A055229 -id:A055229 - cp's OEIS frontend

A056059 GCD of largest square and squarefree part of central binomial coefficients.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 6, 2, 1, 1, 1, 3, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 6, 3, 1, 1, 1, 2, 3, 6, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 3, 6, 6, 3, 1, 2, 2, 1, 2, 1, 3, 6, 1, 1, 1, 2, 1, 2

Offset: 1

Labos Elemer, Jul 26 2000

nonn

			n=14, binomial(14,7) = 3432 = 8*3*11*13. The largest square divisor is 4, squarefree part is 858. So a(14) = gcd(4,858) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000188, A001405, A007913, A008833, A034974, A046098, A055229, A055231.
Cf. A056056, A056057, A056058, A056060, A056061.
A056201 is the cube of this sequence.

Table[GCD[First@ Select[Reverse@ Divisors@ #, IntegerQ@ Sqrt@ # &], Times @@ Power @@@ Map[{#1, Mod[#2, 2]} & @@ # &, FactorInteger@ #]] &@ Binomial[n, Floor[n/2]], {n, 80}] (* Michael De Vlieger, Feb 18 2017, after Zak Seidov at A007913 *)

A001405(n) = binomial(n, n\2);
A055229(n) = { my(c=core(n)); gcd(c, n/c); } \\ Charles R Greathouse IV, Nov 20 2012
A056059(n) = A055229(A001405(n)); \\ Antti Karttunen, Jul 20 2017

from sympy import binomial, gcd
from sympy.ntheory.factor_ import core
def a001405(n): return binomial(n, n//2)
def a055229(n):
    c = core(n)
    return gcd(c, n//c)
def a(n): return a055229(a001405(n))
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Jul 20 2017

a(n) = A055229(A001405(n)), where A055229(n) = gcd(A008833(n), A007913(n)).

Formula clarified by Antti Karttunen, Jul 20 2017

A056057 The largest square which divides n-th central binomial coefficient.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 9, 36, 1, 4, 4, 4, 9, 9, 1, 4, 1, 4, 4, 4, 1, 4, 100, 100, 900, 900, 36, 144, 9, 9, 9, 36, 25, 100, 100, 100, 9, 36, 4, 4, 4, 4, 900, 3600, 225, 900, 1764, 1764, 1764, 1764, 196, 784, 4, 4, 4, 16, 4, 16, 16, 16, 441, 441, 49, 196, 49, 196, 36, 36, 1, 4

Offset: 1

Labos Elemer, Jul 26 2000

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001405, A000188, A008833, A007913, A055229, A055231, A046098, A034974.

Cf. A056056, A056058, A056059, A056060, A056061.

Table[First@ Select[Reverse@ Divisors@ Binomial[n, Floor[n/2]], IntegerQ@ Sqrt@ # &], {n, 72}] (* Michael De Vlieger, Feb 18 2017 *)
a[n_] := Times @@ (First[#]^(2*Floor[Last[#]/2]) & /@ FactorInteger[Binomial[n, Floor[n/2]]]); Array[a, 100] (* Amiram Eldar, Sep 06 2020 *)

a(n) = A008833(A001405(n)).

a(A046098(n)) = 1.

A056626 Number of non-unitary square divisors of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Labos Elemer, Aug 08 2000

nonn

			n = p^u prime power has u+1 square divisors of which 2 (i.e., 1 and n) are unitary but u-1 are not unitary, so a(p^u) = u - 1. E.g., n = 4^4 = 256, has 5 square divisors {1, 4, 16, 64, 256} of which {4, 16, 64} are not unitary, so a(256)=3.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000188, A008833, A034444, A046951, A055229, A056624, A162641.

Table[DivisorSum[n, 1 &, And[IntegerQ@ Sqrt@ #, ! CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Jul 28 2017 *)
f1[p_, e_] := 1 + Floor[e/2]; f2[p_, e_] := 2^(1 - Mod[e, 2]); a[1] = 0; a[n_] := Times @@ f1 @@@ (fct = FactorInteger[n])- Times @@ f2 @@@ fct; Array[a, 100] (* Amiram Eldar, Sep 26 2022 *)

a(n) = {my(f = factor(n), r=0, m = 0); prod(i=1,#f~,f[i,2]>>1 + 1) - 2^(omega(f) - omega(core(f)))} \\ David A. Corneth, Jul 28 2017

a(n) = sumdiv(n, d, if(gcd(d, n/d)!=1, issquare(d))); \\ Michel Marcus, Jul 29 2017

from math import prod
from sympy import factorint
def A056626(n):
    f = factorint(n).values()
    return prod((e>>1)+1 for e in f)-(1<Chai Wah Wu, Aug 04 2024

a(n) = A046951(n) - 2^r(n), where r(n) is the number of distinct prime factors of the largest unitary square divisor of n. [Corrected by Amiram Eldar, Aug 03 2024]
a(n) = A046951(n) - 2^(A162641(n)). -  David A. Corneth, Jul 28 2017
From Amiram Eldar, Sep 26 2022: (Start)
a(n) = A046951(n) - A056624(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(2)*(1 - 1/zeta(3)) = 0.27650128922802056073... . (End)

a(32) and a(96) corrected by Michael De Vlieger, Jul 29 2017

A060476 Let n = 2^e_2 * 3^e_3 * 5^e_5 * ... be the prime factorization of n; sequence gives n such that 1 + max{e_2, e_3, ...} is nonprime.

