|
Search: id:A073545
|
|
|
| A073545 |
|
Least k such that 1/tau(k) + 1/tau(k+1) + 1/tau(k+2) + ... + 1/tau(k+n) is equal to 1 (where tau(k)=A000005(k) is the number of divisors of k). |
|
+0
1
|
|
| 1, 2, 6, 25, 54, 243, 1204, 3549, 19544, 81829, 104663, 663490, 743764, 7925355
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
EXAMPLE
|
a(2)=6 because 1/tau(6)+1/tau(7)+1/tau(8) = 1/4+1/2+1/4 = 1.
|
|
MATHEMATICA
|
a[n_] := For[k=1, True, k++, If[Sum[1/DivisorSigma[0, k+i], {i, 0, n}]==1, Return[k]]]
|
|
CROSSREFS
|
Sequence in context: A089718 A123150 A086591 this_sequence A103063 A030228 A066317
Adjacent sequences: A073542 A073543 A073544 this_sequence A073546 A073547 A073548
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 27 2002
|
|
EXTENSIONS
|
Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Sep 03 2002
2 more terms from Ryan Propper (rpropper(AT)stanford.edu), Sep 04 2005
|
|
|
Search completed in 0.002 seconds
|
