|
|
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).
|
|
0
|
|
|
3, 407, 4081, 1742192177, 1964289620189, 26430927136768997, 12913609418092462447, 14639800647032731764901, 21461951639001843544904995612963, 489697309796854053100609288112563213, 97796057728171000155497946604711651753457
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
2,1
|
|
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, ...
|
|
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
|
|
|
EXAMPLE
|
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.
|
|
MATHEMATICA
|
a[n_] := Numerator[Sum[1/(i- 1)^3, {i, 2, (p = Prime[n])}]]/p; Array[a, 11, 2] (* Amiram Eldar, Nov 27 2019 *)
|
|
PROG
|
(Magma) [(Numerator(&+ [1/(k-1)^3:k in [2..NthPrime(n)]])) / NthPrime(n):n in [2..12]]; // Marius A. Burtea, Nov 27 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|