A133623 Binomial(n+p, n) mod n where p=3.

0, 0, 2, 3, 1, 0, 1, 5, 4, 6, 1, 11, 1, 8, 6, 9, 1, 16, 1, 11, 8, 12, 1, 21, 1, 14, 10, 15, 1, 26, 1, 17, 12, 18, 1, 31, 1, 20, 14, 21, 1, 36, 1, 23, 16, 24, 1, 41, 1, 26, 18, 27, 1, 46, 1, 29, 20, 30, 1, 51, 1, 32, 22, 33, 1, 56, 1, 35, 24, 36, 1, 61, 1, 38, 26, 39, 1, 66, 1, 41, 28, 42, 1

Offset

1,3

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.

Appears to satisfy the recurrence: a(n) = -2*a(n-1) - a(n-2) + 2*a(n-3) + 4*a(n-4) + 2*a(n-5) - a(n-6) - 2*a(n-7) - a(n-8) for n > 14. - Chai Wah Wu, May 25 2016

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (-2,-1,2,4,2,-1,-2,-1).

Formula

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

a(n)=1 if n is a prime > 3, since binomial(n+3,n)==(1+floor(3/n))(mod n), provided n is a prime.

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

a(n) = (n^3 + 5*n + 6)/6 mod n.

For n > 6:

if n mod 6 == 0, then a(n) = 5*n/6 + 1.

if n mod 6 is in {1, 5}, then a(n) = 1.

if n mod 6 is in {2, 4}, then a(n) = n/2 + 1.

if n mod 6 == 3, then a(n) = n/3 + 1.

(End)

Mathematica

Table[Mod[Binomial[n+3, n], n], {n, 90}] (* Harvey P. Dale, Nov 22 2011 *)

Crossrefs

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.

Cf. A133873, A133883, A133880, A133890, A133900, A133910.

Sequence in context: A006705 A031269 A006703 * A065862 A368495 A189117

Adjacent sequences: A133620 A133621 A133622 * A133624 A133625 A133626

Keyword

nonn

Author

Hieronymus Fischer, Sep 30 2007

Status

approved