cp's OEIS Frontend

Multiplicative with a(p^e) = p.

Product of the distinct prime factors of n.

a(k)=k for k=squarefree numbers A005117. - Lekraj Beedassy, Sep 05 2006

A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), b*c = A019554(n) = "outer square root" of n, and a(n) = lcm(a(b),c). Unless n is biquadrateful (A046101), a(n) = lcm(b,c). [Edited by Jeppe Stig Nielsen, Oct 10 2021, and Andrey Zabolotskiy, Feb 12 2025]

a(n) = A128651(A129132(n-1) + 2) for n > 1. - Reinhard Zumkeller, Mar 30 2007

Also the least common multiple of the prime factors of n. - Peter Luschny, Mar 22 2011

The Mobius transform of the sequence generates the sequence of absolute values of A097945. - R. J. Mathar, Apr 04 2011

Appears to be the period length of k^n mod n. For example, n^12 mod 12 has period 6, repeating 1,4,9,4,1,0, so a(12)= 6. - Gary Detlefs, Apr 14 2013

a(n) differs from A014963(n) when n is a term of A024619. - Eric Desbiaux, Mar 24 2014

a(n) is also the smallest base (also termed radix) for which the representation of 1/n is of finite length. For example a(12) = 6 and 1/12 in base 6 is 0.03, which is of finite length. - Lee A. Newberg, Jul 27 2016

a(n) is also the divisor k of n such that d(k) = 2^omega(n). a(n) is also the smallest divisor u of n such that n divides u^n. - Juri-Stepan Gerasimov, Apr 06 2017

The minimal exponent of the symmetric group S_n, i.e., the least positive integer for which x^a(n)=1 for all x in S_n. - Franz Vrabec, Dec 28 2008

Product over all primes of highest power of prime less than or equal to n. a(0) = 1 by convention.

Also smallest number whose set of divisors contains an n-term arithmetic progression. - Reinhard Zumkeller, Dec 09 2002

An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006. (This is wrong for n = 1 and n = 2. Should "for n large enough" be added? - Georgi Guninski, Oct 22 2011)

Corollary 3 of Farhi gives a proof that a(n) >= 2^(n-1). - Jonathan Vos Post, Jun 15 2009

Appears to be row products of the triangle T(n,k) = b(A010766) where b = A130087/A130086. - Mats Granvik, Jul 08 2009

Greg Martin (see link) proved that "the product of the Gamma function sampled over the set of all rational numbers in the open interval (0,1) whose denominator in lowest terms is at most n" equals (2*Pi)^(1/2)*a(n)^(-1/2). - Jonathan Vos Post, Jul 28 2009

a(n) = lcm(A188666(n), A188666(n)+1, ..., n). - Reinhard Zumkeller, Apr 25 2011

a(n+1) is the smallest integer such that all polynomials a(n+1)*(1^i + 2^i + ... + m^i) in m, for i=0,1,...,n, are polynomials with integer coefficients. - Vladimir Shevelev, Dec 23 2011

It appears that A020500(n) = a(n)/a(n-1). - Asher Auel, corrected by Bill McEachen, Apr 05 2024

n-th distinct value = A051451(n). - Matthew Vandermast, Nov 27 2009

a(n+1) = least common multiple of n-th row in A213999. - Reinhard Zumkeller, Jul 03 2012

For n > 2, (n-1) = Sum_{k=2..n} exp(a(n)*2*i*Pi/k). - Eric Desbiaux, Sep 13 2012

First column minus second column of A027446. - Eric Desbiaux, Mar 29 2013

For n > 0, a(n) is the smallest number k such that n is the n-th divisor of k. - Michel Lagneau, Apr 24 2014

Slowest growing integer > 0 in Z converging to 0 in Z^ when considered as profinite integer. - Herbert Eberle, May 01 2016

What is the largest number of consecutive terms that are all equal? I found 112 equal terms from a(370261) to a(370372). - Dmitry Kamenetsky, May 05 2019

Answer: there exist arbitrarily long sequences of consecutive terms with the same value; also, the maximal run of consecutive terms with different values is 5 from a(1) to a(5) (see link Roger B. Eggleton). - Bernard Schott, Aug 07 2019

