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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A001694 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).

Original entry on oeis.org

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
Offset: 1

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Author

N. J. A. Sloane

Keywords

Comments

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
This sequence is closed under multiplication. The primitive elements are A168363. - Franklin T. Adams-Watters, May 30 2011
Complement of A052485. - Reinhard Zumkeller, Sep 16 2011
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(m) mod prime(n) > 0 for m < A258599(n); a(A258599(n)) = A001248(n) = prime(n)^2. - Reinhard Zumkeller, Jun 06 2015
From Des MacHale, Mar 07 2021: (Start)
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

Examples

			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

References

  • 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).

Links

Crossrefs

Disjoint union of A062503 and A320966.
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.
Cf. A052485 (complement), A076446 (first differences), A376361, A376362.

Programs

  • Haskell
    a001694 n = a001694_list !! (n-1)
a001694_list = filter ((== 1) . a112526) [1..]
-- Reinhard Zumkeller, Nov 30 2012
  • Maple
    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:
seq(A001694(n),n=1..20) ; # R. J. Mathar, Jun 07 2011
  • Mathematica
    Join[{1}, Select[ Range@ 1250, Min@ FactorInteger[#][[All, 2]] > 1 &]]
(* Harvey P. Dale, Sep 18 2011; modified by Robert G. Wilson v, Aug 11 2014 *)
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 *)
  • 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
  • PARI
    is=ispowerful \\ Charles R Greathouse IV, Nov 13 2012
  • Python
    from sympy import factorint
A001694 = [1]+[n for n in range(2,10**6) if min(factorint(n).values()) > 1]
# Chai Wah Wu, Aug 14 2014
  • Python
    from math import isqrt
from sympy import mobius, integer_nthroot
def A001694(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, 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
    return bisection(f,n,n) # Chai Wah Wu, Sep 09 2024
  • Sage
    sloane.A001694.list(54) # Peter Luschny, Feb 08 2015

Formula

A112526(a(n)) = 1. - Reinhard Zumkeller, Sep 16 2011
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
a(n) = A224866(n) - 1. - Reinhard Zumkeller, Jul 23 2013
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

Extensions

More terms from Henry Bottomley, Mar 16 2000
Definition expanded by Jonathan Sondow, Jan 03 2016

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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