A133624 Binomial(n+p, n) mod n, where p=4.

0, 1, 2, 2, 1, 0, 1, 7, 4, 1, 1, 8, 1, 8, 6, 13, 1, 7, 1, 6, 8, 12, 1, 3, 1, 1, 10, 8, 1, 26, 1, 25, 12, 1, 1, 22, 1, 20, 14, 31, 1, 15, 1, 12, 16, 24, 1, 5, 1, 1, 18, 14, 1, 46, 1, 43, 20, 1, 1, 36, 1, 32, 22, 49, 1, 23, 1, 18, 24, 36, 1, 7, 1, 1, 26, 20, 1, 66, 1, 61, 28, 1, 1, 50, 1, 44, 30

Offset: 1

Author

Hieronymus Fischer, Sep 30 2007

Keywords

• nonn

Comments

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1).

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133874, A133884, A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Programs

Magma

[Binomial(n+4,4) mod n: n in [1..100]]; // Vincenzo Librandi, Apr 27 2014

Mathematica

Table[Mod[Binomial[n+4,n],n],{n,90}] (* Harvey P. Dale, Apr 26 2014 *)

Formula

a(n) = binomial(n+4,4) mod n.

a(n)=1 if n is a prime > 4, since binomial(n+4,n) == (1+floor(4/n))(mod n), provided n is a prime.

From Chai Wah Wu, May 26 2016: (Start)

a(n) = (n^4 + 10*n^3 + 11*n^2 + 2*n + 24)/24 mod n.

For n > 6:

if n mod 24 == 0, then a(n) = n/12 + 1.

if n mod 24 is in {1, 2, 5, 7, 10, 11, 13, 17, 19, 23}, then a(n) = 1.

if n mod 24 is in {3, 9, 15, 18, 21}, then a(n) = n/3 + 1.

if n mod 24 is in {4, 20}, then a(n) = n/4 + 1.

if n mod 24 == 6, then a(n) = 5*n/6 + 1.

if n mod 24 is in {8, 16}, then a(n) = 3*n/4 + 1.

if n mod 24 == 12, then a(n) = 7*n/12 + 1.

if n mod 24 is in {14, 22}, then a(n) = n/2 + 1.

(End)

For n > 54, a(n) = 2*a(n-24) - a(n-48). - Ray Chandler, Apr 23 2023