A000188 -id:A000188 - cp's OEIS frontend

A053164 4th root of largest 4th power dividing n.

1, 1, 1, 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, 1, 2, 1, 1, 1, 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, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 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, 1, 2

Offset: 1

Henry Bottomley, Feb 29 2000

Multiplicative with a(p^e) = p^[e/4]. - Mitch Harris, Apr 19 2005

			a(32) = 2 since 2 = 16^(1/4) and 16 is the largest 4th power dividing 32.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000188, A000190, A008835, A053150.

A053164 := proc(n) local a,f,e,p ; for f in ifactors(n)[2] do e:= op(2,f) ; p := op(1,f) ; a := a*p^floor(e/4) ; end do ; a ; end proc: # R. J. Mathar, Jan 11 2012

f[list_] := list[[1]]^Quotient[list[[2]], 4]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)

a(n) = A000188(A000188(n)) = A008835(n)^(1/4).
Multiplicative with a(p^e) = p^[e/4].
Dirichlet g.f.: zeta(4s-1)*zeta(s)/zeta(4s). - R. J. Mathar, Apr 09 2011
Sum_{k=1..n} a(k) ~ 90*zeta(3)*n/Pi^4 + 3*zeta(1/2)*sqrt(n)/Pi^2. - Vaclav Kotesovec, Dec 01 2020
a(n) = Sum_{d^4|n} phi(d). - Ridouane Oudra, Dec 31 2020
G.f.: Sum_{k>=1} phi(k) * x^(k^4) / (1 - x^(k^4)). - Ilya Gutkovskiy, Aug 20 2021

More terms from Antti Karttunen, Sep 13 2017

A053150 Cube root of largest cube dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Henry Bottomley, Feb 28 2000

This can be thought as a "lower 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "upper 3rd root" of positive integers is given by A019555. - Petros Hadjicostas, Sep 15 2019

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Vaclav Kotesovec, Graph - the asymptotic ratio.

Cf. A000188 (inner square root), A019554 (outer square root), A019555 (outer third root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A000190, A008834, A008835, A015051, A061704.

f[list_] := list[[1]]^Quotient[list[[2]], 3]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)
Table[SelectFirst[Reverse@ Divisors@ n, IntegerQ[#^(1/3)] &]^(1/3), {n, 105}] (* Michael De Vlieger, Jul 28 2017 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

A053150(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= (f[k, 1]^(f[k, 2]\3)); ); m; } \\ Antti Karttunen, Jul 28 2017

a(n) = my(f = factor(n)); for (k=1, #f~, f[k,2] = f[k,2]\3); factorback(f); \\ Michel Marcus, Jul 28 2017

from math import prod
from sympy import factorint
def A053150(n): return prod(p**(q//3) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

Multiplicative with a(p^e) = p^[e/3]. - Mitch Harris, Apr 19 2005
a(n) = A008834(n)^(1/3) = sqrt(A000189(n)/A000188(A050985(n))).
Dirichlet g.f.: zeta(3s-1)*zeta(s)/zeta(3s). - R. J. Mathar, Apr 09 2011
Sum_{k=1..n} a(k) ~ Pi^2 * n / (6*zeta(3)) + 3*zeta(2/3) * n^(2/3) / Pi^2. - Vaclav Kotesovec, Jan 31 2019
a(n) = Sum_{d^3|n} phi(d). - Ridouane Oudra, Dec 30 2020
G.f.: Sum_{k>=1} phi(k) * x^(k^3) / (1 - x^(k^3)). - Ilya Gutkovskiy, Aug 20 2021

More terms from Antti Karttunen, Jul 28 2017

A069290 Sum of the square roots of the square divisors of n.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 3, 1, 1, 1, 7, 1, 4, 1, 3, 1, 1, 1, 3, 6, 1, 4, 3, 1, 1, 1, 7, 1, 1, 1, 12, 1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 7, 8, 6, 1, 3, 1, 4, 1, 3, 1, 1, 1, 3, 1, 1, 4, 15, 1, 1, 1, 3, 1, 1, 1, 12, 1, 1, 6, 3, 1, 1, 1, 7, 13, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 1, 1, 1, 7, 1, 8, 4, 18, 1, 1

Offset: 1

Reinhard Zumkeller, Mar 14 2002

a(m)=1 iff m is squarefree (A005117).

