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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
a(n) = A001008(p-1)/p^2, p=A000040(n). - R. J. Mathar, Jan 09 2017
a(n) = A120285(n)/A001248(n). - R. J. Mathar, Jan 09 2017
MAPLE
A061002:=proc(n) local p;
p:=ithprime(n);
(1/p^2)*numer(add(1/i, i=1..p-1));
end proc;
[seq(A061002(n), n=3..20)];
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
Sequence in context: A338102 A167445 A302531 * A303412 A209548 A302898
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, May 15 2001
STATUS
approved

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Last modified March 28 10:31 EDT 2024. Contains 371240 sequences. (Running on oeis4.)