A014847 -id:A014847 - cp's OEIS frontend

A120622 Numbers k such that the k-th Catalan number C(2k, k)/(k + 1) is divisible by k/2 but not divisible by k.

4, 24, 40, 72, 304, 480, 560, 1032, 1248, 1376, 1440, 1632, 1696, 2128, 2320, 2400, 2576, 2720, 3360, 3520, 4104, 4176, 4240, 4320, 4368, 4704, 4896, 5056, 5280, 5568, 5664, 5824, 6160, 6240, 6592, 6848, 6976, 7200, 7360, 7744, 8064, 8200, 8240, 8272

Offset: 1

Robert G. Wilson v, Jun 19 2006

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000108, A000217, A014847, A120623.

fQ[n_] := fQ[n_] := IntegerQ[ Binomial[2n, n]/(n(n + 1)/2)] && !IntegerQ[ Binomial[2n, n]/n]; Select[ Range@8719, fQ@# &]

Better description from Joel B. Lewis, Nov 15 2006

A120623 Numbers n such that the n-th Catalan number C(2n, n)/(n + 1) is divisible by n/3 but not divisible by n.

Original entry on oeis.org

3, 12, 18, 84, 99, 120, 198, 216, 273, 342, 432, 522, 540, 792, 828, 945, 975, 984, 990, 1035, 1071, 1323, 1377, 1512, 1548, 1575, 1710, 1755, 1863, 2052, 2106, 2226, 2385, 2442, 2790, 2928, 3009, 3015, 3132, 3198, 3483, 3672, 3807, 3915, 3996, 4212, 4428

Offset: 1

Robert G. Wilson v, Jun 19 2006

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000108, A014847, A120622.

fQ[n_] := fQ[n_] := IntegerQ[ Binomial[2n, n]/(n(n + 1)/3)] && !IntegerQ[ Binomial[2n, n]/n]; Select[ Range@4463, fQ@# &]

A120623_list, b = [], 1
for n in range(1, 10**5):
    if b % n and not (3*b) % n:
        A120623_list.append(n)
    b = b*(4*n+2)//(n+2) # Chai Wah Wu, Mar 30 2016

Better description from Joel B. Lewis, Nov 15 2006

A120624 Numbers n such that the n-th Catalan number C(2n,n)/(n+1) is divisible by 2n.

Original entry on oeis.org

6, 28, 42, 45, 66, 77, 91, 110, 126, 140, 153, 156, 170, 187, 190, 204, 209, 210, 220, 228, 231, 238, 266, 276, 299, 308, 312, 315, 322, 325, 330, 345, 378, 414, 420, 429, 435, 440, 442, 450, 459, 460, 468, 476, 483, 493, 496, 510, 527, 551, 558, 561, 570

Offset: 1

Robert G. Wilson v, Jun 19 2006

nonn

Equivalently, numbers such that the n-th central binomial coefficient C(2n, n) is divisible by 2n(n + 1). - Joel B. Lewis, Jan 07 2008

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Subset of A014847.

Cf. A120622.

fQ[n_] := fQ[n_] := IntegerQ[ Binomial[2n, n]/(2n(n + 1))]; Select[ Range@8719, fQ@# &]
Select[Range[600],Divisible[CatalanNumber[#],2#]&] (* Harvey P. Dale, Aug 30 2016 *)

from _future_ import division
A120624_list, b = [], 1
for n in range(1,10**5):
    if not b % (2*n):
        A120624_list.append(n)
    b = b*(4*n+2)//(n+2) # Chai Wah Wu, Mar 25 2016

Definition corrected by Joel B. Lewis, Apr 30 2009

A260636 a(n) = binomial(3n, n) mod n.

Original entry on oeis.org

0, 1, 0, 3, 3, 0, 3, 7, 3, 5, 3, 0, 3, 8, 9, 15, 3, 15, 3, 15, 0, 4, 3, 12, 3, 2, 3, 12, 3, 24, 3, 15, 18, 15, 0, 9, 3, 34, 6, 31, 3, 21, 3, 0, 15, 38, 3, 36, 3, 40, 33, 40, 3, 42, 0, 16, 27, 44, 3, 0, 3, 46, 45, 47, 39, 51, 3, 53, 15, 0, 3, 45, 3, 15, 9, 20, 76, 0, 3, 7, 3

Offset: 1

M. F. Hasler, Nov 11 2015

Motivated by A080469: C(3n,n)=3^n (mod n), A109641, A109642 and other sequences.