			Square divisors for n=48: {1, 2^2, 4^2}, so a(48) = 1+2+4 = 7.

Nick Hobson, Table of n, a(n) for n = 1..1000
A. Dixit, B. Maji, and A. Vatwani, Voronoi summation formula for the generalized divisor function sigma_z^k(n), arXiv:2303.09937 [math.NT], 2023, sigma_(z=1,k=2,n).

Cf. A000005 (tau), A000188, A000203 (sigma), A001620, A005117, A035316, A037213, A046951.

nn = 102;f[list_, i_] := list[[i]]; a =Table[If[IntegerQ[n^(1/2)], n^(1/2), 0], {n, 1, nn}]; b =Table[1, {n, 1, nn}]; Table[DirichletConvolve[f[a, n], f[b, n], n, m], {m, 1, nn}] (* Geoffrey Critzer, Feb 21 2015 *)
f[p_, e_] := (p^(Floor[e/2] + 1) - 1)/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 20 2020 *)

vector(102, n, sumdiv(n, d, issquare(d)*sqrtint(d)))

a(n)={my(s=0);fordiv(n,d,if(issquare(d),s+=sqrtint(d)));s;} \\ Joerg Arndt, Feb 22 2015

from math import prod
from sympy import factorint
def A069290(n): return prod((p**(q//2+1)-1)//(p-1) for p, q in factorint(n).items()) # Chai Wah Wu, Jun 14 2021

Multiplicative with a(p^e) = (p^(floor(e/2)+1)-1)/(p-1). - Vladeta Jovovic, Apr 23 2002
G.f.: Sum_{k>=1} k*x^k^2/(1-x^k^2). - Ralf Stephan, Apr 21 2003
Dirichlet g.f.: zeta(2s-1)*zeta(s). Inverse Mobius transform of A037213. - R. J. Mathar, Oct 31 2011
Sum_{k=1..n} a(k) ~ n/2 * (log(n) - 1 + 3*gamma), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jan 31 2019
a(n) = Sum_{k=1..n} (1 - ceiling(n/k^2) + floor(n/k^2)) * k. - Wesley Ivan Hurt, Jan 28 2021
a(n) = A000203(A000188(n)). - Amiram Eldar, Sep 01 2023
a(n) = Sum_{d|n} d^(1/2)*(1-(-1)^tau(d))/2, [See Mathar comment]. - Wesley Ivan Hurt, Jul 09 2025

More terms from Larry Reeves (larryr(AT)acm.org), Jul 01 2002

A008835 Largest 4th power dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 81

Offset: 1

N. J. A. Sloane

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000188, A000190, A008833, A008834, A053164, A053165.

with(numtheory): [ seq( expand(nthpow(i,4)),i=1..200) ];

Max@ Select[Divisors@ #, IntegerQ@ Power[#, 1/4] &] & /@ Range@ 81 (* Michael De Vlieger, Mar 18 2015 *)
f[p_, e_] := p^(e - Mod[e, 4]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 15 2023 *)

a(n) = {f = factor(n); for (i=1, #f~, f[i,2] = 4*(f[i,2]\4);); factorback(f);} \\ Michel Marcus, Mar 16 2015

from math import prod
from sympy import factorint
def A008835(n): return prod(p**(e&-4) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 08 2024

a(n) = A000188(A000188(n))^4.
Multiplicative with a(p^e) = p^(4[e/4]). - Mitch Harris, Apr 19 2005
Dirichlet g.f.: zeta(s) * zeta(4s-4) / zeta(4s). - Álvar Ibeas, Feb 12 2015
Sum_{k=1..n} a(k) ~ zeta(5/4) * n^(5/4) / (5*zeta(5)) - 45*n/Pi^4. - Vaclav Kotesovec, Feb 03 2019
a(n) = n/A053165(n). - Amiram Eldar, Aug 15 2023
a(n) = A053164(n)^4. - Amiram Eldar, Sep 01 2024

Entry improved by comments from Henry Bottomley, Feb 29 2000

A055653 Sum of phi(d) [A000010] over all unitary divisors d of n (that is, gcd(d,n/d) = 1).