Original entry on oeis.org

1, 8, 24, 27, 32, 40, 54, 56, 72, 88, 96, 104, 108, 120, 125, 128, 135, 136, 152, 160, 168, 184, 189, 200, 216, 224, 232, 243, 248, 250, 256, 264, 270, 280, 288, 296, 297, 312, 328, 343, 344, 351, 352, 360, 375, 376, 378, 384, 392, 408, 416, 424, 440, 456, 459, 472, 480

Offset: 1

N. J. A. Sloane, Sep 18 2008

nonn

The old entry with this sequence number was a duplicate of A005171.

The asymptotic density of this sequence is Sum_{c composite} (1/zeta(c) - 1/zeta(c-1)) = 0.1182437806... - Amiram Eldar, Oct 18 2020

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A096432, A074661.

Cf. A010051, A051903, A055229.

a060476 n = a060476_list !! (n-1)
a060476_list = filter ((== 0) . a010051' . (+ 1) . a051903) [1..]
-- Reinhard Zumkeller, Nov 30 2015

Join[{1}, Select[Range[500], !PrimeQ[1+Max[FactorInteger[#][[All, 2]]]]&]] (* Jean-François Alcover, Aug 02 2018 *)

isA060476(n) = if(n<2,1,!isprime(vecmax(factor(n)[,2])+1))

From Reinhard Zumkeller, Nov 30 2015: (Start)
A010051(A051903(a(n)+1)) = 1.
a(A055229(n)) > 1 for n > 1. (End)

A056060 The powerfree part of the central binomial coefficients.

Original entry on oeis.org

1, 2, 3, 6, 10, 5, 35, 70, 14, 7, 462, 231, 429, 429, 715, 1430, 24310, 12155, 92378, 46189, 88179, 88179, 1352078, 676039, 52003, 52003, 7429, 7429, 1077205, 1077205, 33393355, 66786710, 43214930, 21607465, 181502706, 90751353, 176726319, 176726319, 7658140490

Offset: 1

Labos Elemer, Jul 26 2000

nonn

			n=14, binomial(14,7) = 3432 = 8*3*11*13. The largest square divisor is 4, and the squarefree part is 858. So GCD(4,858) = 2 and a(14) = 858/2 = 429.

Amiram Eldar, Table of n, a(n) for n = 1..3388

Cf. A001405, A000188, A008833, A007913, A055229, A055231, A046098, A034974, A056056, A056057, A056058, A056059, A056060, A056061.

a[n_] := Denominator[(b = Binomial[n, Floor[n/2]])/(Times @@ First /@ FactorInteger[b])^2]; Array[a, 36] (* Amiram Eldar, Sep 05 2020 *)

a(n) = A055231(A001405(n)).

New name and more terms from Amiram Eldar, Sep 05 2020

A056623 If n=LLgggf (see A056192) and a(n) = LL, then its complementary divisor n/LL = gggf and gcd(L^2, n/LL) = 1.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 1, 25, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 4, 1, 49, 9

Offset: 1

Labos Elemer, Aug 08 2000

The part of the name "Largest unitary square divisor of n" was removed because it is correct only for numbers whose odd exponents in their prime factorization are all smaller than 5. For the correct largest unitary square divisor of n see A350388. - Amiram Eldar, Jul 26 2024

			a(200) = A008833(200)/A055229(200)^2 = 100/2^2 = 25.
a(250) = A008833(250)/A055229(250)^2 = 25/5^2 = 1.

