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.

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A374460 Indices of the nonsquarefree terms in the sequence of exponentially odd numbers (A268335).

Original entry on oeis.org

7, 18, 20, 24, 31, 39, 41, 63, 69, 74, 86, 89, 91, 97, 98, 109, 115, 121, 131, 135, 154, 161, 167, 174, 177, 179, 189, 194, 200, 211, 212, 223, 234, 243, 244, 249, 250, 265, 266, 268, 273, 290, 296, 302, 314, 325, 328, 338, 343, 348, 350, 366, 367, 373, 382, 388, 393
Offset: 1

Views

Author

Amiram Eldar, Jul 09 2024

Keywords

Comments

The asymptotic density of this sequence is 1 - A059956 / A065463 = 0.13700925215474602945... .

Examples

			The first 7 exponentially odd numbers are 1, 2, 3, 5, 6, 7, and 8. A268335(7) = 8 = 3^3 is the least nonsquarefree term. Therefore a(1) = 7.

Links

Crossrefs

Cf. A005117, A059956, A065463, A268335.
Similar sequences: A361936, A363189, A371186, A371188.

Programs

  • Mathematica
    Position[Select[Range[120], AllTrue[FactorInteger[#][[;; , 2]], OddQ] &], _?(!SquareFreeQ[#] &), Heads -> False] // Flatten
  • PARI
    isexpodd(k) = {my(e = factor(k)[, 2]); for(i = 1, #e, if(!(e[i] % 2), return(0))); 1;}
lista(kmax) = {my(f, c = 0); for(k = 1, kmax, if(isexpodd(k), c++; if(!issquarefree(k), print1(c, ", "))));}

Formula

A268335(a(n)) = A374459(n).

A371185 Indices of the cubefull numbers in the sequence of powerful numbers.

Original entry on oeis.org

1, 3, 5, 7, 8, 11, 13, 17, 18, 23, 25, 26, 30, 34, 38, 41, 42, 45, 49, 54, 55, 61, 63, 72, 77, 78, 80, 82, 83, 87, 89, 93, 99, 105, 106, 113, 115, 116, 127, 128, 130, 137, 140, 148, 151, 153, 161, 164, 166, 179, 185, 186, 188, 192, 196, 201, 206, 221, 227, 234
Offset: 1

Views

Author

Amiram Eldar, Mar 14 2024

Keywords

Examples

			The first 5 powerful numbers are 1, 4, 8, 9, and 16. The 1st, 3rd, and 5th, 1, 8, and 16, are also cubefull numbers. Therefore, the first 3 terms of this sequence are 1, 3, and 5.

Links

Crossrefs

Cf. A001694, A036966.
Cf. A090699, A362974.
Similar sequences: A361936, A371186.

Programs

  • Mathematica
    powQ[n_, emin_] := n == 1 || AllTrue[FactorInteger[n], Last[#] >= emin &]; Position[Select[Range[20000], powQ[#, 2] &], _?(powQ[#1, 3] &), Heads -> False] // Flatten
  • PARI
    ispow(n, emin) = n == 1 || vecmin(factor(n)[, 2]) >= emin;
lista(kmax) = {my(f, c = 0); for(k = 1, kmax, if(ispow(k, 2), c++; if(ispow(k, 3), print1(c, ", "))));}
  • Python
    from math import isqrt, gcd
from sympy import mobius, integer_nthroot, factorint
def A371185(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 = n+x
        for w in range(1,integer_nthroot(x,5)[0]+1):
            if all(d<=1 for d in factorint(w).values()):
                for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                    if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c -= integer_nthroot(z//y**4,3)[0]
        return c
    c, l, m = 0, 0, bisection(f,n,n)
    j = isqrt(m)
    while j>1:
        k2 = integer_nthroot(m//j**2,3)[0]+1
        w = squarefreepi(k2-1)
        c += j*(w-l)
        l, j = w, isqrt(m//k2**3)
    c += squarefreepi(integer_nthroot(m,3)[0])-l
    return c # Chai Wah Wu, Sep 12 2024

Formula

A001694(a(n)) = A036966(n).
a(n) ~ c * n^(3/2), where c = A090699 / A362974^(3/2) = 0.216089803749...

A371187 Numbers k such that there are no cubefull numbers between k^3 and (k+1)^3.

Original entry on oeis.org

1, 11, 16, 23, 72, 84, 140, 144, 197, 208, 223, 252, 286, 296, 300, 306, 313, 353, 477, 500, 502, 525, 528, 620, 671, 694, 721, 734, 737, 751, 785, 802, 827, 858, 900, 913, 916, 976, 1026, 1056, 1059, 1074, 1080, 1143, 1182, 1197, 1230, 1268, 1281, 1284, 1324
Offset: 1

Views

Author

Amiram Eldar, Mar 14 2024

Keywords

Comments

Positions of 0's in A337736.
This sequence has a positive asymptotic density (Shiu, 1991).

Links

Crossrefs

Cf. A036966, A336175, A337736, A371186.

Programs

  • Mathematica
    cubQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 2; Select[Range[1000], ! AnyTrue[Range[#^3 + 1, (# + 1)^3 - 1], cubQ] &]
(* or *)
seq[max_] := Module[{cubs = Union[Flatten[Table[i^5*j^4*k^3, {i, 1, Surd[max, 5]}, {j, 1, Surd[max/i^5, 4]}, {k, Surd[max/(i^5*j^4), 3]}]]], s = {}}, Do[If[IntegerQ[Surd[cubs[[k]], 3]], AppendTo[s, k]], {k, 1, Length[cubs]}]; Position[Differences[s], 1] // Flatten]; seq[10^10]
  • PARI
    iscub(n) = n == 1 || vecmin(factor(n)[, 2]) >= 3;
is(n) = for(k = n^3+1, (n+1)^3-1, if(iscub(k), return(0))); 1;

Formula

1 is a term since the two numbers between 1^2 = 1 and (1+1)^2 = 4, 2 and 3, are not cubefull.
Showing 1-3 of 3 results.

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