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 5, 7, 10, 11, 9, 13, 14, 15, 9, 17, 14, 19, 15, 21, 22, 23, 15, 21, 26, 19, 21, 29, 30, 31, 17, 33, 34, 35, 21, 37, 38, 39, 25, 41, 42, 43, 33, 35, 46, 47, 27, 43, 42, 51, 39, 53, 38, 55, 35, 57, 58, 59, 45, 61, 62, 49, 33, 65, 66, 67, 51, 69, 70, 71, 35, 73

Offset: 1

Labos Elemer, Jun 07 2000

Phi-summation over d-s if runs over all divisors is n, so these values do not exceed n. Compare also other "Phi-summations" like A053570, A053571, or distinct primes dividing n, etc.
a(n) is also the number of solutions of x^(k+1)=x mod n for some k>=1. - Steven Finch, Apr 11 2006
An integer a is called regular (mod n) if there is an integer x such that a^2 x == a (mod n). Then a(n) is also the number of regular integers a (mod n) such that 1 <= a <= n. - Laszlo Toth, Sep 04 2008
Equals row sums of triangle A157361 and inverse Mobius transform of A114810. - Gary W. Adamson, Feb 28 2009
a(m) = m iff m is squarefree, a(A005117(n)) = A005117(n). - Reinhard Zumkeller, Mar 11 2012
Apostol & Tóth call this ϱ(n), i.e., varrho(n). - Charles R Greathouse IV, Apr 23 2013

			n=1260 has 36 divisors of which 16 are unitary ones: {1,4,5,7,9,20,28,35,36,45,63,140,180,252,315,1260}.
EulerPhi values of these divisors are: {1,2,4,6,6,8,12,24,12,24,36,48,48,72,144,288}.
The sum is 735, thus a(1260)=735.
Or, 1260=2^2*3^2*5*7, thus a(1260) = (1 + 2^2 - 2)*(1 + 3^2 - 3)*(1 + 5 - 5^0)*(1 + 7 - 7^0) = 735.

J. Morgado, Inteiros regulares módulo n, Gazeta de Matematica (Lisboa), 33 (1972), no. 125-128, 1-5. [From Laszlo Toth, Sep 04 2008]
J. Morgado, A property of the Euler phi-function concerning the integers which are regular modulo n, Portugal. Math., 33 (1974), 185-191.

Antti Karttunen, Table of n, a(n) for n = 1..65537 (first 1000 terms from T. D. Noe)
Osama Alkam and Emad Abu Osba, On the regular elements in Zn, Turk J Math, 32 (2008), 31-39.
B. Apostol and L. Petrescu, Extremal Orders of Certain Functions Associated with Regular Integers (mod n), Journal of Integer Sequences, 2013, # 13.7.5.
Brăduţ Apostol and László Tóth, Some remarks on regular integers modulo n, arXiv:1304.2699 [math.NT], 2013.
Klaus Dohmen, On the Number of Regular Elements in Zn, arXiv:2304.02471 [math.CO], 2023.
S. R. Finch, Idempotents and nilpotents modulo n, arXiv:1304.2699 [math.NT], 2013.
V. S. Joshi, Order-free integers (mod m), Number Theory (Mysore, 1981), Lect. Notes in Math. 938, Springer-Verlag, 1982, pp. 93-100.
Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / (Pi^2 * n^2 / 12) for n = 1..100000
Sagar Mandal, Divisibility and Sequence Properties of sigma+ and phi+, arXiv:2508.11660 [math.GM], 2025.
József Sándor and Krassimir Atanassov, Some new arithmetic functions, Notes on Number Theory and Discrete Mathematics, Volume 30, 2024, Number 4, Pages 851-856. See phi+ function.
L. Tóth, Regular integers modulo n, arXiv:0710.1936 [math.NT], 2007-2008; Annales Univ. Sci. Budapest., Sect. Comp., 29 (2008), 263-275.
L. Tóth, A gcd-sum function over regular integers modulo n, JIS 12 (2009) 09.2.5.

