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.

A322382 a(n) = p*a(n/p) + 1, where p is the smallest prime divisor of n; a(1)=0.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 1, 7, 4, 3, 1, 7, 1, 3, 4, 15, 1, 9, 1, 7, 4, 3, 1, 15, 6, 3, 13, 7, 1, 9, 1, 31, 4, 3, 6, 19, 1, 3, 4, 15, 1, 9, 1, 7, 13, 3, 1, 31, 8, 13, 4, 7, 1, 27, 6, 15, 4, 3, 1, 19, 1, 3, 13, 63, 6, 9, 1, 7, 4, 13, 1, 39, 1, 3, 19, 7, 8, 9, 1, 31, 40, 3, 1, 19, 6
Offset: 1

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Author

David James Sycamore, Dec 05 2018

Keywords

Comments

An equivalent definition of this sequence is: a(1) = 0; and for n > 1, a(n) = n*w(n), where if p1, ..., pk are the prime divisors of n (with repetition) and p1 <= p2 <= ... <= pk, then w(n) = 1/pk + 1/(pk-1*pk) + ... + 1/(p2*p3*...*pk) + ... + 1/(p1*p2*...*pk). Since 2 is smallest prime w(n) <= 1/2 + 1/(2^2) + ... + 1/(2^k), a partial sum of a series which -->1 as n-->oo. Therefore w(n) < 1 and n-a(n) is a sequence of positive numbers (1,1,2,1,4,3,6,...). For n=p^k, p a prime and n >= 1, a(n) = a(p^k) = p^(n-1) + p^(n-2) + ... + p^2 + p + 1 = (p^k-1)/(p-1); e.g., a(2^k) = 2^k - 1.

Examples

			For any prime p, a(p) = p*a(p/p)+1 = p*a(1)+1 = 1, because a(1) = 0.
For n = 6, the least prime divisor is 2, so a(6) = 2*a(6/2)+1 = 2*a(3)+1 = 3.
Using the equivalent definition we get w(6) = 1/3 + 1/6 = 1/2, so a(6) = 6*w(6) = 6*1/2 = 3. For n=32, a(32) = a(2^5) = 2^5 - 1 = 32 - 1 = 31.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := #*a[n/#] + 1 &[FactorInteger[n][[1, 1]] ]; a[1] = 0; Array[a, 1000] (* Michael De Vlieger, Jun 20 2025 *)
  • PARI
    a(n) = if (n==1, 0, my(p = vecmin(factor(n)[,1])); p*a(n/p)+1); \\ Michel Marcus, Jan 25 2019

Formula

From Antti Karttunen, Feb 28 2019 , Mar 04 2019: (Start)
a(1) = 0; for n > 1, a(n) = 1 + A020639(n)*a(A032742(n)).
If n is of the form p^k, k >= 1, then a(n) = A000203(A003557(n)). [Based on author's comments above] (End)
a(n) = Sum_{k=1..bigomega(n)} F^k(n), where F^k(n) is the k-th iterate of F(n) = A052126(n). - Ridouane Oudra, Aug 17 2024

Extensions

More terms from Michel Marcus, Jan 25 2019