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 A212324 (1/p)*Sum_{k=0..n}binomial(n,k)^4 where p is a prime in the interval ]n, 4n/3] 0
 362, 37692, 55185580, 758665388, 641947636, 8948910312, 126203947828, 1374427747336, 19738612219080, 285537263797392, 255480709713456, 4156846359584754, 3719283584891622, 54448930157994828, 801081711581734764, 661763153045780892, 9779554604050169400 (list; graph; refs; listen; history; text; internal format)
 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 Sequence in context: A020539 A157442 A200558 * A007565 A098253 A031517 Adjacent sequences:  A212321 A212322 A212323 * A212325 A212326 A212327 KEYWORD nonn AUTHOR Michel Lagneau, May 14 2012 STATUS approved

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Last modified May 21 07:22 EDT 2022. Contains 353889 sequences. (Running on oeis4.)