A110493 -id:A110493 - cp's OEIS frontend

A110494 Least k such that prime(n)^2 divides binomial(2k,k).

3, 5, 13, 25, 61, 85, 145, 181, 265, 421, 481, 685, 841, 925, 1105, 1405, 1741, 1861, 2245, 2521, 2665, 3121, 3445, 3961, 4705, 5101, 5305, 5725, 5941, 6385, 8065, 8581, 9385, 9661, 11101, 11401, 12325, 13285, 13945, 14965, 16021, 16381, 18241, 18625, 19405

Offset: 1

T. D. Noe, Jul 22 2005

For prime p > sqrt(2n), p^2 does not divide binomial(2n,n).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 60 terms from T. D. Noe)

Cf. A110493 (largest prime p such that p^2 divides binomial(2n, n)).

t=Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[ -1, 1]]], {n, 100}]; Table[p=Prime[i]; First[Flatten[Position[t, p]]], {i, PrimePi[Max[t]]}]
lk[n_]:=Module[{k=1,c=Prime[n]^2},While[!Divisible[Binomial[2k,k],c], k=k+2]; k]; Array[lk,40] (* Harvey P. Dale, Oct 10 2012 *)

fv(n,p)=my(s); while(n\=p, s+=n); s
a(n)=my(p=prime(n),k=p^2\2+1); while(fv(2*k,p)-2*fv(k,p)<2,k++); k \\ Charles R Greathouse IV, Mar 27 2014

a(n)=prime(n)^2\2+1 \\ Charles R Greathouse IV, Mar 27 2014

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

a(n) = A066885(n), n > 1. - R. J. Mathar, Aug 18 2008

A239622 Conjecturally, the irregular triangle of numbers k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

Original entry on oeis.org

0, 1, 2, 4, 3, 6, 7, 9, 10, 11, 12, 21, 22, 28, 29, 30, 31, 36, 37, 54, 55, 57, 58, 110, 171, 784, 786, 5, 8, 15, 16, 17, 20, 35, 42, 45, 50, 51, 52, 53, 56, 59, 60, 77, 80, 133, 134, 135, 136, 156, 157, 158, 159, 160, 161, 170, 210, 211, 212, 400, 401, 402, 651, 652, 785

Offset: 0

T. D. Noe, Mar 27 2014

Row 0 lists the numbers k such that binomial(2k,k) is squarefree. Sequence A110494 lists the first term of each row; A239623 lists the conjectured last term; A239624 lists the conjectured length of each row.

			The irregular triangle begins:
0, 1, 2, 4
3, 6, 7, 9,..., 784, 786
5, 8, 15, 16,..., 652, 785
13, 14, 18, 19,..., 445, 2080
25, 26, 27, 32,..., 783, 902
61, 62, 63, 64,..., 2033, 2034

T. D. Noe, Rows n = 0..40 of irregular triangle, flattened

Cf. A059097 (union of first two rows), A110493, A110494, A239623, A239624.

b = 1; t = Table[b = b*(4 - 2/n); last = 0; Do[If[Mod[b, p^2] == 0, last = p], {p, Prime[Range[PrimePi[Sqrt[2*n]]]]}]; last, {n, 20000}]; t = Join[{0}, t]; Table[Flatten[Position[t, p]] - 1, {p, Join[{0}, Prime[Range[20]]]}]

A239623 Conjecturally, the largest k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

Original entry on oeis.org

4, 786, 785, 2080, 902, 2034, 2079, 1086, 2081, 2090, 1652, 2562, 3905, 8185, 4987, 3507, 5562, 2713, 3584, 4191, 8285, 9319, 12237, 12117, 12248, 9311, 8180, 8399, 9308, 20123, 11977, 11683, 12261, 14365, 15403, 20114, 16867, 19938, 19559, 20316, 24706

Offset: 0

T. D. Noe, Mar 27 2014

nonn

The last number in row n of A239622. The 0th term is the largest number k such that binomial(2k,k) is squarefree. The first 41 terms were checked by computing binomial(2k,k) for k <= 10^5. See the plot in A110493.

Vincenzo Librandi, Table of n, a(n) for n = 0..125

Cf. A110493, A110494, A239622, A239624.

b = 1; t = Table[b = b*(4 - 2/n); last = 0; Do[If[Mod[b, p^2] == 0, last = p], {p, Prime[Range[PrimePi[Sqrt[2*n]]]]}]; last, {n, 25000}]; t = Join[{0}, t]; Table[Flatten[Position[t, p]][[-1]] - 1, {p, Join[{0}, Prime[Range[20]]]}]

A239624 Conjecturally, the number of numbers k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

Original entry on oeis.org

4, 23, 38, 50, 51, 54, 65, 70, 107, 127, 127, 165, 155, 150, 239, 287, 280, 179, 336, 314, 230, 453, 423, 600, 612, 419, 246, 454, 455, 892, 1117, 624, 916, 432, 1115, 363, 934, 1061, 763, 1073, 1203, 524, 1523, 559, 1278, 735, 2221, 1987, 929, 475, 1179, 1605

Offset: 0

T. D. Noe, Mar 27 2014

nonn

The 0th term is the largest number k such that binomial(2k,k) is squarefree.

Cf. A110493, A110494, A239622, A239623.

b = 1; t = Table[b = b*(4 - 2/n); last = 0; Do[If[Mod[b, p^2] == 0, last = p], {p, Prime[Range[PrimePi[Sqrt[2*n]]]]}]; last, {n, 20000}]; t = Join[{0}, t]; Table[Length[Position[t, p]], {p, Join[{0}, Prime[Range[20]]]}]

A227902 Numbers n such that triangular(n) divides binomial(2n,n).

Original entry on oeis.org

1, 2, 4, 6, 15, 20, 24, 28, 40, 42, 45, 66, 72, 77, 88, 91, 104, 110, 126, 140, 153, 156, 170, 187, 190, 204, 209, 210, 220, 228, 231, 238, 240, 266, 276, 299, 304, 308, 312, 315, 322, 325, 330, 345, 368, 378, 414, 420, 429, 435, 440, 442, 450, 459, 460, 464, 468, 476, 480

Offset: 1

Alex Ratushnyak, Oct 14 2013

A014847 is a subsequence.

			triangular(6)=21, A000984(6)=924. Because 21 divides 924, 6 is in the sequence.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A000984, A014847, A081766, A081767, A110493-A110496, A226047.

Select[Range[480], Mod[Binomial[2 #, #], # (# + 1)/2] == 0 &] (* T. D. Noe, Oct 16 2013 *)

is(n) = { my(f = factor(binomial(n+1, 2))); for(i = 1, #f~, if(val(2*n, f[i, 1]) - 2*val(n, f[i, 1]) < f[i, 2], return(0) ) ); 1 }
val(n, p) = my(r=0); while(n, r+=n\=p);r \\ David A. Corneth, Apr 03 2021

from sympy import binomial
for n in range(1, 444):
    CBC = binomial(2 * n, n)
    if not CBC % binomial(n + 1, 2):
       print(n, end=",")

cp's OEIS Frontend

A110494 Least k such that prime(n)^2 divides binomial(2k,k).

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A239622 Conjecturally, the irregular triangle of numbers k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

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A239623 Conjecturally, the largest k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

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A239624 Conjecturally, the number of numbers k such that prime(n)^2 is the largest squared prime divisor of binomial(2k,k).

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A227902 Numbers n such that triangular(n) divides binomial(2n,n).

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