Cf. A000010, A053570, A053571, A000188, A008833, A055654, A157361, A114810, A000010, A077610, A318661, A318662.

a055653 = sum . map a000010 . a077610_row
-- Reinhard Zumkeller, Mar 11 2012

A055653 := proc(n) local ans, i:ans := 1: for i from 1 to nops(ifactors(n)[ 2 ]) do ans := ans*(1+ifactors(n)[ 2 ][ i ] [ 1 ]^ifactors(n)[ 2 ] [ i ] [ 2 ]-ifactors(n)[ 2 ][ i ] [ 1 ]^(ifactors(n)[ 2 ] [ i ] [ 2 ]-1)): od: RETURN(ans) end:

a[n_] := Total[EulerPhi[Select[Divisors[n], GCD[#, n/#] == 1 &]]]; Array[a, 73] (* Jean-François Alcover, May 03 2011 *)
f[p_, e_] := p^e - p^(e-1) + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 10 2020 *)

a(n) = sumdiv(n, d, if(gcd(n/d, d)==1, eulerphi(d))); \\ Charles R Greathouse IV, Feb 19 2013, corrected by Antti Karttunen, Sep 03 2018

a(n)=my(f=factor(n));prod(i=1,#f[,1],f[i,1]^f[i,2]-f[i,1]^(f[i,2]-1)+1) \\ Charles R Greathouse IV, Feb 19 2013

If n = product p_i^e_i, a(n) = product (1+p_i^e_i-p_i^(e_i-1)). - Vladeta Jovovic, Apr 19 2001
Dirichlet g.f.: zeta(s)*zeta(s-1)*product_{primes p} (1+p^(-2s)-p^(1-2s)-p^(-s)). - R. J. Mathar, Oct 24 2011
Dirichlet convolution square of A318661(n)/A318662(n). - Antti Karttunen, Sep 03 2018
Sum_{k=1..n} a(k) ~ c * Pi^2 * n^2 / 12, where c = Product_{primes p} (1 - 1/p^2 - 1/p^3 + 1/p^4) = A330523 = 0.535896... - Vaclav Kotesovec, Dec 17 2019

A056595 Number of nonsquare divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 2, 1, 3, 1, 4, 1, 3, 3, 2, 1, 4, 1, 4, 3, 3, 1, 6, 1, 3, 2, 4, 1, 7, 1, 3, 3, 3, 3, 5, 1, 3, 3, 6, 1, 7, 1, 4, 4, 3, 1, 7, 1, 4, 3, 4, 1, 6, 3, 6, 3, 3, 1, 10, 1, 3, 4, 3, 3, 7, 1, 4, 3, 7, 1, 8, 1, 3, 4, 4, 3, 7, 1, 7, 2, 3, 1, 10, 3, 3, 3, 6, 1, 10, 3, 4, 3, 3, 3, 9, 1, 4, 4, 5, 1, 7, 1

Offset: 1

Labos Elemer, Jul 21 2000

nonn

a(A000430(n))=1; a(A030078(n))=2; a(A030514(n))=2; a(A006881(n))=3; a(A050997(n))=3; a(A030516(n))=3; a(A054753(n))=4; a(A000290(n))=A055205(n). - Reinhard Zumkeller, Aug 15 2011

			a(36)=5 because the set of divisors of 36 has tau(36)=nine elements, {1, 2, 3, 4, 6, 9, 12, 18, 36}, five of which, that is {2, 3, 6, 12, 18}, are not perfect squares.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000005, A000188, A046951.
See A194095 and A194096 for record values and where they occur.
Cf. A001620, A013661.

a056595 n = length [d | d <- [1..n], mod n d == 0, a010052 d == 0]
-- Reinhard Zumkeller, Aug 15 2011

A056595 := proc(n)
    local a,d ;
    a := 0 ;
    for d in numtheory[divisors](n) do
        if not issqr(d) then
            a := a+1 ;
        end if;
    end do:
    a;
end proc:
seq(A056595(n),n=1..40) ; # R. J. Mathar, Aug 18 2024

Table[Count[Divisors[n],?(#!=Floor[Sqrt[#]]^2&)],{n,110}] (* _Harvey P. Dale, Jul 10 2013 *)
a[1] = 0; a[n_] := Times @@ (1 + (e = Last /@ FactorInteger[n])) - Times @@ (1 + Floor[e/2]); Array[a, 100] (* Amiram Eldar, Jul 22 2019 *)

a(n)=sumdiv(n,d,!issquare(d)) \\ Charles R Greathouse IV, Aug 28 2016

a(n) = A000005(n) - A046951(n) = tau(n) - tau(A000188(n)).

