|
|
A061002
|
|
As p runs through the primes >= 5, sequence gives { numerator of Sum_{k=1..p-1} 1/k } / p^2.
|
|
25
|
|
|
1, 1, 61, 509, 8431, 39541, 36093, 375035183, 9682292227, 40030624861, 1236275063173, 6657281227331, 2690511212793403, 5006621632408586951, 73077117446662772669, 4062642402613316532391, 46571842059597941563297, 8437878094593961096374353
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
3,3
|
|
COMMENTS
|
This is an integer by a theorem of Waring and Wolstenholme.
Conjecture: If p is the n-th prime and H(n) is the n-th harmonic number, then denominator(H(p)/H(p-1))/numerator(H(p-1)/p^2) = p^3. A193758(p)/a(n) = p^3, p > 3. - Gary Detlefs, Feb 20 2013
The sequence which gives the numerators of H_{p-1} = Sum_{k=1..p-1} 1/k } for p prime >= 5 is A076637. - Bernard Schott, Dec 02 2018
|
|
REFERENCES
|
Z. I. Borevich and I. R. Shafarevich, Number Theory. Academic Press, NY, 1966, p. 388 Problem 5.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 115.
|
|
LINKS
|
|
|
FORMULA
|
|
|
MAPLE
|
p:=ithprime(n);
(1/p^2)*numer(add(1/i, i=1..p-1));
end proc;
|
|
MATHEMATICA
|
Table[Function[p, Numerator[Sum[1/k, {k, p - 1}]/p^2]]@ Prime@ n, {n, 3, 20}] (* Michael De Vlieger, Feb 04 2017 *)
|
|
PROG
|
(GAP) List(List(Filtered([5..80], p->IsPrime(p)), i->Sum([1..i-1], k->1/k)/i^2), NumeratorRat); # Muniru A Asiru, Dec 02 2018
(PARI) a(n) = my(p=prime(n)); numerator(sum(k=1, p-1, 1/k))/p^2; \\ Michel Marcus, Dec 03 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|