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.
LINKS
Ray Chandler, Table of n, a(n) for n = 1..1000
FORMULA
a(n)=binomial(n+5,5) mod n.
a(n)=1 if n is a prime > 5, since binomial(n+5,n)==(1+floor(5/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^5 + 15*n^4 + 85*n^3 + 105*n^2 + 34*n + 120)/120 mod n.
For n > 6:
if n mod 120 == 0, then a(n) = 17*n/60 + 1.
if n mod 120 is in {1, 2, 7, 11, 13, 17, 19, 23, 26, 29, 31, 34, 37, 41, 43, 47, 49, 53, 58, 59, 61, 67, 71, 73, 74, 77, 79, 82, 83, 89, 91, 97, 98, 101, 103, 106, 107, 109, 113, 119}, then a(n) = 1.
if n mod 120 is in {3, 9, 18, 21, 27, 33, 39, 42, 51, 57, 63, 66, 69, 81, 87, 93, 99, 111, 114, 117}, then a(n) = n/3 + 1.
if n mod 120 is in {4, 28, 44, 52, 68, 76, 92, 116}, then a(n) = n/4 + 1.
if n mod 120 is in {5, 10, 25, 35, 50, 55, 65, 85, 95, 115}, then a(n) = n/5 + 1.
if n mod 120 is in {6, 54, 78, 102}, then a(n) = 5*n/6 + 1.
if n mod 120 is in {8, 16, 32, 56, 64, 88, 104, 112}, then a(n) = 3*n/4 + 1.
if n mod 120 is in {12, 36, 84, 108}, then a(n) = 7*n/12 + 1.
if n mod 120 is in {14, 22, 38, 46, 62, 86, 94, 118}, then a(n) = n/2 + 1.
if n mod 120 is in {15, 45, 75, 90, 105}, then a(n) = 8*n/15 + 1.
if n mod 120 is in {20, 100}, then a(n) = 9*n/20 + 1.
if n mod 120 is in {24, 48, 72, 96}, then a(n) = n/12 + 1.
if n mod 120 == 30, then a(n) = n/30 + 1.
if n mod 120 is in {40, 80}, then a(n) = 19*n/20 + 1.
if n mod 120 == 60, then a(n) = 47*n/60 + 1.
if n mod 120 is in {70, 110}, then a(n) = 7*n/10 + 1.
(End)
For n > 246, a(n) = 2*a(n-120) - a(n-240). - Ray Chandler, Apr 23 2023
MATHEMATICA
Table[Mod[Binomial[n+5, n], n], {n, 90}] (* Harvey P. Dale, Oct 02 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Sep 30 2007
STATUS
approved