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A279859
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a(n) is the integer r with |r| < prime(n)/2 such that (prime(n)*Sum_{k=1..prime(n)-1} (3*H(k-1)^2 + 4*H(k-1)/k)/(k^2*binomial(2k,k)) + 3*H(prime(n)-1)/prime(n)^2)/prime(n)^2 == r (mod prime(n)), where H(m) denotes the harmonic number Sum_{k=1..m} 1/k.
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1
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3, -1, 2, 6, 8, -11, 10, -14, 0, -16, 15, 14, -8, -26, -9, 29, -26, -27, -1, 37, -34, 47, 40, 20, -36, 26, 6, -57, 29, -55, -23, 53, -9, 58, 52, -65, 33, -37, -83, 3, 24, 73, -72, -66, -76, 105, 45, -108, 84, -84, 19, 109, -84, 21, -28, -19, -139
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OFFSET
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4,1
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COMMENTS
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Conjecture: (i) We have the new identity
Sum_{k>0} (3*H(k-1)^2 + 4*H(k-1)/k)/(k^2*binomial(2k,k)) = Pi^4/360.
(ii) Let B_0, B_1, B_2, ... be the Bernoulli numbers. For any prime p > 3, we have the congruences
p*Sum_{k=1..p-1} (3*H(k-1)^2 + 4*H(k-1)/k)/(k^2*binomial(2k,k)) == -3*H(p-1)/p^2 - p^2/5*B_{p-5} (mod p^3)
and
Sum_{k=1..p-1} (3*H(k)^2 - 4*H(k)/k)*binomial(2k,k)/k
== 6*H(p-1)/p^2 + (8/5)*p^2*B_{p-5} (mod p^3).
It is known that H(p-1) == -(p^2/3)*B_{p-3} (mod p^3) for any prime p > 3. The first congruence in part (ii) of the conjecture implies that a(n) == -B_{prime(n)-5}/5 (mod prime(n)) for all n = 4,5,....
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LINKS
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EXAMPLE
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a(4) = 3 since prime(4) = 7 and (7*Sum_{k=1..6} (3*H(k-1)^2 + 4*H(k-1)/k)/(k^2*binomial(2k,k)) + 3*H(6)/7^2)/7^2 = 3 - 7*657251/1555200 == 3 (mod 7).
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MATHEMATICA
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rMod[m_, n_]:=rMod[m, n]=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2];
R[n_]:=R[n]=rMod[(Prime[n]*Sum[(3*HarmonicNumber[k-1]^2+4*HarmonicNumber[k-1]/k)/(k^2*Binomial[2k, k]), {k, 1, Prime[n]-1}]+3*HarmonicNumber[Prime[n]-1]/Prime[n]^2)/Prime[n]^2, Prime[n]];
Table[R[n], {n, 4, 60}]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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