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A247246
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Differences of consecutive Achilles numbers.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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?
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LINKS
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Carlos Rivera, Problem 53, The Prime Puzzles and Problems Connection.
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FORMULA
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MAPLE
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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
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MATHEMATICA
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achillesQ[n_] := With[{ee = FactorInteger[n][[All, 2]]}, Min[ee] > 1 && GCD @@ ee == 1];
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PROG
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(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 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
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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STATUS
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approved
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