login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A246132 Binomial(2n, n) - 2 mod n^2. 7
0, 0, 0, 4, 0, 22, 0, 4, 18, 54, 0, 122, 0, 102, 43, 68, 0, 274, 0, 18, 361, 246, 0, 538, 250, 342, 504, 166, 0, 722, 0, 580, 865, 582, 5, 50, 0, 726, 18, 818, 0, 1510, 0, 310, 493, 1062, 0, 538, 1029, 2254, 2041, 406, 0, 922, 855, 1206, 379, 1686, 0, 3454, 0, 1926, 3538, 580, 3123, 922, 0, 4114, 547, 1298, 0, 4930, 0, 2742, 2518, 790, 3309, 2950, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

When e=2, the numbers binomial(2n, n)-2 mod n^e are 0 whenever n is a prime (see A246130 for introductory comments). This follows from Wolstenholme's theorem or, in a simpler way, from the identity binomial(2n, n)-2 = sum_{k=1..(n-1)} binomial(n,k)^2, in which every RHS term is divisible by n^2 whenever n is a prime. No composite number n for which a(n)=0 was found up to n=431500; nevertheless, the existence of such a composite is likely (personal opinion, based on the combinatorial nature of the problem).

LINKS

Stanislav Sykora, Table of n, a(n) for n = 1..10000

Wikipedia, Wolstenholme's theorem

FORMULA

For any prime p, a(p)=0.

EXAMPLE

a(7) = (binomial(14,7)-2) mod 7^2 = (3432-2) mod 49 = 70*49 mod 49 = 0.

PROG

(PARI) a(n) = (binomial(2*n, n)-2)%n^2

CROSSREFS

Cf. A000984, A246130 (e=1), A246133 (e=3), A246134 (e=4).

Sequence in context: A078630 A178671 A179270 * A229827 A243270 A242027

Adjacent sequences:  A246129 A246130 A246131 * A246133 A246134 A246135

KEYWORD

nonn

AUTHOR

Stanislav Sykora, Aug 16 2014

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy .

Last modified July 25 20:28 EDT 2017. Contains 289797 sequences.