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A001694
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Powerful numbers, definition (1): if a prime p divides n then p^2 must also divide n (also called squareful, square full, square-full or 2-powerful numbers).
(Formerly M3325 N1335)
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358
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1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000
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OFFSET
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1,2
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COMMENTS
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Numbers of the form a^2*b^3, a >= 1, b >= 1.
In other words, if the prime factorization of n is Product_k p_k^e_k then all e_k are greater than 1.
Numbers n such that Sum_{d|n} phi(d)*phi(n/d)*mu(d) > 0; places of nonzero A300717. - Benoit Cloitre, Nov 30 2002
The number of terms less than or equal to 10^k beginning with k = 0: 1, 4, 14, 54, 185, 619, 2027, 6553, 21044, ...: A118896. - Robert G. Wilson v, Aug 11 2014
a(10^n): 1, 49, 3136, 253472, 23002083, 2200079025, 215523459072, 21348015504200, 2125390162618116, ... . - Robert G. Wilson v, Aug 15 2014
A number m is powerful if and only if |R/Z(R)| = m, for some finite non-commutative ring R.
A number m is powerful if and only if |G/Z(G)| = m, for some finite nilpotent class two group G (Reference Aine Nishe). (End)
Numbers n such that Sum_{k=1..n} phi(gcd(n,k))*mu(gcd(n,k)) > 0. - Richard L. Ollerton, May 09 2021
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REFERENCES
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G. E. Hardy and M. V. Subbarao, Highly powerful numbers, Congress. Numer. 37 (1983), 277-307.
Aleksandar Ivić, The Riemann Zeta-Function, Wiley, NY, 1985, see p. 407.
Richard A. Mollin, Quadratics, CRC Press, 1996, Section 1.6.
Aine NiShe, Commutativity and Generalisations in Finite Groups, Ph.D. Thesis, University College Cork, 2000.
Paulo Ribenboim, Meine Zahlen, meine Freunde, 2009, Springer, 9.1 Potente Zahlen, pp. 241-247.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Gérald Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge University Press, 1995, p. 54, exercise 10 (in the third edition 2015, p. 63, exercise 70).
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LINKS
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Eric Weisstein's World of Mathematics, Squareful.
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FORMULA
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Bateman & Grosswald prove that there are zeta(3/2)/zeta(3) x^{1/2} + zeta(2/3)/zeta(2) x^{1/3} + O(x^{1/6}) terms up to x; see section 5 for a more precise error term. - Charles R Greathouse IV, Nov 19 2012
Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6). - Ivan Neretin, Aug 30 2015
Sum_{n>=1} 1/a(n)^s = zeta(2*s)*zeta(3*s)/zeta(6*s), s > 1/2 (Golomb, 1970). - Amiram Eldar, Oct 02 2022
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EXAMPLE
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1 is a term because for every prime p that divides 1, p^2 also divides 1.
2 is not a term since 2 divides 2 but 2^2 does not.
4 is a term because 2 is the only prime that divides 4 and 2^2 does divide 4. - N. J. A. Sloane, Jan 16 2022
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MAPLE
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isA001694 := proc(n) for p in ifactors(n)[2] do if op(2, p) = 1 then return false; end if; end do; return true; end proc:
A001694 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do if isA001694(a) then return a; end if; end do; end if; end proc:
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MATHEMATICA
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Join[{1}, Select[ Range@ 1250, Min@ FactorInteger[#][[All, 2]] > 1 &]]
max = 10^3; Union@ Flatten@ Table[a^2*b^3, {b, max^(1/3)}, {a, Sqrt[max/b^3]}] (* Robert G. Wilson v, Aug 11 2014 *)
nextPowerfulNumber[n_] := Block[{r = Range[ Floor[1 + n^(1/3)]]^3}, Min@ Select[ Sort[ r*Floor[1 + Sqrt[n/r]]^2], # > n &]]; NestList[ nextPowerfulNumber, 1, 55] (* Robert G. Wilson v, Aug 16 2014 *)
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PROG
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(PARI) isA001694(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]==1, return(0))); 1 \\ Charles R Greathouse IV, Feb 11 2011
(PARI) list(lim, mn=2)=my(v=List(), t); for(m=1, sqrtnint(lim\1, 3), t=m^3; for(n=1, sqrtint(lim\t), listput(v, t*n^2))); Set(v) \\ Charles R Greathouse IV, Jul 31 2011; edited Sep 22 2015
(Haskell)
a001694 n = a001694_list !! (n-1)
a001694_list = filter ((== 1) . a112526) [1..]
(Python)
from sympy import factorint
A001694 = [1]+[n for n in range(2, 10**6) if min(factorint(n).values()) > 1]
(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, 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
return c
(Sage)
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CROSSREFS
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Cf. A007532 (Powerful numbers, definition (2)), A005934, A005188, A003321, A014576, A023052 (Powerful numbers, definition (3)), A046074, A013929, A076871, A258599, A001248, A112526, A168363, A224866, A261883, A300717.
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KEYWORD
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nonn,nice,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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