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.

A056168 Duplicate of A054041.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 12, 15, 17, 20, 23, 25, 28, 32, 35, 38, 41, 45, 49, 52, 56, 60, 64, 68
Offset: 0

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Author

Keywords

A056813 Largest non-unitary prime factor of LCM(1,...,n); that is, the largest prime which occurs to power > 1 in prime factorization of LCM(1,..,n).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
Offset: 1

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Author

Labos Elemer, Aug 28 2000

Keywords

Comments

For n>0, prime(n) appears {(prime(n+1))^2 - (prime(n))^2} times [from n=(prime(n))^2 to n=(prime(n+1))^2 - 1], that is, A000040(n) appears A069482(n) times (from n=A001248(n) to n=A084920(n+1)). - Lekraj Beedassy, Mar 31 2005
a(n) is the largest prime factor of A045948(n). [Matthew Vandermast, Oct 29 2008]
Alternative definition: a(n) = largest prime <= sqrt(n) (considering 1 as prime for this occasion, see A008578 for the 19th century definition of primes). - Jean-Christophe Hervé, Oct 29 2013

Examples

			The j-th prime appears at the position of its square, at n = prime(j)^2.
		

Crossrefs

Programs

Formula

a(n) = prime(w) if prime(w)^2 <= n < prime(w+1)^2.
To get the sequence, repeat 1 three times, and then for any k >= 1, repeat A000040(k) A069482(k) times; or equivalently, for any k >= 1, repeat A008578(k) a number of times equal to A008578(k+1)^2 - A008578(k)^2. - Jean-Christophe Hervé, Oct 29 2013

Extensions

Corrected offset by Jean-Christophe Hervé, Oct 29 2013

A231405 Least integer j such that Sum_{i=1..j} 1/i^(1/3) >= n.

Original entry on oeis.org

1, 1, 3, 4, 6, 8, 10, 12, 15, 17, 20, 23, 25, 28, 32, 35, 38, 41, 45, 49, 52, 56, 60, 64, 68, 72, 76, 81, 85, 89, 94, 98, 103, 108, 113, 117, 122, 127, 132, 138, 143, 148, 153, 159, 164, 170, 175, 181, 187, 192, 198, 204, 210, 216, 222, 228, 234, 240, 247, 253
Offset: 0

Views

Author

Carmine Suriano, Nov 08 2013

Keywords

Examples

			a(7)=12 since Sum_{i=1..12} 1/i^(1/3) = 7.106248... and Sum_{i=1..11} 1/i^(1/3) = 6.669458... .
		

Crossrefs

Cf. A004080, A054040, A054041 (condition >n).
Cf. A067086.

Programs

  • JavaScript
    s=0;n=1;
    for (i=1;i<30;i++) {
    s+=1/Math.pow(i,1/3);
    if (s>=n) {n++;document.write(Math.floor(i)+", ");}
    }
  • Mathematica
    s = 0; i = 0; Table[i++; While[s = s + 1/(i^(1/3)); s < n, i++]; i, {n, 100}] (* T. D. Noe, Nov 09 2013 *)
    Module[{nn=300,c},c=Accumulate[1/Surd[Range[nn],3]];Table[Position[ c,?(#>=n&),1,1],{n,0,60}]]//Flatten (* _Harvey P. Dale, Aug 14 2021 *)

Extensions

a(0) added by Jon Perry, Nov 10 2013
Showing 1-3 of 3 results.