A056175 -id:A056175 - cp's OEIS frontend

A056173 Number of unitary prime divisors of central binomial coefficient C(n, floor(n/2)) (A001405).

0, 1, 1, 2, 2, 1, 2, 3, 2, 1, 4, 3, 3, 3, 3, 4, 5, 4, 5, 4, 5, 5, 6, 5, 4, 4, 3, 3, 5, 5, 6, 7, 7, 6, 8, 7, 7, 7, 9, 8, 9, 9, 9, 9, 6, 6, 8, 7, 7, 7, 7, 7, 8, 8, 11, 11, 12, 12, 11, 11, 11, 11, 10, 11, 13, 12, 13, 12, 12, 12, 14, 13, 13, 13, 13, 13, 11, 11, 14, 13, 12, 12, 14, 14, 13, 13, 13

Offset: 1

Labos Elemer, Jul 27 2000

nonn

A prime divisor is unitary iff its exponent equals 1.

			For n = 10: binomial(10,5) = 252 = 2*2*3*3*7 has 3 prime factors of which only one, p = 7, is unitary. So a(10) = 1.

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

Cf. A001221, A001405, A034973, A056169, A056175.

Array[Function[k, Count[FactorInteger[k][[All, 1]], ?(CoprimeQ[#, k/#] &)]]@ Binomial[#, Floor[#/2]] &, 87]  (* _Michael De Vlieger, Oct 26 2017 *)

a(n) = vecsum(apply(x -> if(x > 1, 0, 1), factor(binomial(n, n\2))[,2])); \\ Amiram Eldar, Jul 22 2024

a(n) = A056169(A001405(n)). - Michel Marcus, Oct 27 2017 [corrected by Amiram Eldar, Jul 22 2024]

A056651 Numbers k such that binomial(k,floor(k/2)) has no non-unitary square divisors: all of their square divisors are unitary ones.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 31, 32, 35, 36, 37, 39, 40, 41, 43, 47, 48, 49, 55, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 79, 80, 95, 96, 97, 111, 129, 130, 131, 132, 133, 143, 144, 151, 161, 163, 167, 191, 192, 193

Offset: 1

Labos Elemer, Aug 09 2000

nonn

This property is weaker than "squarefreedom", but shows how central binomial coefficients are "poor of squares".

Numbers k such that binomial(k,floor(k/2)) is cubefree (A004709). - Amiram Eldar, Jul 22 2024

			223 is a term because x = binomial(223,111) has 35 prime divisors. 33 arises at power 1. Only 2 and 13 has powers 2 > 1. So square divisors of x are {1, 4, 169, 676} ={s}. All of them are also unitary divisors since GCD(s,x/s) = 1 holds for them.

T. D. Noe, Table of n, a(n) for n = 1..129 (no others < 10^8)

Cf. A001405, A004709, A046098, A056175, A110495.

Select[Range[0, 11000], AllTrue[FactorInteger[Binomial[#, Floor[#/2]]][[;;, 2]], #1 <= 2 &] &] (* Amiram Eldar, Jul 22 2024 *)

is(n) = if(n <= 1, 1, vecmax(factor(binomial(n, floor(n/2)))[, 2]) < 3); \\ Amiram Eldar, Jul 22 2024

A081394 a(n) is the smallest k such that number of non-unitary prime divisors of central binomial coefficient, A001405(k) = C(k, floor(k/2)) equals n.

Original entry on oeis.org

1, 6, 10, 27, 96, 147, 363, 627, 959, 1547, 1919, 2641, 2645, 3339, 6241, 6909, 6913, 6943, 6923, 6937, 16405, 19981, 24325, 31675, 31679, 35329, 36959, 36963, 38915, 38927, 73563, 39729, 73577, 80095, 87205, 87309, 95035, 123307, 123305, 123369, 123367, 174239, 185915, 186361, 186369, 186373, 186381

Offset: 0

Labos Elemer, Mar 27 2003

nonn

			n=8: a(8)=959, C(959,479) has 8 non-unitary prime divisors: {2,3,5,7,11,13,23,29} and 959 is the smallest.

