A110495 -id:A110495 - cp's OEIS frontend

A110496 Least k such that prime(n)^3 divides binomial(2k,k).

7, 14, 63, 172, 666, 1099, 2457, 3430, 6084, 12195, 14896, 25327, 34461, 39754, 51912, 74439, 102690, 113491, 150382, 178956, 194509, 246520, 285894, 352485, 456337, 515151, 546364, 612522, 647515, 721449, 1024192, 1124046, 1285677

Offset: 1

T. D. Noe, Jul 22 2005

nonn

For prime p > (2n)^(1/3), p^3 does not divide binomial(2n,n).

Chai Wah Wu, Table of n, a(n) for n = 1..100

Cf. A000984, A030078.

Cf. A110495 (binomial(2k, k) is cubefree).

t3=Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>2&]; If[s=={}, 0, s[[ -1, 1]]], {n, 15000}]; Table[p=Prime[i]; First[Flatten[Position[t3, p]]], {i, PrimePi[Max[t3]]}]
lst={7};Do[AppendTo[lst,(DivisorSigma[3,Prime[n]])/2],{n,2,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)

a(n) = (prime(n)^3 + 1)/2 = (1+A030078(n))/2 for n>1.

Product_{n>=1} (1 - 1/a(n)) = (54/49)*zeta(6)/zeta(3)^2. - Amiram Eldar, Jun 08 2022

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

A110499 Largest k such that n is the highest power in the factorization of binomial(2k,k).

Original entry on oeis.org

4, 1056, 540928

Offset: 1

T. D. Noe, Jul 22 2005

Checked all k < 10^8. a(1) has been proved. a(2) and a(3) are conjectured.

Checked all k < 10^9. a(4) is at least 537927682. - T. D. Noe, Jul 27 2005

			a(1)=4 because binomial(8,4) is the largest squarefree central binomial coefficient. a(2)=1056 because binomial(2112,1056) is the largest cubefree.

Cf. A110495 (n such that binomial(2n, n) is cubefree).

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A110496 Least k such that prime(n)^3 divides binomial(2k,k).

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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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A110499 Largest k such that n is the highest power in the factorization of binomial(2k,k).

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