|
Search: id:A016017
|
|
|
| A016017 |
|
Smallest k such that 1/k can be written as a sum of exactly 2 unit fractions in n ways. |
|
+0 7
|
|
| 1, 2, 4, 8, 6, 32, 64, 12, 256, 512, 24, 2048, 36, 30, 16384, 32768, 96, 72, 262144, 192, 1048576, 2097152, 60, 8388608, 216, 768, 67108864, 288, 1536, 536870912, 1073741824, 120, 576, 8589934592, 6144, 34359738368, 68719476736, 180, 864
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Square root of n-th term of A061283. a(n)=Sqrt[Min{k| A000005(k)=2n-1}]. If n = p(i) = p, a prime, then a(p) = 2^[(p-1)/2] = 2^A005097(i). - Labos E. (labos(AT)ana.sote.hu), May 22 2001
a(n+1)<=2^n.
|
|
FORMULA
|
Least k such that (tau(k^2)+1)/2=n. - Vladeta Jovovic (vladeta(AT)Eunet.yu), Aug 01 2001
|
|
EXAMPLE
|
a(1)=1 and a(2)=2 because 1/2 = 1/3+1/6 = 1/4+1/4.
a(3)=4 because 1/4 = 1/5+1/20 = 1/6+1/12 = 1/8+1/8.
a(4)=8 because 1/8 = 1/9+1/72 = 1/10+1/40 = 1/12+1/24 = 1/16+1/16.
a(5)=6 because 1/6 = 1/7+1/42 = 1/8+1/24 = 1/9+1/18 = 1/10+1/15 = 1/12+1/12.
|
|
MATHEMATICA
|
f[j_, n_] := (Times @@ (j(Last /@ FactorInteger[n]) + 1) + j - 1)/j; t = Table[0, {50}]; Do[a = f[2, n]; If[a < 51 && t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 2^30}] (from Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 03 2005)
|
|
CROSSREFS
|
Identical to A071571 shifted right.
Cf. A000005, A000290, A005408, A005179, A003680, A037992, A016013, A016017, A055079, A048691.
Sequence in context: A124510 A131886 A061284 this_sequence A071571 A029898 A021406
Adjacent sequences: A016014 A016015 A016016 this_sequence A016018 A016019 A016020
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Robert G. Wilson v (rgwv(AT)rgwv.com)
|
|
EXTENSIONS
|
Entry revised by njas, Aug 14 2005
|
|
|
Search completed in 0.003 seconds
|