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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A052409 a(n) = largest integer power m for which a representation of the form n = k^m exists (for some k).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Eric W. Weisstein

Keywords

Comments

Greatest common divisor of all prime-exponents in canonical factorization of n for n>1: a(n)>1 iff n is a perfect power; a(A001597(k))=A025479(k). - Reinhard Zumkeller, Oct 13 2002
a(1) set to 0 since there is no largest finite integer power m for which a representation of the form 1 = 1^m exists (infinite largest m). - Daniel Forgues, Mar 06 2009
A052410(n)^a(n) = n. - Reinhard Zumkeller, Apr 06 2014
Positions of 1's are A007916. Smallest base is given by A052410. - Gus Wiseman, Jun 09 2020

Examples

			n = 72 = 2*2*2*3*3: GCD[exponents] = GCD[3,2] = 1. This is the least n for which a(n) <> A051904(n), the minimum of exponents.
For n = 10800 = 2^4 * 3^3 * 5^2, GCD[4,3,2] = 1, thus a(10800) = 1.

Links

Crossrefs

Cf. A052410, A005361, A051903, A072411-A072414, A124010, A075802, A072776, A270492.
Apart from the initial term essentially the same as A253641.
Differs from A051904 for the first time at n=72, where a(72) = 1, while A051904(72) = 2.
Differs from A158378 for the first time at n=10800, where a(10800) = 1, while A158378(10800) = 2.
Cf. A000005, A000961, A001597, A052410, A303386, A327501.

Programs

Formula

a(1) = 0; for n > 1, a(n) = gcd(A067029(n), a(A028234(n))). - Antti Karttunen, Aug 07 2017

Extensions

More terms from Labos Elemer, Jun 17 2002

A157754 a(1) = 0, a(n) = lcm(A051904(n), A051903(n)) for n >= 2.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2
Offset: 1

Views

Author

Jaroslav Krizek, Mar 05 2009

Keywords

Comments

a(n) for n >= 2 equals LCM of minimum and maximum exponents in the prime factorization of n.
a(n) for n >= 2 deviates from A072411, first different term is a(360), a(360) = 3, A072411(360) = 6.

Examples

			For n = 12 = 2^2 * 3^1 we have a(12) = lcm(2,1) = 2.
For n = 144 = 2^4 * 3^2 we have a(144) = lcm(4,2) = 4.

Links

Crossrefs

Cf. A000040, A003990, A006881, A033150, A120944, A000961, A000027, A072411, A051904, A051903, A158378.

Programs

  • Mathematica
    Table[LCM @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* Michael De Vlieger, Jul 12 2017 *)
  • PARI
    a(n) = if(n == 1, 0, my(e = factor(n)[,2]); lcm(vecmin(e), vecmax(e))); \\ Amiram Eldar, Sep 11 2024

Formula

a(1) = 0, a(p) = 1, a(pq) = 1, a(pq...z) = 1, a(p^k) = k, for p = primes (A000040), pq = product of two distinct primes (A006881), pq...z = product of k (k > 2) distinct primes p, q, ..., z (A120944), p^k = prime powers (A000961(n) for n > 1) k = natural numbers (A000027).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A033150. - Amiram Eldar, Sep 11 2024

A327503 Number of steps to reach a fixed point starting with n and repeatedly taking the quotient by the maximum divisor that is 1 or not a perfect power (A327501, A327502).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Sep 16 2019

Keywords

Comments

First differs from A052409 and A158378 at a(216) = 2, A052409(216) = A158378(216) = 3.
A multiset is aperiodic if its multiplicities are relatively prime. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). Heinz numbers of aperiodic multisets are numbers that are not perfect powers (A007916).
a(n) does not depend only on the prime signature of n. A351948 gives the positions where a(A046523(n)) <> a(n). n = 125000 is the first time this happens, see the examples. - Antti Karttunen, Apr 03 2022

Examples

			The transformation A327502 takes 144 -> 2 -> 1, so a(144) = 2.
From _Antti Karttunen_, Apr 03 2022: (Start)
For n = 1728 = 2^6 * 3^3, A327501(1728) = 864 = 2^5 * 3^3, and therefore A327502(1728) = 1728/864 = 2. A327501(2) = 2, thus A327502(2) = 2/2 = 1, so we reached 1 (= A327502(1)) in two steps, and therefore a(1728) = 2.
For n = 125000 = 2^3 * 5^6, A327501(125000) = 31250 = 2^1 * 5^6, and therefore A327502(125000) = 125000/31250 = 4. A327501(4) = 2, thus A327502(4) = 4/2 = 2, from which we reach 1 in one more step, therefore a(125000) = 3.
(End)

Links

Crossrefs

See link for additional cross-references.
Cf. A000961, A001597, A007916 (positions of 1's), A052409, A158378, A303386, A327501, A327502, A351948.
Cf. also A327500.

Programs

  • Mathematica
    Table[Length[FixedPointList[#/Max[Select[Divisors[#],GCD@@Last/@FactorInteger[#]==1&]]&,n]]-2,{n,100}]
  • PARI
    A327502(n) = if(n==1, 1, n/vecmax(select(x->((x>1) && !ispower(x)), divisors(n))));
A327503(n) = { my(u=A327502(n)); if(u==n, 0, 1+A327503(u)); }; \\ Antti Karttunen, Apr 02 2022

Formula

a(2^n) = n.

Extensions

Data section extended up to 105 terms by Antti Karttunen, Apr 02 2022
Showing 1-3 of 3 results.

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