See A059288 for the "2n" analog. Sequence A260640 yields the indices of zeros (analog to A014847 for 2n).

			n=1: C(3,1) = 3 = 0 (mod 1).
n=2: C(3*2,2) = 15 = 1 (mod 2).
n=3: C(3*3,3) = 84 = 0 (mod 3).
n=4: C(3*4,4) = 495 = 3 (mod 4).

Chai Wah Wu, Table of n, a(n) for n = 1..10000
M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (Vers. 1.4, 11/2015).

Cf. A080469, A109641, A109642; A260640 (indices of zeros); A059288, A014847 (analogs for 2n).

[Binomial(3*n,n) mod n : n in [1..100]]; // Wesley Ivan Hurt, Nov 12 2015

A260636:=n->binomial(3*n,n) mod n: seq(A260636(n), n=1..100); # Wesley Ivan Hurt, Nov 12 2015

Array[Mod[Binomial[3 #, #], #] &, 112] (* Michael De Vlieger, Nov 12 2015 *)

PARI
```
a(n)=binomial(3*n,n)%n
```

A260636(n)=lift(binomod(3*n,n,n)) \\ using binomod.gp by M. Alekseyev, cf. Links.

from _future_ import division
A260636_list, b = [], 3
for n in range(1,10001):
    A260636_list.append(b % n)
    b = b*3*(3*n+2)*(3*n+1)//((2*n+2)*(2*n+1)) # Chai Wah Wu, Jan 27 2016

A260640 Numbers n such that binomial(3*n,n) == 0 (mod n).

Original entry on oeis.org

1, 3, 6, 12, 21, 35, 44, 55, 60, 70, 78, 88, 90, 99, 102, 110, 117, 119, 120, 133, 156, 171, 176, 180, 184, 204, 207, 220, 225, 230, 231, 234, 238, 240, 247, 252, 255, 285, 286, 300, 312, 341, 342, 348, 360, 368, 372, 391, 403, 408, 414, 425, 434, 460, 462, 465, 468, 481, 483, 494, 495, 504, 506, 510, 550, 555, 561, 572, 574, 585, 600

Offset: 1

M. F. Hasler, Nov 11 2015

nonn

See A014847 for the analog for 2n.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (v.1.4, 11/2015).

Cf. A260636, A080469, A109641, A109642, A014847.

[n: n in [1..600] |Binomial(3*n,n) mod n eq 0]; // Vincenzo Librandi, Jan 29 2016

Select[Range@ 600, Mod[Binomial[3 #, #], #] == 0 &] (* Michael De Vlieger, Nov 12 2015 *)

for(n=1,999,binomod(3*n,n,n)==0&&print1(n",")) \\ Using binomod.gp by M. Alekseyev, cf. links.

from _future_ import division
A260640_list, b = [], 3
for n in range(1,10**3):
    if not b % n:
        A260640_list.append(n)
    b = b*3*(3*n+2)*(3*n+1)//((2*n+2)*(2*n+1)) # Chai Wah Wu, Jan 27 2016

A048618 Even numbers n such that binomial(n,n/2) is divisible by n/2.

Original entry on oeis.org

2, 4, 12, 30, 40, 56, 84, 90, 132, 154, 176, 182, 208, 220, 252, 280, 306, 312, 340, 374, 380, 408, 418, 420, 440, 456, 462, 476, 480, 532, 552, 598, 616, 624, 630, 644, 650, 660, 690, 736, 756, 828, 840, 858, 870, 880, 884, 900, 918, 920, 928, 936, 952, 966

Offset: 1

Labos Elemer

nonn

			For n=30, binomial(30,15) = 155117520 = 15^10341168, so 30 is a term.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (V. 1.4, 11/2015).