Cf. A081387, A081393, A081395, A056175.

seq[len_, kmax_] := Module[{s = Table[0, {len}], k = 1, c = 0, i}, While[c < len && k < kmax, i = Count[FactorInteger[Binomial[k, Floor[k/2]]][[;; , 2]], ?(# > 1 &)] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = k]; k++]; TakeWhile[s, # > 0 &]]; seq[20, 10^4] (* _Amiram Eldar, May 15 2023 *)

a(n) = Min{k; A056175(k) = n}.

a(9)-a(19) from Michel Marcus, Sep 01 2019

a(20)-a(46) from Amiram Eldar, May 15 2023

A344272 a(n) is the least k such that the average number of nonunitary divisors of {1..k} is >= n.

Original entry on oeis.org

54, 816, 10530, 135200, 1733760, 22216752, 284685408, 3647978320, 46745561100, 599002268832, 7675674748560

Offset: 1

Amiram Eldar, May 13 2021

			a(1) = 54 since the average of the number of nonunitary divisors of {1..54} is (Sum_{k=1..54} A056175(k))/54 = 1.

Cf. A013661, A056175.
The nonunitary version of A085829.
Similar sequences: A328331, A336304, A338891, A338943, A344273, A344274.

nd[n_] := DivisorSigma[0,n] - 2^PrimeNu[n]; seq={}; s = 0; k = 1; Do[While[s = s + nd[k]; s < k*n, k++]; AppendTo[seq, k]; k++, {n, 1, 5}]; seq

Lim_{n->oo} a(n+1)/a(n) = exp(1/(1-1/zeta(2))) = exp(Pi^2/(Pi^2-6)) = 12.8140996101...

a(10)-a(11) from Martin Ehrenstein, May 23 2021

A056670 Largest non-unitary prime factor of A001405(n) = binomial(n, floor(n/2)), or 1 if no such prime exists.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 2, 2, 2, 3, 3, 1, 2, 1, 2, 2, 2, 1, 2, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 3, 3, 2, 2, 2, 2, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 2, 2, 2, 2, 2, 2, 2, 2, 7, 7, 7, 7, 7, 7, 3, 3, 1, 2, 2, 2, 5, 5, 7, 7, 7, 7, 7, 7, 3, 3, 5, 5, 5, 5, 3, 3, 7, 7, 7, 7, 7, 7, 5, 5, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Labos Elemer, Aug 10 2000

nonn

The largest prime divisor of A056057(n), the largest square divisor of binomial(n, floor(n/2)), or 1 if no such prime exists.

			For n = 28: binomial(28,14) = 2*2*2*3*3*3*5*5*17*19*23, so a(28) = 5.
For n = 342: binomial(342,171) = 32*F, where F is squarefree, so a(341) = 2.

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

Cf. A001405, A006530, A056057, A056175.

a[n_] := Module[{f = Select[FactorInteger[Binomial[n, Floor[n/2]]], Last[#] > 1 &]}, If[f == {}, 1, f[[-1, 1]]]]; Array[a, 100] (* Amiram Eldar, Oct 05 2024 *)

a(n) = A006530(A056057(n)).

cp's OEIS Frontend

A056173 Number of unitary prime divisors of central binomial coefficient C(n, floor(n/2)) (A001405).

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A056651 Numbers k such that binomial(k,floor(k/2)) has no non-unitary square divisors: all of their square divisors are unitary ones.

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A081394 a(n) is the smallest k such that number of non-unitary prime divisors of central binomial coefficient, A001405(k) = C(k, floor(k/2)) equals n.

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A344272 a(n) is the least k such that the average number of nonunitary divisors of {1..k} is >= n.

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A056670 Largest non-unitary prime factor of A001405(n) = binomial(n, floor(n/2)), or 1 if no such prime exists.

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