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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A186970 The oex analog of the Euler phi-function for the oex prime power factorization of positive integers.

Original entry on oeis.org

1, 1, 2, 3, 4, 3, 6, 4, 8, 5, 10, 7, 12, 8, 9, 12, 16, 11, 18, 12, 14, 14, 22, 9, 24, 16, 18, 18, 28, 13, 30, 16, 22, 21, 25, 24, 36, 24, 27, 17, 40, 17, 42, 30, 33, 29, 46, 27, 48, 32, 36, 36, 52, 24, 42, 25, 40, 37, 58, 28, 60, 40, 49, 48, 50, 30, 66, 48, 49, 35, 70, 32, 72, 48, 54, 54, 61, 36
Offset: 1

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Author

Vladimir Shevelev, Mar 01 2011

Keywords

Comments

Oex divisors d of an integer n are defined in A186443: those divisors d which are either 1 or numbers such that d^k || n (the highest power of d dividing n) has odd exponent k.
A positive number is called an oex prime if it has only two oex divisors; since every n >= 2 has at least two oex divisors, 1 and n, an oex prime q has only oex divisors 1 and q. A000430 is the sequence of oex primes q, i.e., A186643(q) = 2 iff q is an entry in A000430.
A unique factorization, called an oex prime power factorization, of integers n is introduced as follows: each factor p^e in the conventional prime power factorization n = Product(p^e) is written as (p^2)^(e/2) if e is even, and as (p^2)^floor(e/2)*p if e is odd. This represents n as a product of oex primes of the type q=p^2, with unconstrained exponents e/2, and of oex primes of the type q=p with exponents 0 or 1. (This is similar to splitting n into its squarefree part A007913(n) times A008833(n), followed by an ordinary prime factorization in both parts separately.)
Let n = q_1^a_1*q_2^a_2*... and m = q_1^b_1*q_2^b_2*..., a_i,b_i >= 0 be the oex prime power factorizations of n and m. Define the oex GCD of n and m as [n,m] = q_1^min(a_1,b_1) * q_2^min(a_2,b_2) * .... Then a(n) = Sum_{m=1..n, [m,n]=1} 1, the oex analog of the Euler-phi function.

Examples

			The oex prime power factorization of 16 is 4^2. Since [16,i]=1 for i=1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, and 15, a(16)=12.
The oex prime power factorization of 9 is 9. Thus a(9)=8.
		

Crossrefs

Programs

  • Maple
    highpp := proc(n,d) local nshf,a ; if n mod d <> 0 then 0; else nshf := n ; a := 0 ; while nshf mod d = 0 do nshf := nshf /d ; a := a+1 ; end do: a; end if; end proc:
    oexgcd := proc(n,m) local a,p,kn,km ; a := 1 ; for p in numtheory[factorset](n) do kn := highpp(n,p) ; km := highpp(m,p) ; if type(kn,'even') = type(km,'even') then ; else kn := 2*floor(kn/2) ; km := 2*floor(km/2) ; end if; a := a*p^min(kn,km) ; end do: a ; end proc:
    A186970 := proc(n) local a,i; a := 0 ; for i from 1 to n do if oexgcd(n,i) = 1 then a := a+1 ; end if; end do: a; end proc:
    seq(A186970(n),n=1..80) ; # R. J. Mathar, Mar 18 2011

Formula

Let core(n) = p_1*...*p_r = A007913(n), n/core(n) = A008833(n) = q_1^c_1*...*q_t^c_t, where q_i are squares of primes.
If core(n)=1, then a(n) = n*Product_{j=1..r} (1-1/q_i); if core(n) tends to infinity, then a(n) ~ n * core(n) * Product_{i=1..t} (1-1/q_i) / Product_{j=1..r} (1+p_j).
a(n) <= A064380(n).

A186889 Oex perfect numbers: n such that A186644(n) = 2*n.

Original entry on oeis.org

6, 18, 20, 100, 1888, 2044928, 33099776, 35021696, 45335936, 533020672
Offset: 1

Views

Author

Vladimir Shevelev, Feb 28 2011

Keywords

Comments

There are no squarefree infinitary perfect numbers > 6 (cf. A007357). Therefore, the second and all further terms of the sequence are infinitary deficient (A129657).
No further term between 1888 and 1440000. - R. J. Mathar, Mar 18 2011
a(11) > 3*10^10. 1471763808896 is also a term. - Donovan Johnson, Jan 30 2013

Examples

			Let n = 100 with divisors 1, 2, 4, 5, 10, 20, 25, 50, and 100. By the definition in A186643, only 1, 4, 20, 25, 50, 100 among these are oex divisors. Since 1+4+20+25+50+100 = 2*100, 100 is in the sequence.
		

Crossrefs

Programs

  • PARI
    for(n=4, 10^9, if(isprime(n), next); d=divisors(n); s=n+1; for(j=2, numdiv(n)-1, for(k=2, 30, if(n%d[j]^k<>0, if(k%2==0, s=s+d[j]); k=30))); if(s==2*n, print(n))) /* Donovan Johnson, Jan 28 2013 */

Extensions

a(6)-a(10) from Donovan Johnson, Jan 28 2013
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