Sum_{k=1..n} a(k) ~ n*log(n) + (2*gamma - zeta(2) - 1)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Dec 01 2023

A055773 a(n) = Product_{p in P_n} where P_n = {p prime, n/2 < p <= n }.

Original entry on oeis.org

1, 1, 2, 6, 3, 15, 5, 35, 35, 35, 7, 77, 77, 1001, 143, 143, 143, 2431, 2431, 46189, 46189, 46189, 4199, 96577, 96577, 96577, 7429, 7429, 7429, 215441, 215441, 6678671, 6678671, 6678671, 392863, 392863, 392863, 14535931, 765049, 765049, 765049

Offset: 0

Labos Elemer, Jul 12 2000

nonn

Old name: Product of primes p for which p divides n! but p^2 does not (i.e. ord_p(n!)=1). - Dion Gijswijt (gijswijt(AT)science.uva.nl), Jan 07 2007
Squarefree part of n! divided by gcd(Q,F), where Q is the largest square divisor and F is the squarefree part of n!. - Labos Elemer, Jul 12 2000
a(1) = 1, a(n) = n*a(n-1) if n is a prime else a(n) = least integer multiple of a(n-1)/n. - Amarnath Murthy, Apr 29 2004
Let P(i) denote the primorial number A034386(i). Then a(n) = P(n)/P(floor(n/2)). - Peter Luschny, Mar 05 2011
Letting H(n) = 1 + 1/2 + ... + 1/n denote the n-th harmonic number, it is known that a(n) is equal to the denominator (in lowest terms) of H(n)^2*n! for n >= 6 (see below example). - John M. Campbell, Mar 27 2016
For all n satisfying 6 <= n < 897, a(n) = A130087(n). - John M. Campbell, Mar 27 2016
It is also known that a(n) is equal to lcm^2(1, 2, ..., n)/gcd(lcm^2(1, 2, ..., n), n!). - John M. Campbell, Apr 04 2016

			n = 13, P_n = {7, 11, 13}, a(13) = 7*11*13 = 1001.
Letting n = 14, the denominator (in lowest terms) of H(n)^2*n! = 131803989435744/143 is a(14)=143. - _John M. Campbell_, Mar 27 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200 (corrected by Michel Marcus, Jan 19 2019)
J. M. Campbell et al., A problem involving the product prod_{k=1..n} k^mu(k), where mu denotes the Möbius function, Mathematics Stack Exchange (2016).
F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
Index entries for sequences related to Gijswijt's sequence

Cf. A000188, A008833, A007913, A055229, A055231 (for n), A055071, A055204, A055230, A094299, A094302, A193477, A130087.

a := n -> mul(k,k=select(isprime,[$iquo(n,2)+1..n])); # Peter Luschny, Jun 20 2009
A055773 := n -> numer(n!/iquo(n,2)!^4); # Peter Luschny, Jul 30 2011

Table[Numerator[n!/Floor[n/2]!^4], {n, 0, 40}] (* Michael De Vlieger, Mar 27 2016 *)

q=1;for(n=2,41,print1(q,",");q=if(isprime(n),q*n,q/gcd(q,n))) \\ Klaus Brockhaus, May 02 2004

a(n) = k=1;forprime(p=nextprime(n\2+1),precprime(n),k=k*p);k \\ Klaus Brockhaus, May 02 2004

a(n) = prod(i=primepi(n/2)+1,primepi(n),prime(i)) \\ John M. Campbell, Mar 27 2016

from math import prod
from sympy import primerange
def A055773(n): return prod(primerange((n>>1)+1,n+1)) # Chai Wah Wu, Apr 13 2024

a(n) = numerator(A056040(n)^2/n!).
a(n) = numerator(A056040(n)/floor(n/2)!^2).
a(n) = numerator(n!/floor(n/2)!^4). - Peter Luschny, Jul 30 2011
a(n) = product of primes p such that n/2 < p <= n. - Klaus Brockhaus, May 02 2004
a(n) = A055204(n)/A055230(n) = A055231(n!) = A007913(n!)/A055229(n!).
a(n) = Product_{i=pi(n/2)+1..pi(n)} prime(i), where pi denotes the prime counting function and prime(i) denotes the i-th prime number. - John M. Campbell, Mar 27 2016

