A247246 Differences of consecutive Achilles numbers.

36, 92, 88, 104, 40, 68, 148, 27, 125, 64, 104, 4, 153, 27, 171, 29, 20, 196, 232, 144, 56, 312, 280, 108, 188, 199, 113, 67, 189, 72, 344, 16, 112, 232, 268, 63, 45, 392, 292, 32, 76, 8, 80, 587, 50, 147, 456, 184, 288, 488, 115, 772, 137, 36, 40, 212, 248

Offset

1,1

Comments

29 is the first prime in this sequence, and it equals 1352 - 1323. Clearly, if the difference is prime, then these two Achilles numbers must be relatively prime, so primes appear in this sequence rarely. However, are there infinitely many n such that a(n) is prime?

The number 1 can also appear in this sequence, because it equals 5425069448 - 5425069447 = (2^3 * 26041^2) - (7^3 * 41^2 * 97^2). Does every natural number appear in this sequence? If so, do they appear infinitely often?

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Carlos Rivera, Problem 53, The Prime Puzzles and Problems Connection.

Eric Weisstein's World of Mathematics, Achilles number.

Wikipedia, Achilles number.

Formula

a(n) = A052486(n+1) - A052486(n).

Maple

f:= proc(n) local E; E:= map(t->t[2], ifactors(n)[2]); min(E)>1 and igcd(op(E))=1 end proc:
Achilles:= select(f, [$1..10^5]):
seq(Achilles[i+1]-Achilles[i], i=1..nops(Achilles)-1); # Robert Israel, Dec 13 2014

Mathematica

achillesQ[n_] := With[{ee = FactorInteger[n][[All, 2]]}, Min[ee] > 1 && GCD @@ ee == 1];
Select[Range[10^4], achillesQ] // Differences (* Jean-François Alcover, Sep 26 2020 *)

Prog

(PARI) isA052486(n) = { n>1 & vecmin(factor(n)[, 2])>1 & !ispower(n); }
lista(nn) = {v = select(n->isA052486(n), vector(nn, i, i)); vector(#v-1, n, v[n+1] - v[n]); } \\ Michel Marcus, Nov 29 2014
(Python)
from math import isqrt
from sympy import mobius, integer_nthroot
def A247246(n):
    def squarefreepi(n):
        return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
    def bisection(f, kmin=0, kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, l = n+x+1, 0
        j = isqrt(x)
        while j>1:
            k2 = integer_nthroot(x//j**2, 3)[0]+1
            w = squarefreepi(k2-1)
            c -= j*(w-l)
            l, j = w, isqrt(x//k2**3)
        c -= squarefreepi(integer_nthroot(x, 3)[0])-l+sum(mobius(k)*(integer_nthroot(x, k)[0]-1) for k in range(2, x.bit_length()))
        return c
    return -(a:=bisection(f, n, n))+bisection(lambda x:f(x)+1, a, a) # Chai Wah Wu, Sep 10 2024

Crossrefs

Cf. A052486, A076446, A053289.

Sequence in context: A305068 A182467 A060936 * A242238 A200866 A324297

Adjacent sequences: A247243 A247244 A247245 * A247247 A247248 A247249

Keyword

nonn,easy

Author

Eric Chen, Nov 28 2014

Status

approved