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COMMENTS
| Prime p>2 divides a(p). p^3 divides a(p) for prime p>3. p divides a((p+1)/2) for prime p = {7,11,17,19,23,31,41,43,47,59,67,71,73,79,83,89,97,103,...} = all primes excluding 2 and 3 from A045323[n] Primes congruent to {1, 2, 3, 7} mod 8.
a(n) = Numerator[ H(n-1,n) ], where H(k,r)= Sum[ 1/i^r, {i,1,k} ] is generalized harmonic number. Numerators of Sum[ 1/k^p, {k,1,p-1} ]], where p = Prime[n], are listed in A120352(n) = {1, 9, 257875, 940908897061, 26038773205374138944970092886340352227, ...}. a(p)/p^3 for prime p>3 are listed in A119722(n) = {2063, 2743174627, 19563315706517008974432827112201617, ...}.
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