A080469 -id:A080469 - cp's OEIS frontend

A109641 Composite n such that binomial(3n, n) == 3^k (mod n) for some integer k > 0.

4, 9, 15, 25, 27, 34, 36, 49, 51, 57, 63, 68, 75, 81, 87, 93, 111, 121, 125, 129, 132, 138, 141, 153, 155, 159, 169, 177, 237, 249, 258, 261, 264, 267, 274, 276, 279, 289, 298, 303, 324, 339, 343, 357, 361, 375, 381, 387, 393, 411, 417, 423, 441, 447, 453, 477

Offset: 1

Ryan Propper, Aug 05 2005

nonn

Includes p^k for k >= 2 and p > 2 in A019334 but not in A014127, as binomial(3n,n) is coprime to p and 3 is a primitive root mod p^k. - Robert Israel, Nov 12 2017

			Binomial(3*34,34) == 3^6 (mod 34), so 34 is a member.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A019334, A080469, A014127, A109642.

filter:= proc(n) local p,m,k,t;
  if isprime(n) then return false fi;
  p:= padic:-ordp(n,3);
  p:= p + numtheory:-order(3, n/3^p);
  m:= binomial(3*n,n) mod n;
  t:= 1;
  for k from 1 to p do
    t:= t*3 mod n;
    if t = m then return true fi;
  od:
false
end proc;
select(filter, [$2..1000]); # Robert Israel, Nov 12 2017

okQ[n_] := Module[{p, m}, If[PrimeQ[n], Return[False]]; p = IntegerExponent[n, 3]; p = p + MultiplicativeOrder[3, n/3^p]; m = Mod[Binomial[3n, n], n]; AnyTrue[Range[p], m == PowerMod[3, #, n]&]];
Select[Range[2, 500], okQ] (* Jean-François Alcover, Mar 27 2019, after Robert Israel *)

Corrected and extended by Max Alekseyev, Sep 13 2009

Edited by Max Alekseyev, Sep 20 2009

A109642 a(n) = least composite m such that binomial(3m,m) mod m = 3^n.

Original entry on oeis.org

4, 15, 57, 765, 1025, 2097, 4947, 9189, 103599, 216927, 4346128, 1558269, 1977777, 208510373, 14493123, 493262541, 144228033

Offset: 1

Ryan Propper, Aug 05 2005

Subsequence of A109641.

Cf. A080469, A109641.

In[1]:= n = 1; Do[If[ !PrimeQ[k] && Mod[Binomial[3*k, k], k] == 3^n, Print[k]; n++ ], {k, 1, 10^4}]

Edited by Max Alekseyev, Nov 03 2009

a(14)-a(17) from Chai Wah Wu, Jul 30 2025

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

A109760 Composite n such that binomial(5*n,n) == 5^n (mod n).

Original entry on oeis.org

4, 365, 400, 685, 3200, 6400, 12550, 12800, 16525, 25600, 51200, 225125, 70463125, 271094125, 431434441

Offset: 1

Ryan Propper, Aug 12 2005

No other terms below 10^9.

			4 is a term because binomial(5*4, 4) = 4845, 5^4 = 625 and 4845 mod 4 = 625 mod 4 = 1.

Max Alekseyev, PARI/GP scripts for various problems

Cf. A080469.

Do[If[ !PrimeQ[n], If[Mod[Binomial[5*n, n], n] == Mod[5^n, n], Print[n]]], {n, 1, 50000}]
Select[Range[250000],CompositeQ[#]&&Mod[Binomial[5#,#],#]==PowerMod[5,#,#]&] (* The program generates the first 12 terms of the sequence. *) (* Harvey P. Dale, Jul 28 2025 *)

a(12) from D. S. McNeil, Mar 15 2009
225125 from Max Alekseyev, Sep 13 2009
Three more terms from Max Alekseyev, Nov 06 2009

A109769 Composite n such that binomial(7*n, n) == 7^n (mod n).

Original entry on oeis.org

18, 25, 133, 2107, 4676, 226037, 4477739, 827867201

Offset: 1

Ryan Propper, Aug 13 2005

No other terms below 10^9.

			18 is a term because binomial(7*18, 18) = 2797093093529137508875, 7^18 = 1628413597910449 and 2797093093529137508875 mod 18 = 1628413597910449 mod 18 = 1.

Max Alekseyev, PARI/GP scripts for various problems

Cf. A080469.

Do[If[ !PrimeQ[n], If[Mod[Binomial[7*n, n], n] == Mod[7^n, n], Print[n]]], {n, 2, 20000}]

226037 from Max Alekseyev, Sep 13 2009

Two more terms from Max Alekseyev, Nov 06 2009

A084699 Composite integers j such that binomial(2*j,j) == 2^j (mod j).

Original entry on oeis.org

12, 30, 56, 424, 992, 16256, 58288, 119984, 356992, 1194649, 9973504, 12327121, 13141696, 22891184, 67100672, 233850649

Offset: 1

Benoit Cloitre, Oct 15 2003

If p is prime, binomial(2*p,p) == 2^p (mod p).
a(17) > 10^9.
From Gabriel Guedes and Ricardo Machado, Nov 16 2023: (Start)
Theorem. Let j = (2^k)*p, where p is an odd prime and k is in N; then binomial(2*j,j) == 2^j (mod j) if and only if p satisfies the following conditions:
a) p divides binomial(2^(k+1),2^k) - 2^(2^k);
b) p has at least k 1's in its binary expansion.
Theorem. If m is an even perfect number then j = 2m satisfies the congruence binomial(2*j,j) == 2^j (mod j). See A000396.
Theorem. Let j = p^2 with p a prime number. Then p is a Wieferich prime if and only if binomial(2*j,j) == 2^j (mod j). See A001220. (End)
Contains 17179738112 and 274877382656 (from Guedes-Machado paper). - Michael De Vlieger, Nov 22 2023
Contains 3386741824, 750984028672, 33029195197184, 1145067923695616, 422612863956511744. - Ricardo Machado, Nov 23 2023
Contains 84385517065596416, 62648180117928433664, 273984397779878971648, 36506097537257040703232. - Max Alekseyev, Dec 07 2023

Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (V. 1.4, 11/2015).
Gabriel Guedes and Ricardo Machado, Perfect numbers, Wieferich primes and the solutions of binomial(2n, n) congruent to 2^n mod n, Notes Num. Theor. Disc. Math. (NNTDM 2023) Vol. 29, No. 4, 705-712.

Contains A139256 as a subsequence.

Cf. A015910, A059288, A080469, A082180, A109760, A109769, A228562, A328497.

lista(nn) = {forcomposite(n=1, nn, if (binomod(2*n, n, n) == Mod(2, n)^n, print1(n, ", ")));} \\ Michel Marcus, Dec 06 2013 and Dec 03 2023

More terms from David Wasserman, Jan 03 2005

a(11)-a(16) from Max Alekseyev, Aug 05 2011

cp's OEIS Frontend

A109641 Composite n such that binomial(3n, n) == 3^k (mod n) for some integer k > 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A109642 a(n) = least composite m such that binomial(3m,m) mod m = 3^n.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

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

A109760 Composite n such that binomial(5*n,n) == 5^n (mod n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A109769 Composite n such that binomial(7*n, n) == 7^n (mod n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A084699 Composite integers j such that binomial(2*j,j) == 2^j (mod j).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI