|
Search: id:A066169
|
|
|
| A066169 |
|
a(n) = least k such that EulerPhi(k) >= n. |
|
+0
3
|
|
| 1, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13, 17, 17, 17, 17, 19, 19, 23, 23, 23, 23, 29, 29, 29, 29, 29, 29, 31, 31, 37, 37, 37, 37, 37, 37, 41, 41, 41, 41, 43, 43, 47, 47, 47, 47, 53, 53, 53, 53, 53, 53, 59, 59, 59, 59, 59, 59, 61, 61, 67, 67, 67, 67, 67, 67, 71, 71, 71, 71, 73
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
Thinking of n as time, a(n) represents the first time EulerPhi catches up with i(n), where i is the identity function. a(n) - n can be seen as the lag of EulerPhi behind i at time n. The sequence of these lags begins 0 1 2 1 2 1 4 3 2 1 2 1 4 3 2 1 2 1 4 3 2 1
a(n) is the smallest number for which the reduced residue system (=RRS[a(n)]) contains {1,2,...,n} as a subset; a(m) jumps at a(p)-1 and a(p) from value of p to nextprime[p]; a(x)=p[n] holds {p[n-1]....p[n]-1}; p(n) is repeated p(n)-p(n-1) times. For n>1, a(n)=p[Pi(n)+1], while a(1)=1. - Labos E. (labos(AT)ana.sote.hu), May 14 2003
|
|
FORMULA
|
a(1) = 1 a(n) = p(s+1) for n in [p(s), p(s+1) - 1], where p(s) denotes the s-th prime.
For n>1 a(n)=A007918(n+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 04 2002
For n>1, a(n)=A000040[A000720(n)+1], while a(1)=1. - Labos E. (labos(AT)ana.sote.hu), May 14 2003
|
|
EXAMPLE
|
a(5) = 7 since EulerPhi(7) = 6 is at least 5 and 7 is the smallest k satisfying EulerPhi(k) is greater than or equal to 5.
|
|
MATHEMATICA
|
a(1)=1; Table[Prime[PrimePi[w]+1], {w, 1, 100}]
|
|
CROSSREFS
|
Cf. A000040, A000720, A057237.
Sequence in context: A087243 A112276 A079578 this_sequence A087821 A109258 A088081
Adjacent sequences: A066166 A066167 A066168 this_sequence A066170 A066171 A066172
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Dec 13 2001
|
|
EXTENSIONS
|
More terms from Benoit Cloitre (benoit7848c(AT)orange.fr), May 04 2002
|
|
|
Search completed in 0.002 seconds
|
