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

Formula

a(n) = A001597(n+1) - A001597(n). - Jonathan Vos Post, Feb 18 2008

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