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A008833, A055229, A000188, A046951, A034444, A056192, A056622, A350388.

f[p_, 1] := 1; f[p_, e_] := If[EvenQ[e], p^e, p^(e-3)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 11 2020 *)

A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ Charles R Greathouse IV, Nov 20 2012
A008833(n) = n/core(n) \\ Michael B. Porter, Oct 17 2009
A056623(n) = (A008833(n)/(A055229(n)^2)); \\ Antti Karttunen, Nov 19 2017

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(e == 1, 1, if(e%2, p^(e-3), p^e)));} \\ Amiram Eldar, Dec 25 2023

(definec (A056623 n) (if (= 1 n) n (let ((e (A067029 n)) (rest (A056623 (A028234 n)))) (cond ((even? e) (* (A028233 n) rest)) ((= 1 e) rest) (else (* (expt (A020639 n) (- e 3)) rest)))))) ;; After Jovovic's multiplicative formula, using memoization-macro definec - Antti Karttunen, Nov 19 2017

a(n) = A008833(n)/A055229(n)^2 = K^2/g^2, which coincides with the largest square divisor iff the g-factor is 1.
Multiplicative with a(p^e)=p^e for even e, a(p)=1, a(p^e)=p^(e-3) for odd e > 1. - Vladeta Jovovic, Apr 30 2002
From Amiram Eldar, Dec 25 2023 (Start)
Dirichlet g.f.: zeta(2*s-2) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s-2) + 1/p^(3*s)).
Sum_{k=1..n} a(k) ~ c * n^(3/2) / 3, where c = Product_{p prime} (1 + 1/p^(3/2) - 1/p^(5/2) + 1/p^(9/2)) = 1.81133051934001073532... . (End)
a(n) = A056622(n)^2. - Amiram Eldar, Jul 26 2024

Name edited by Amiram Eldar, Jul 26 2024

A056191 Characteristic cube divisor of n: cube of g = gcd(K,F), where K is the largest square root divisor of n (A000188) and F = n/(KK) = A007913(n) is its squarefree part; g^2 divides K^2 = A008833(n) = g^2L^2 and g divides F = gf.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Labos Elemer, Aug 02 2000

This is not the largest cube which divides n. It is canonical, since the decomposition n = KKgggf is unique (factors are defined above and dependent on n).

			If n=24, largest square divisor is 4, squarefree part is 6, g=2, a(24)=8; n=81, largest square divisor is 81, both F and g is 1, a(81)=1.

Antti Karttunen, Table of n, a(n) for n = 1..19683

Cf. A055229, A000188, A008833, A007913, A055231, A056192.

a[n_]:=With[{sf=Times@@Power@@@({#[[1]], Mod[#[[2]], 2]}&/@FactorInteger[n])}, GCD[sf, n/sf]]; Table[a[n]^3, {n, 1, 100}] (* Vincenzo Librandi, Oct 08 2017 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 1 || !(f[i,2]%2), 1,  f[i,1]^3));} \\ Amiram Eldar, Sep 05 2023

a(n) = A055229(n)^3 = g^3 = ggg; n = (KK)*(ggg)*f = K^2*g^3*f = KK*a(n)^3*f.

Multiplicative with a(p^e)=1 for even e, a(p)=1, a(p^e)=p^3 for odd e > 1. - Vladeta Jovovic, May 01 2002

A056192 a(n) = n divided by its characteristic cube divisor A056191.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 1, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 3, 25, 26, 1, 28, 29, 30, 31, 4, 33, 34, 35, 36, 37, 38, 39, 5, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 2, 55, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 9, 73, 74, 75

Offset: 1

Labos Elemer, Aug 02 2000

Different from A056552: e.g. a(16) = 16, while A056552(16) = 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000188, A008833, A007913, A055229, A055231, A056552, A056191.

f[p_, e_] := If[EvenQ[e], p^e, If[e == 1, p, p^(e - 3)]]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2020 *)

a(n) = n/A055229(n)^3 = n/g^3=n/ggg and n=(LL)*(ggg)*f=L^2*g^3*f=LL*a(n)^3*f, so n=L^2*(g*3)*f, where L=A000188(n)/A055229(n), f=A055231(n), g=A055231(n).
Multiplicative with a(p^e)=p^e for even e, a(p)=p, a(p^e)=p^(e-3) for odd e>1. - Vladeta Jovovic, Apr 30 2002
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4 + 1/p^6 - 1/p^7) = 0.4462648652... . - Amiram Eldar, Nov 13 2022

A056674 Number of squarefree divisors which are not unitary. Also number of unitary divisors which are not squarefree.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 2, 0, 0, 0, 2, 1, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 2, 0, 0, 0, 2, 2, 0, 0, 2, 1, 2, 0, 2, 0, 2, 0, 2, 0, 0, 0, 4, 0, 0, 2, 1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 2, 2, 0, 0, 0, 2, 1, 0, 0, 4, 0, 0, 0, 2, 0, 4, 0, 2, 0, 0, 0, 2, 0, 2, 2, 3, 0, 0, 0, 2, 0

Offset: 1

Labos Elemer, Aug 10 2000

Numbers of unitary and of squarefree divisors are identical, although the 2 sets are usually different, so sizes of parts outside overlap are also equal to each other.

