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).

3, 407, 4081, 1742192177, 1964289620189, 26430927136768997, 12913609418092462447, 14639800647032731764901, 21461951639001843544904995612963, 489697309796854053100609288112563213, 97796057728171000155497946604711651753457

Offset: 2

Author

Bernard Schott, Nov 27 2019

Keywords

• nonn

Comments

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].

The first fractions N(n)/D(n) are 9/8, 2035/1728, 28567/24000, 19164113947/16003008000, 25535765062457/21300003648000, ...

Examples

			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.
		

References

• 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.

Links

• Index to sequences related to Olympiads.

Crossrefs

Cf. A076637, A061002, A076637 (Wolstenholme's Theorem).

Programs

Magma

[(Numerator(&+ [1/(k-1)^3:k in [2..NthPrime(n)]])) / NthPrime(n):n in [2..12]]; // Marius A. Burtea, Nov 27 2019

Mathematica

a[n_] := Numerator[Sum[1/(i- 1)^3, {i, 2,(p = Prime[n])}]]/p; Array[a, 11, 2] (* Amiram Eldar, Nov 27 2019 *)