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A133624
Binomial(n+p, n) mod n, where p=4.
3
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
1,3
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
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).
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
MATHEMATICA
Table[Mod[Binomial[n+4, n], n], {n, 90}] (* Harvey P. Dale, Apr 26 2014 *)
PROG
(Magma) [Binomial(n+4, 4) mod n: n in [1..100]]; // Vincenzo Librandi, Apr 27 2014
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Sep 30 2007
STATUS
approved