Cf. A001405, A020475, A014847, A067348 (binomial(2*n,n) is divisible by 2*n).

a:=[];
for n from 1 to 1000 do if ( binomial(2*n,n) mod n ) = 0 then a:=[op(a),2*n]; fi; od;
a;   # N. J. A. Sloane, Aug 03 2017

Select[Range[2,1000,2],Mod[Binomial[#,#/2],#/2]==0&] (* Harvey P. Dale, Jan 23 2025 *)

a(n) = 2 * A014847(n). - Rémy Sigrist, Aug 27 2017

Definition corrected by N. J. A. Sloane, Aug 03 2017

A278961 Triangle read by rows: row n consists of k, 1<=k<=n, such that binomial(n,k) is divisible by gcd(n,k).

Original entry on oeis.org

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

Offset: 1

Robert Israel, Dec 02 2016

All k coprime to n are always included, in particular 1 and n-1.
Row n contains k if and only if it contains n-k.
A014847 consists of k such that row 2k contains k.
If n is prime or the square of a prime, row n contains all numbers from 1 to n-1. This is not true for higher powers: row p^r does not contain any multiples of p^(r-1) if r > 2.
Prime p is in row n>p if and only if the p-adic order of n is not 1.

			Row 8 contains 2 because gcd(8,2)=2 divides binomial(8,2) = 28, but not 4 because gcd(8,4)=4 does not divide binomial(8,4)= 70.

Robert Israel, Table of n, a(n) for n = 1..10029(rows 1 to 155 flattened)

Cf. A014847.

f:= proc(n,m) if binomial(n,m) mod igcd(n,m) = 0 then m else NULL fi end proc:
seq(seq(f(n,m),m=1..n),n=1..40);

Table[If[Divisible[Binomial[n,k],GCD[n,k]],k,Nothing],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Dec 04 2022 *)

A325630 Numbers k such that A000110(k) is divisible by k.

Original entry on oeis.org

1, 2, 35, 16833, 16989, 23684

Offset: 1

Vaclav Kotesovec, Sep 07 2019

No other terms below 50000.
From Amiram Eldar, Jun 20 2024: (Start)
Numbers k such that A166226(k) = 0.
All the terms above 2 are composites since A166226(p) == 2 (mod p) for prime p. (End)
No other terms below 90000. - Michael S. Branicky, Jan 09 2025

			35 is in the sequence because A000110(35) = 35 * 8045720086273150473238297902.

Cf. A000110, A051130, A066562, A073877, A113883, A166226, A280727, A325935.

Cf. A014847, A016089, A023172, A051177, A056848.

Select[Range[1000], Divisible[BellB[#], #] &]

A073077 Numbers k that divide C(2k,k) and C(4k,2k).

Original entry on oeis.org

1, 2, 20, 110, 156, 210, 220, 238, 240, 312, 460, 468, 483, 510, 561, 600, 624, 665, 684, 696, 720, 744, 770, 806, 812, 816, 868, 936, 966, 1001, 1012, 1045, 1064, 1100, 1110, 1122, 1144, 1155, 1170, 1200, 1295, 1309, 1320, 1326, 1332, 1360, 1368, 1394, 1404

Offset: 1

Benoit Cloitre, Aug 17 2002

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

Subsequence of A014847.

Cf. A000984, A001448.

Select[Range[1500], Divisible[Binomial[2*#, #], #] && Divisible[Binomial[4*#, 2*#], #] &] (* Amiram Eldar, Apr 26 2025 *)

isok(n) = !(binomial(2*n, n) % n) && !(binomial(4*n, 2*n) % n); \\ Michel Marcus, Nov 28 2013

More terms from Michel Marcus, Nov 28 2013

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

A120622 Numbers k such that the k-th Catalan number C(2k, k)/(k + 1) is divisible by k/2 but not divisible by k.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Extensions

A120623 Numbers n such that the n-th Catalan number C(2n, n)/(n + 1) is divisible by n/3 but not divisible by n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Extensions

A120624 Numbers n such that the n-th Catalan number C(2n,n)/(n+1) is divisible by 2n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Extensions

A260636 a(n) = binomial(3n, n) mod n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

A260640 Numbers n such that binomial(3*n,n) == 0 (mod n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

A048618 Even numbers n such that binomial(n,n/2) is divisible by n/2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A278961 Triangle read by rows: row n consists of k, 1<=k<=n, such that binomial(n,k) is divisible by gcd(n,k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple