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A330014 When prime(n) is an odd prime (n >= 2) and N(n) / D(n) = Sum_{k=1..prime(n)-1} 1/k^3, then prime(n) divides N(n) and a(n) = N(n) / prime(n).

%I A330014 #34 Sep 08 2022 08:46:24
%S A330014 3,407,4081,1742192177,1964289620189,26430927136768997,
%T A330014 12913609418092462447,14639800647032731764901,
%U A330014 21461951639001843544904995612963,489697309796854053100609288112563213,97796057728171000155497946604711651753457
%N A330014 When prime(n) is an odd prime (n >= 2) and N(n) / D(n) = Sum_{k=1..prime(n)-1} 1/k^3, then prime(n) divides N(n) and a(n) = N(n) / prime(n).
%C A330014 The idea of this sequence comes from the 1st exercise of "sélection de la délégation française" in 2005 for IMO 2006 where it was asked to prove that prime(n) divides N(n) [See reference].
%C A330014 The first fractions N(n)/D(n) are 9/8, 2035/1728, 28567/24000, 19164113947/16003008000, 25535765062457/21300003648000, ...
%D A330014 Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 10, Exercices de sélection de la délégation française en Octobre 2005 pour OIM 2006, Exercice 1, p. 169, p. 179.
%H A330014 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%e A330014 For prime(4) = 7 then 1 + 1/2^3 + 1/3^3 + 1/4^3 + 1/5^3 + 1/6^3 = 28567/24000 and 28567/7 = 4081, a(4) = 4081.
%t A330014 a[n_] := Numerator[Sum[1/(i- 1)^3, {i, 2,(p = Prime[n])}]]/p; Array[a, 11, 2] (* _Amiram Eldar_, Nov 27 2019 *)
%o A330014 (Magma) [(Numerator(&+ [1/(k-1)^3:k in [2..NthPrime(n)]])) / NthPrime(n):n in [2..12]]; // _Marius A. Burtea_, Nov 27 2019
%Y A330014 Cf. A076637, A061002, A076637 (Wolstenholme's Theorem).
%K A330014 nonn
%O A330014 2,1
%A A330014 _Bernard Schott_, Nov 27 2019