Related to the inequality (54) in Ramanujan's paper about highly composite numbers A002182, also used in A199337: a(A329570(m))^2 is a (not minimal) bound above which all highly composite numbers are divisible by m, according to the right part of that inequality. - M. F. Hasler, Jan 04 2020

For n > 2, a(n) is of the form 2^e_1 * p_2^e_2 * ... * p_m^e_m, where e_m = 1 and e = floor(log_2(p_m)) <= e_1. Therefore, 2^e * p_m^e_m is a primitive Zumkeler number (A180332). Therefore, 2^e_1 * p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 2, a(n) = 2^e_1 * p_m^e_m * r, where r is relatively prime to 2*p_m, is a Zumkeller number (see my proof at A002182 for details). - Ivan N. Ianakiev, May 10 2020

For n > 1, 2|(a(n)+2) ... n|(a(n)+n), so a(n)+2 .. a(n)+n are all composite and (part of) a prime gap of at least n. (Compare n!+2 .. n!+n). - Stephen E. Witham, Oct 09 2021

The elements are called prime powers in contrast to the powers of primes which are the numbers of the same form but with k >= 0, cf. A000961.

Every nonzero integer is the product of elements of this sequence which are relatively prime and an element of {-1, 1}. This product is up to a rearrangement of the factors unique. (This statement is the fundamental theorem of arithmetic.)

These numbers are the numbers such that the von Mangoldt function is nonzero.

These numbers are the numbers of elements in finite fields. - Franz Vrabec, Aug 11 2004

A positive integer n is a prime power if and only if nZ is a primary ideal of Z. - John Cremona, Sep 02 2014

Also, numbers n divisible by their cototients A051953(n). - Ivan Neretin, May 29 2016

Numbers n such that (theta_3(q) - theta_3(q^n)) / 2 is the g.f. of a multiplicative sequence. - Michael Somos, Oct 17 2016

Numbers that are evenly divisible by exactly one prime number. - Lee A. Newberg, May 07 2018

Ram proved that these are precisely the numbers n such that the binomial coefficients n!/(m!(n-m)!) for m = 1..n-1 have a common factor greater than 1 (which is the unique prime dividing n). See Joris, Oestreicher & Steinig for a generalization. - Charles R Greathouse IV, Apr 24 2019

Blagojević & Ziegler prove that for these n and for any convex polygon in the plane, the polygon can be partitioned into n polygons with equal area and equal perimeter. The result is conjectured (by Nandakumar & Rao, who proved the case n = 2) to hold for all n. - Charles R Greathouse IV, Apr 24 2019

Numbers n such that A367064(n) < 0. - Chai Wah Wu, Nov 06 2023

This sequence represents all positive high amplitude peaks of the inverse Riemann spectrum R(x)= Sum_{k=1..oo} -cos(log(x)*Im(z_k)) going over the imaginary part of the nontrivial zeros "Im(z_k)" in the Riemann zeta function. - Marc Morgenegg, Jul 29 2025

The sequence of numbers divisible by a prime number of primes coincides with this up to 210, which has 4 prime factors. - Lior Manor, Aug 23 2001

A085970(n) = Max{k: a(k)<=n}.

Numbers n such that LCM of proper divisors of n equals neither 1 nor n. - Labos Elemer, Dec 01 2004

a(n) provides bases b in which automorphic numbers m^2 ending with m in base b exist. In the complement there aren't any automorphic numbers. - Martin Renner, Dec 07 2011

Numbers with at least 2 distinct prime factors. - Jonathan Sondow, Oct 17 2013

There exists an equiangular n-gon whose edge lengths form a permutation of 1, 2, ..., n if and only if n is in the sequence (see Woeginger's survey and Munteanu & Munteanu). - Jonathan Sondow, Oct 17 2013

Numbers that are the product of two relatively prime factors. These numbers are used in testing a sequence for multiplicativity. - Michael Somos, Jun 02 2015

A theorem from Donald McCarthy: Let d be any positive integer which is not a prime power; then there exists a finite group whose order is divisible by d but which contains no subgroup of order d (see link and A340511). - Bernard Schott, Dec 04 2021

