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%I #6 May 16 2012 12:54:07
%S 362,37692,55185580,758665388,641947636,8948910312,126203947828,
%T 1374427747336,19738612219080,285537263797392,255480709713456,
%U 4156846359584754,3719283584891622,54448930157994828,801081711581734764,661763153045780892,9779554604050169400
%N (1/p)*Sum_{k=0..n}binomial(n,k)^4 where p is a prime in the interval ]n, 4n/3]
%C For any p in the interval ]n, 4n/3], p divides Sum_{k=0..n}binomial(n,k)^4 (see the link).
%H Peter Vandendriessche and Hojoo Lee, <a href="http://www.scribd.com/doc/24487088/Hojoo-Lee-Peter-Vandendriessche-Problems-in-Elementary-Number-Theory">Problems in elementary number theory</a>, Problem E16
%e 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.
%p 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:
%K nonn
%O 1,1
%A _Michel Lagneau_, May 14 2012
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