Entry revised by N. J. A. Sloane, Jan 07 2007

Simpler definition based on a comment of Klaus Brockhaus, set offset to 0 and prepended 1 to data. - Peter Luschny, Mar 09 2013

A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

Original entry on oeis.org

2, 6, 18, 30, 90, 270, 450, 210, 630, 6750, 20250, 9450, 15750, 47250, 22050, 2310, 6930, 330750, 3543750, 1653750, 4961250, 53156250, 24806250, 727650, 1212750, 57881250, 173643750, 18191250, 8489250, 25467750, 2668050, 30030, 90090, 40020750, 1910081250, 891371250, 9550406250, 455814843750, 212713593750

Offset: 0

Antti Karttunen, Oct 10 2016

nonn

From David A. Corneth, Oct 22 2016: (Start)
The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
The exponent of 2 in the prime factorization of a(n) is 1. (End)

			A method to find terms of this sequence, explained by an example to find a(7). To find k = a(7), we find k such that A048675(k) = 2*7+1 = 15. 7 has the binary partitions: {[7, 0, 0], [5, 1, 0], [3, 2, 0], [1, 3, 0], [3, 0, 1], [1, 1, 1]}. To each of those, we prepend a 1. This gives the binary partitions of 15 starting with a 1. For example, for the first we get [1, 7, 0, 0]. We see that only [1, 5, 1, 0], [1, 3, 2, 0] and [1, 1, 1, 1] start nondecreasing, then nonincreasing, so we only check those. These numbers will be the exponents in a prime factorization. [1, 5, 1, 0] corresponds to prime(1)^1 * prime(2)^5 * prime(3)^1 * prime(4)^0 = 2430. We find that [1, 1, 1, 1] gives k = 210 for which A048675(k) = 15 so a(7) = 210. - _David A. Corneth_, Oct 22 2016

Antti Karttunen, Table of n, a(n) for n = 0..1024

Cf. A005811, A005940, A007306, A007949, A156552, A260443, A277189, A277323, A283484, A283975, A284267, A284268, A284563, A284564, A284573.

Cf. A277200 (same sequence sorted into ascending order).

a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a[2 n + 1], {n, 0, 38}] (* Michael De Vlieger, Apr 05 2017 *)

from sympy import factorint, prime, primepi
from operator import mul
def a003961(n):
    F=factorint(n)
    return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
def a260443(n): return n + 1 if n<2 else a003961(a260443(n//2)) if n%2==0 else a260443((n - 1)//2)*a260443((n + 1)//2)
def a(n): return a260443(2*n + 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017

a(n) = A260443((2*n)+1).
a(0) = 2; for n >= 1, a(n) = A260443(n) * A260443(n+1).
Other identities. For all n >= 0:
A007949(a(n)) = A005811(n). [See comments in A125184.]
A156552(a(n)) = A277189(n), a(n) = A005940(1+A277189(n)).
A048675(a(n)) = 2n + 1. - David A. Corneth, Oct 22 2016
A001222(a(n)) = A007306(1+n).
A056169(a(n)) = A284267(n).
A275812(a(n)) = A284268(n).
A248663(a(n)) = A283975(n).
A000188(a(n)) = A283484(n).
A247503(a(n)) = A284563(n).
A248101(a(n)) = A284564(n).
A046523(a(n)) = A284573(n).
a(n) = A277198(n) * A284008(n).
a(n) = A284576(n) * A284578(n) = A284577(n) * A000290(A284578(n)).

More linking formulas added by Antti Karttunen, Apr 16 2017

A055772 Square root of largest square dividing n!.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 24, 72, 720, 720, 1440, 1440, 10080, 30240, 120960, 120960, 725760, 725760, 7257600, 7257600, 79833600, 79833600, 958003200, 4790016000, 62270208000, 186810624000, 2615348736000, 2615348736000, 15692092416000

Offset: 1

Labos Elemer, Jul 12 2000

nonn

			For n=6, 6! = 720 = 144*5 so a(6) = sqrt(144) = 12.