Also nonprime n such that sigma(n)*phi(n) > (n-1)^2. - Benoit Cloitre, Apr 12 2002

If p is a term of the sequence, then the index n for which a(n) = p is given by n := b(p) := 1 + Sum_{k>=2} PrimePi(p^(1/k)). Here, the sum has floor(log_2(p)) positive terms. For any m > 0, the greatest number n such that a(n) <= m is also given by b(m), thus, b(m) is the number of such prime powers <= m. - Hieronymus Fischer, May 31 2013

That 8 and 9 are the only two consecutive integers in this sequence is known as Catalan's Conjecture and was proved in 2002 by Preda Mihăilescu. - Geoffrey Critzer, Nov 15 2015

a(n) = 1 iff A000961(n) = A006549(k) for some k. - Reinhard Zumkeller, Aug 25 2002

Also run lengths of distinct terms in A070198. - Reinhard Zumkeller, Mar 01 2012

Does this sequence contain all positive integers? - Gus Wiseman, Oct 09 2024

The original name was "Bit-reverse of n, including as many leading as trailing zeros." - Antti Karttunen, Dec 25 2024

A permutation of integers consisting only of fixed points and pairs. a(n)=n when n is a binary palindrome (including as many leading as trailing zeros), otherwise a(n)=A003010(n) (i.e. n has no axis of symmetry). A057890 gives the palindromes (fixed points, akin to A006995) while A057891 gives the "antidromes" (pairs). See also A280505.

This is multiplicative in domain GF(2)[X], i.e. with carryless binary arithmetic. A193231 is another such permutation of natural numbers. - Antti Karttunen, Dec 25 2024

Numbers k such that k + (0, 1) is a prime power pair.

Consecutive prime powers.

k + (0, 2m), m >= 1, being an admissible pattern for prime pairs, since (0, 2m) == (0, 0) (mod 2), has high density.

k + (0, 2m-1), m >= 1, being a non-admissible pattern for prime pairs, since (0, 2m-1) == (0, 1) (mod 2), has low density [the only possible pairs are (2^a - 2m-1, 2^a) or (2^a, 2^a + 2m-1), a >= 0].

Numbers k such that k and k+1 are primes would give only 2, for the prime pair (2, 3).

This sequence corresponds to the least member of each one of the following prime power pairs, ordered by increasing value of least member: (1, 2), (2^3, 3^2), (Fermat primes - 1, Fermat primes), (Mersenne primes, Mersenne primes + 1).

It is not known whether this sequence is infinite, but is conjectured to be since:

(*) 2^3, 3^2 are the only consecutive prime powers with exponents >= 2

(as a consequence of Mihailescu's theorem -- Mihailescu proved Catalan's conjecture in 2002);

(*) Only the first 5 Fermat numbers f_0 to f_4 are known to be prime

(it is conjectured that there might be no others, f_5 to f_32 are all composite);

(*) It has been conjectured that there exist an infinite number of Mersenne primes.

Numbers k such that A003418(k) appears only once in the sequence A003418. This may suggest that k is also characterized by the pairs formed by a 2 whose direct neighbor is a prime number in the sequence A014963. - Eric Desbiaux, Feb 11 2015

The power graph and enhanced power graph of the groups PGL(2,q) have the same clique number iff q>1 is a term of this sequence (Peter Cameron's link). - Bernard Schott, Dec 14 2021

Characterization: Let e_{p}(m) denote the exponent of the prime p in the prime factorization of m and [.] denote the Iverson bracket, then

e_{p}(a(n)) = Sum_{k>=1} [floor(n/p^k) is even].

This implies, among other things, that no prime > floor(n/2) can divide a(n). The prime exponents e_{2}(a(2n)) give Guy Steele's sequence GS(5,3) A080100.

Asymptotics: log a(n) ~ n(1 - log 2). It is conjectured that log a(n) ~ n(1 - log 2) + O(n^{1/2+eps}) for all eps > 0.

Bounds: A056040(floor(n/3)) <= a(n) <= A056040(floor(n/2)) if n >= 285.

