OFFSET
1,1
COMMENTS
For any p in the interval ]n, 4n/3], p divides Sum_{k=0..n}binomial(n,k)^4 (see the link).
LINKS
Peter Vandendriessche and Hojoo Lee, Problems in elementary number theory, Problem E16
EXAMPLE
362 is in the sequence because, for n=4, 5 is in the interval ]4, 16/3] and (1/5)*Sum_{k=0..4}binomial(4,k)^4 =(1/5)*(1^4 + 4^4 + 6^4 + 4^4 + 1^4) = 1810/5 = 362.
MAPLE
with(numtheory): for n from 1 to 20 do: for p from n+1 to floor(4*n/3) do: if type(p, prime)=true then s:=sum(binomial(n, k)^4, k=0..n):s:=s/p: printf(`%d, `, s):else fi:od:od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 14 2012
STATUS
approved