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