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 1-4 of 4 results.

A307134 Terms of A216427 that are the sum of two coprime terms of A216427.

Original entry on oeis.org

7688, 70688, 95048, 120125, 131072, 186003, 219488, 219501, 265837, 286443, 304175, 371293, 412232, 464648, 465125, 596183, 628864, 699867, 729632, 732736, 834632, 860672, 1104500, 1119371, 1162213, 1173512, 1257728, 1290496, 1318707, 1431125, 1438208, 1472207, 1527752, 1597696, 1601613
Offset: 1

Views

Author

Robert Israel, Mar 26 2019

Keywords

Comments

It is possible for a term of the sequence to be such a sum in more than one way, e.g., 1119371 = 215168 + 904203 = 366368 + 753003.
There are parametric solutions, and in particular the sequence is infinite. For example, 3^3*(-44100*k^2 - 21140*k + 471)^2 + 5^3*(-26460*k^2 + 4788*k + 865)^2 = 2^3*(132300*k^2 + 8820*k + 3527)^2, and these are coprime unless k==3 (mod 13).

Examples

			a(3)=95048 is in the sequence because 95048 = 2^3*109^2 = 45125 + 49923 = 5^3*19^2 + 3^3*43^2, and gcd(45125,49923)=1.
		

Crossrefs

Cf. A216427.

Programs

  • Maple
    N:= 10^6: # to get terms <= N
    A23:= {seq(seq(x^2*y^3, x= 2.. floor(sqrt(N/abs(y)^3))),y=2..floor(N^(1/3)))}: n:=nops(A23):
    Res:= NULL:
    for k from 1 to n do
      z:= A23[k];
      for i from 1 to n do
        x:= A23[i];
        if 2*x > z then break fi;
        if member(z-x,A23) and igcd(z,x)=1 then  Res:= Res, z; break fi
    od od:
    Res;

A320965 Squares divisible by a single cube > 1.

Original entry on oeis.org

16, 81, 144, 324, 400, 625, 784, 1936, 2025, 2401, 2500, 2704, 3600, 3969, 4624, 5625, 5776, 7056, 8100, 8464, 9604, 9801, 13456, 13689, 14641, 15376, 15876, 17424, 19600, 21609, 21904, 22500, 23409, 24336, 26896, 28561, 29241, 29584, 30625, 35344, 39204, 41616, 42849
Offset: 1

Views

Author

Hugo Pfoertner, Oct 25 2018

Keywords

Comments

Numbers whose prime factorization has a single exponent that is equal to 4 and all the rest, if they exist, are equal to 2. - Amiram Eldar, Jun 25 2022

Crossrefs

Programs

Formula

From Amiram Eldar, Jun 25 2022: (Start)
a(n) = A060687(n)^2.
Sum_{n>=1} 1/a(n) = (15/Pi^2) * Sum_{k>=2} (-1)^k * P(2*k) = 0.09603403868516162554..., where P(s) is the prime zeta function. (End)

A216426 Numbers of the form a^2*b^3, where a != b and a, b > 1.

Original entry on oeis.org

72, 108, 128, 200, 256, 288, 392, 432, 500, 512, 576, 648, 675, 800, 864, 968, 972, 1125, 1152, 1323, 1352, 1372, 1568, 1600, 1728, 1800, 1944, 2000, 2048, 2187, 2304, 2312, 2592, 2700, 2888, 2916, 3087, 3136, 3200, 3267, 3456, 3528, 3872, 3888, 4000
Offset: 1

Views

Author

V. Raman, Sep 07 2012

Keywords

Comments

Terms of A216427 that are not 5th powers of squarefree numbers (A113850) and not 10th powers of primes (A030629). - Amiram Eldar, Feb 07 2023

Crossrefs

Cf. A143610.
Subsequence of A216427. - Zak Seidov, Jan 03 2014

Programs

  • Mathematica
    With[{upto=4000},Select[Union[Flatten[{#[[1]]^2 #[[2]]^3,#[[2]]^2 #[[1]]^3}& /@ Subsets[Range[2,Surd[upto,2]],{2}]]],#<=upto&]](* Harvey P. Dale, Jan 04 2014 *)
    pMx = 25; mx = 2^3 pMx^2; t = Flatten[Table[If[a != b, a^2 b^3, 0], {a, 2, mx^(1/2)}, {b, 2, mx^(1/3)}]]; Union[Select[t, 0 < # <= mx &]] (* T. D. Noe, Jan 02 2014 *)
  • PARI
    list(lim)=my(v=List()); for(b=2, sqrtnint(lim\4,3), for(a=2, sqrtint(lim\b^3), if(a!=b, listput(v, a^2*b^3)))); Set(v) \\ Charles R Greathouse IV, Jan 02 2014
    
  • Python
    from math import isqrt
    from sympy import integer_nthroot, mobius, primepi
    def A216426(n):
        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):
            j, b, a, d = isqrt(x), integer_nthroot(x,6)[0], integer_nthroot(x,5)[0], integer_nthroot(x,10)[0]
            l, c = 0, n+x-2+primepi(b)+sum(mobius(k)*(j//k**3) for k in range(d+1, b+1))+primepi(d)+sum(mobius(k)*(a//k**2+j//k**3) for k in range(1, d+1))
            while j>1:
                k2 = integer_nthroot(x//j**2,3)[0]+1
                w = sum(mobius(k)*((k2-1)//k**2) for k in range(1, isqrt(k2-1)+1))
                c -= j*(w-l)
                l, j = w, isqrt(x//k2**3)
            return c+l
        return bisection(f,n,n) # Chai Wah Wu, Sep 13 2024

Formula

Sum_{n>=1} 1/a(n) = 2 + ((zeta(2)-1)*(zeta(3)-1)-1)/zeta(6) - zeta(5)/zeta(10) - P(6) - P(10) = 0.09117811499514578262..., where P(s) is the prime zeta function. - Amiram Eldar, Feb 07 2023

Extensions

Name corrected by Charles R Greathouse IV, Jan 02 2014

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 1-4 of 4 results.