cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A320966 Powerful numbers A001694 divisible by a cube > 1.

Original entry on oeis.org

8, 16, 27, 32, 64, 72, 81, 108, 125, 128, 144, 200, 216, 243, 256, 288, 324, 343, 392, 400, 432, 500, 512, 576, 625, 648, 675, 729, 784, 800, 864, 968, 972, 1000, 1024, 1125, 1152, 1296, 1323, 1331, 1352, 1372, 1568, 1600, 1728, 1800, 1936, 1944, 2000, 2025, 2048, 2187, 2197, 2304, 2312, 2401, 2500
Offset: 1

Views

Author

Hugo Pfoertner, Oct 25 2018

Keywords

Comments

Powerful numbers that are not squares of squarefree numbers. - Amiram Eldar, Jun 25 2022

Crossrefs

Intersection of A001694 and A046099.

Programs

  • Mathematica
    Select[Range[2500], (m = MinMax[FactorInteger[#][[;; , 2]]])[[1]] > 1 && m[[2]] > 2 &] (* Amiram Eldar, Jun 25 2022 *)
  • PARI
    isA001694(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]==1, return(0))); 1 \\ from Charles R Greathouse IV
    isA046099(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]>2, return(1)));0
    for (k=1,2500,if(isA001694(k)&&isA046099(k),print1(k,", ")))
    
  • Python
    from math import isqrt
    from sympy import mobius, integer_nthroot
    def A320966(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+squarefreepi(isqrt(x))-squarefreepi(integer_nthroot(x,3)[0]), 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)
            return c+l
        return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024

Formula

Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - 15/Pi^2 = 0.4237786821... . - Amiram Eldar, Jun 25 2022

A062320 Nonsquarefree numbers squared. A013929(n)^2.

Original entry on oeis.org

16, 64, 81, 144, 256, 324, 400, 576, 625, 729, 784, 1024, 1296, 1600, 1936, 2025, 2304, 2401, 2500, 2704, 2916, 3136, 3600, 3969, 4096, 4624, 5184, 5625, 5776, 6400, 6561, 7056, 7744, 8100, 8464, 9216, 9604, 9801, 10000, 10816, 11664, 12544, 13456
Offset: 1

Views

Author

Jason Earls, Jul 05 2001

Keywords

Comments

A008966(A037213(a(n))) = 0. - Reinhard Zumkeller, Sep 03 2015

Crossrefs

Squares in A046099.

Programs

  • Haskell
    a062320 = (^ 2) . a013929 -- Reinhard Zumkeller, Sep 03 2015
    
  • PARI
    for(n=1,55, if(issquarefree(n), n+1,print(n^2)))
    
  • PARI
    n=-1; for (m=1, 10^9, if (!issquarefree(m), write("b062320.txt", n++, " ", m^2); if (n==1000, break))) \\ Harry J. Smith, Aug 04 2009
    
  • PARI
    is(n)=issquare(n,&n) && !issquarefree(n) \\ Charles R Greathouse IV, Sep 18 2015
    
  • Python
    from math import isqrt
    from sympy import mobius
    def A062320(n):
        def f(x): return n+1+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+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
        return bisection(f)**2 # Chai Wah Wu, Aug 31 2024

Formula

Sum_{n>=1} 1/a(n) = Pi^2/6 - 15/Pi^2. - Amiram Eldar, Jul 16 2020

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jul 11 2001
Offset corrected by Andrew Howroyd, Sep 18 2024

A321070 Squares divisible by more than one cube > 1.

Original entry on oeis.org

64, 256, 576, 729, 1024, 1296, 1600, 2304, 2916, 3136, 4096, 5184, 6400, 6561, 7744, 9216, 10000, 10816, 11664, 12544, 14400, 15625, 16384, 18225, 18496, 20736, 23104, 25600, 26244, 28224, 30976, 32400, 33856, 35721, 36864, 38416, 40000, 43264, 46656, 50176
Offset: 1

Views

Author

Hugo Pfoertner, Oct 27 2018

Keywords

Examples

			a(1) = 64 because 16^2 is divisible by 2^3 and by 4^3.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[225]^2, Max[(e = FactorInteger[#][[;; , 2]])] > 4 || (Length[e] > 1 && Sort[e, Greater][[2]] > 2) &] (* Amiram Eldar, Jun 25 2022 *)
  • PARI
    iscubes(n) = {my(nb = 0); fordiv(n, d, if ((d>1) && ispower(d, 3), nb++; if (nb > 1, return(1))););}
    isok(n) = issquare(n) && iscubes(n); \\ Michel Marcus, Oct 27 2018

Formula

From Amiram Eldar, Jun 25 2022: (Start)
Equals A000290 \ (union of A062503 and A320965).
Sum_{n>=1} 1/a(n) = Pi^2/6 - (15/Pi^2) * (1 + Sum_{k>=2} (-1)^k * P(2*k)) = 0.029082273527998239268... . (End)
Showing 1-3 of 3 results.