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A133620
Binomial(n+p,n) mod n where p=10.
77
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
1,5
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.
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
MATHEMATICA
Table[Mod[Binomial[n + 10, n], n], {n, 90}] (* Harvey P. Dale, Apr 04 2015 *)
PROG
(PARI) a(n) = binomial(n+10, n) % n \\ Michel Marcus, Jul 15 2013
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Sep 30 2007
STATUS
approved

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