Charles R Greathouse IV, Table of n, a(n) for n = 1..500

Cf. A000188, A008833, A007913, A055229, A055231, A055071, A055204, A055230.

a:= proc(n)
local r,F,t;
r:= n!;
F:= ifactors(r)[2];
mul(t[1]^floor(t[2]/2),t=F)
end proc:
seq(a(n), n= 1 .. 100); # Robert Israel, Oct 19 2014

Table[Last[Select[Sqrt[#]&/@Divisors[n!],IntegerQ]],{n,30}] (* Harvey P. Dale, Oct 08 2012 *)
(Sqrt@Factorial@Range@30)/.Sqrt[]->1 (* _Morgan L. Owens, May 04 2016 *)
f[p_, e_] := p^Floor[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Jul 26 2024 *)

a(n)=core(n!,2)[2] \\ Charles R Greathouse IV, Apr 03 2012

a(n) = A000188(n!) = sqrt(A008833(n!)) = sqrt(A055071(n)).
n! = a(n)^2*A055204(n) = a(n)^2*A007913(n!).
n! = (A000188(n!)^2)*A055229(n!)*A055231(n!).
log(a(n)) ~ n*log(n)/2. - David Radcliffe, Oct 17 2014

A072905 a(n) is the least k > n such that k*n is a square.

Original entry on oeis.org

4, 8, 12, 9, 20, 24, 28, 18, 16, 40, 44, 27, 52, 56, 60, 25, 68, 32, 76, 45, 84, 88, 92, 54, 36, 104, 48, 63, 116, 120, 124, 50, 132, 136, 140, 49, 148, 152, 156, 90, 164, 168, 172, 99, 80, 184, 188, 75, 64, 72, 204, 117, 212, 96, 220, 126, 228, 232, 236, 135, 244, 248

Offset: 1

Benoit Cloitre, Aug 10 2002

From Peter Kagey, Jun 22 2015: (Start)
a(n) is a bijection from the positive integers to A013929 (numbers that are not squarefree). Proof:
(1) Injection: Suppose that b(2) Surjection: Given some number k in A013929, a(A007913(k)*(A000188(k)-1)^2.) = k (End)

			12 is the smallest integer > 3 such that 3*12 = 6^2 is a perfect square, hence a(3) = 12.

Peter Kagey, Table of n, a(n) for n = 1..5000

Cf. A000188, A007913, A010052, A067722.

Cf. A182448, A253905.

a072905 n = head [k | k <- [n + 1 ..], a010052 (k * n) == 1]
-- Reinhard Zumkeller, Feb 07 2015

f:= proc(n) local F,f,x,y;
     F:= ifactors(n)[2];
     x:= mul(`if`(f[2]::odd,f[1],1),f=F);
     y:= mul(f[1]^floor(f[2]/2),f=F);
     x*(y+1)^2
end proc:
map(f, [$1..100]); # Robert Israel, Jun 23 2015

a[n_] := For[k = n+1, True, k++, If[IntegerQ[Sqrt[k*n]], Return[k]]]; Array[a, 100] (* Jean-François Alcover, Jan 26 2018 *)

a(n)=if(n<0,0,s=n+1; while(issquare(s*n)==0,s++); s)

a(n)=my(c=core(n)); (sqrtint(n/c)+1)^2*c \\ Charles R Greathouse IV, Jun 23 2015

def a(n)
  k = Math.sqrt(n).to_i
  k -= 1 until n % k**2 == 0
  n + 2*n/k + n/(k**2)
end # Peter Kagey, Jul 27 2015

a(n) = n + A067722(n). - Peter Kagey, Feb 05 2015
a(n) = A007913(n)*(A000188(n)+1)^2. - Peter Kagey, Feb 06 2015
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + 2*zeta(3)/zeta(2) + Pi^2/15 = 3.11949956554216757204... . - Amiram Eldar, Feb 17 2024

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A053164 4th root of largest 4th power dividing n.

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A053150 Cube root of largest cube dividing n.

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A069290 Sum of the square roots of the square divisors of n.

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A008835 Largest 4th power dividing n.

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A055653 Sum of phi(d) [A000010] over all unitary divisors d of n (that is, gcd(d,n/d) = 1).

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A056595 Number of nonsquare divisors of n.

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