Rows equal columns and are periodic. No zero elements are found (conjecture). The recurrence is related to the recurrence for the Mahonian numbers. The main diagonal is the Dirichlet inverse of the Euler totient function A023900 (conjecture). [The 2nd and 3rd formulas state that the conjecture is correct. R. J. Mathar, Sep 16 2017]

The sums from n=1 to infinity of T(n,k)/n converge to the Mangoldt function for column k (conjecture).

If gcd(n,k)=1 then T(n,k)=1 and if T(n,k)=1 then gcd(n,k)=1 (conjecture).

The Dirichlet generating functions for s > 1 for the columns appear to be (see A054535):

Zeta(s)*(1 + (Sum over all possible combinations of products of negative distinct prime factors of k, up to rearrangement, 1/((-1* first distinct prime factor)*(-1*second distinct prime factor)*(-1*third distinct prime factor * ...))^(s-1))).

Examples:

k=1: Zeta(s)

k=2: Zeta(s)*(1 - 1/2^(s-1))

k=3: Zeta(s)*(1 - 1/3^(s-1))

k=4: Zeta(s)*(1 - 1/2^(s-1))

k=5: Zeta(s)*(1 - 1/5^(s-1))

k=6: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) + 1/6^(s-1))

k=7: Zeta(s)*(1 - 1/7^(s-1))

...

k=30: Zeta(s)*(1 - 1/2^(s-1) - 1/3^(s-1) - 1/5^(s-1) + 1/6^(s-1) + 1/10^(s-1) + 1/15^(s-1) - 1/30^(s-1))

...

(conjecture)

See triangle A142971 for negative distinct prime factors.

This could probably be checked by matrix multiplication.

The signs of the eigenvalues of this matrix are a rearrangement of the Mobius function A008683 (conjecture). The first few eigenvalues are:

{1.0000}

{-1.4142, 1.4142}

{-2.6554, 1.8662, -1.2108}

{-3.4393, 2.1004, -1.6611, 0}

{-4.7711, -3.3867, 2.5910, -1.4332, 0}

{-5.2439, -3.4641, 3.4641, 2.5169, -2.2730, 0}

The relation to Dirichlet characters for the entries in this matrix appears appears to be formulated in terms of the sequence A089026 which is equal to n if n is a prime, otherwise equal to 1. See Mathematica program below. [Mats Granvik, Nov 23 2013]

From Mats Granvik, Jun 19 2016: (Start)

Remark about the Dirichlet generating function for the whole matrix: Subtracting the first column (in the form of zeta(c)) of the matrix gives us the limit: lim_{c->1} zeta(s)*zeta(c)/zeta(c+s-1)-zeta(c) = -zeta'(s)/zeta(s) which is the classical Dirichlet generating function for the von Mangoldt function.

For n >= k, see A231425, this matrix has row sums equal to zero except for the first row:

1=1

1-1=0

1+1-2=0

1-1+1-1=0

1+1+1+1-4=0

...

log(A014963(n)) = Sum_{k>=1} A191898(n,k)/k, for n>1.

log(A014963(k)) = Sum_{n>=1} A191898(n,k)/n, for k>1.

log(A014963(n)) = limit of zeta(s)*(Sum_{d divides n} A008683(d)/d^(s-1)) as s->1, for n>1.

A008683(n) = Sum_{k=1..n} A191898(n,k)*exp(-i*2*Pi*k/n)/n.

A008683(n) = Sum_{n=1..k} A191898(n,k)*exp(-i*2*Pi*n/k)/k.

(End)

From Mats Granvik, Aug 09 2016: (Start)

The Dirichlet generating function for the matrix is zero for any pair c and s = 2 - c and any pair s and c = 2 - s except at the pole c = 1 and s = 1 where it is indeterminate.

In the Mathematica program section, in the expression for the matrix as Dirichlets characters, the variables s and c can apparently be any pair of positive integers.

Limits related to the Dirichlet generating function for the matrix: Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s-1) = ZetaZero(n). Let s = ZetaZero(n), then lim_{c->1} zeta(s*c)/zeta(c+s*c-1) = ZetaZero(n)/(1+ZetaZero(n)).

(End)