A109641 -id:A109641 - cp's OEIS frontend

A080469 Composite n such that binomial(3*n,n)==3^n (mod n).

36, 57, 121, 132, 552, 8397, 7000713, 9692541, 36294723, 564033861

Offset: 1

Benoit Cloitre, Oct 15 2003

If p is prime, binomial(3*p,p)==3^p (mod p)
No other terms below 10^9.
A subsequence of A109641. The terms a(n) with n=2, 6, 7, 8, 9, 10 are of the form 3^k*p where p is prime and k=1, 3, 2, 5, 6, 7, respectively. It is tempting to conjecture that there are (infinitely many?) more terms of that form. - M. F. Hasler, Nov 11 2015

			57 is a term because binomial(3*57, 57) = 12039059761216294940321619222324879408784636200 mod 57 = 27 == 3^57 mod 57.

Max Alekseyev, PARI scripts for various problems

Cf. A109641, A109642; A109760, A109769.

Do[If[ !PrimeQ[n], k = Binomial[3*n, n]; m = 3^n; If[Mod[k, n] == Mod[m, n], Print[n]]], {n, 1, 70000}] (* Ryan Propper, Aug 12 2005 *)

forcomposite(n=1,1e9, binomod(3*n,n,n)==Mod(3,n)^n && print1(n",")) \\ Cf. Alekseyev link. - M. F. Hasler, Nov 14 2015

One more term a(6) from Ryan Propper, Aug 12 2005

Four new terms a(7)-a(10) added by Max Alekseyev, Nov 05 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

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A080469 Composite n such that binomial(3*n,n)==3^n (mod n).

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

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

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

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Author

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Programs

Magma

Mathematica

PARI

Python