			For n = 252, it has 18 divisors, 8 are unitary, 8 are squarefree, 2 belong to both classes, so 6 are squarefree but not unitary, thus a(252) = 6. The set {2,3,14,21,42} forms squarefree but non-unitary while the set {4,9,36,28,63,252} of same size gives the set of not squarefree but unitary divisors.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n.

Cf. A000005, A007913, A034444, A055229, A055231, A056671.

Table[DivisorSum[n, 1 &, And[SquareFreeQ@ #, ! CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Jul 19 2017 *)
f[p_, e_] := If[e == 1, 2, 1]; a[1] = 0; a[n_] := 2^Length[fct = FactorInteger[n]] - Times @@ (f @@@ fct); Array[a, 100] (* Amiram Eldar, Jul 24 2024 *)

A034444(n) = (2^omega(n));
A057521(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)); } \\ Charles R Greathouse IV, Aug 13 2013
A055231(n) = n/A057521(n);
A056674(n) = (A034444(n) - numdiv(A055231(n)));
\\ Or:
A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ Charles R Greathouse IV, Nov 20 2012
A056674(n) = ((2^omega(n)) - numdiv(core(n)/A055229(n)));
\\ Antti Karttunen, Jul 19 2017

a(n) = {my(f = factor(n), e = f[,2]); 2^omega(f) - prod(i = 1, #e, if(e[i] == 1, 2, 1));} \\ Amiram Eldar, Jul 24 2024

from sympy import gcd, primefactors, divisor_count
from sympy.ntheory.factor_ import core
def a055229(n):
    c=core(n)
    return gcd(c, n//c)
def a056674(n): return 2**len(primefactors(n)) - divisor_count(core(n)//a055229(n))
print([a056674(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 19 2017

a(n) = A034444(n) - A056671(n) = A034444(n) - A000005(A055231(n)) = A034444(n) - A000005(A007913(n)/A055229(n)).

A056627 a(n) = A056622(n!).

Original entry on oeis.org

1, 1, 1, 1, 1, 12, 12, 12, 36, 720, 720, 480, 480, 1680, 3024, 12096, 12096, 145152, 145152, 7257600, 345600, 1900800, 1900800, 136857600, 684288000, 4447872000, 4447872000, 435891456000, 435891456000, 3138418483200, 3138418483200, 6276836966400, 190207180800

Offset: 1

Labos Elemer, Aug 08 2000

nonn

Previous name "Square root of largest unitary square divisor of n!" was incorrect. See A374989 for the correct sequence with this name. - Amiram Eldar, Jul 26 2024

			a(12) = A056622(12!) = A000188(12!)/A055229(12!) = 1440/3 = 480.

Cf. A055772, A055230, A000188, A055229, A056622, A056628, A374989.

f[p_, 1] := 1; f[p_, e_] := If[EvenQ[e], p^(e/2), p^((e-3)/2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 33] (* Amiram Eldar, Jul 08 2024 *)

A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ Charles R Greathouse IV, Nov 20 2012
A008833(n) = n/core(n) \\ Michael B. Porter, Oct 17 2009
A056623(n) = (A008833(n)/(A055229(n)^2)); \\ Antti Karttunen, Nov 19 2017
a(n) = sqrtint(A056623(n!)); \\ Michel Marcus, Aug 16 2020

a(n) = A055772(n)/A055230(n) = A000188(n!)/A055229(n!).
a(n) = A056622(n!). - Michel Marcus, Aug 16 2020
a(n) = sqrt(A056628(n)). - Amiram Eldar, Jul 08 2024

More terms from Michel Marcus, Aug 16 2020

Incorrect name replaced with a formula by Amiram Eldar, Jul 26 2024

cp's OEIS Frontend

A056059 GCD of largest square and squarefree part of central binomial coefficients.

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A056057 The largest square which divides n-th central binomial coefficient.

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A056626 Number of non-unitary square divisors of n.

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A060476 Let n = 2^e_2 * 3^e_3 * 5^e_5 * ... be the prime factorization of n; sequence gives n such that 1 + max{e_2, e_3, ...} is nonprime.

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A056060 The powerfree part of the central binomial coefficients.

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A056623 If n=LLgggf (see A056192) and a(n) = LL, then its complementary divisor n/LL = gggf and gcd(L^2, n/LL) = 1.

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A056191 Characteristic cube divisor of n: cube of g = gcd(K,F), where K is the largest square root divisor of n (A000188) and F = n/(K*K) = A007913(n) is its squarefree part; g^2 divides K^2 = A008833(n) = g^2*L^2 and g divides F = gf.

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A056191 Characteristic cube divisor of n: cube of g = gcd(K,F), where K is the largest square root divisor of n (A000188) and F = n/(KK) = A007913(n) is its squarefree part; g^2 divides K^2 = A008833(n) = g^2L^2 and g divides F = gf.