A133620 Binomial(n+p,n) mod n where p=10.

0, 0, 1, 1, 3, 4, 2, 6, 2, 6, 1, 2, 1, 10, 5, 7, 1, 12, 1, 15, 18, 12, 1, 12, 21, 14, 4, 12, 1, 28, 1, 29, 1, 18, 6, 5, 1, 20, 14, 10, 1, 14, 1, 34, 15, 24, 1, 3, 8, 16, 18, 27, 1, 34, 23, 16, 1, 30, 1, 16, 1, 32, 17, 57, 40, 56, 1, 1, 47, 60, 1, 54, 1, 38, 36, 58, 12, 66, 1, 63, 10, 42, 1

Offset: 1

Author

Hieronymus Fischer, Sep 30 2007

Keywords

• nonn
• easy

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

• Ray Chandler, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, order 58060800.

Crossrefs

Cf. A000040, A133621-A133625, A133630, A038509, A133634, A133635, A133636.

Cf. A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Programs

Mathematica

Table[Mod[Binomial[n + 10, n], n], {n, 90}] (* Harvey P. Dale, Apr 04 2015 *)

PARI

a(n) = binomial(n+10, n) % n \\ Michel Marcus, Jul 15 2013

Formula

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

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

a(n) = A001287(n+10) mod n. - Michel Marcus, Jul 15 2013; corrected by Michel Marcus, Jan 27 2020

For n > 58060802, a(n) = 2*a(n-29030400) - a(n-58060800). - Ray Chandler, Apr 29 2023