|
Search: id:A002541
|
|
|
| A002541 |
|
Sum_{k=1..n-1} floor((n-k)/k).
(Formerly M0970 N0362)
|
|
+0
5
|
|
| 0, 1, 2, 4, 5, 8, 9, 12, 14, 17, 18, 23, 24, 27, 30, 34, 35, 40, 41, 46, 49, 52, 53, 60, 62, 65, 68, 73, 74, 81, 82, 87, 90, 93, 96, 104, 105, 108, 111, 118, 119, 126, 127, 132, 137, 140, 141, 150, 152, 157, 160, 165, 166, 173, 176, 183, 186, 189, 190, 201, 202, 205
(list; graph; listen)
|
|
|
OFFSET
|
1,3
|
|
|
COMMENT
|
Number of pairs (a, b) with 1 <= a < b <= n, a | b.
The sequence shows how many digit "skips" there have been from 2 to n, a skip being either a prime factor or product thereof. Everytime you have a place where you have X skips and the next skip value is X+1, you will have a prime number - since a prime number will only add exactly one more skip to get to it. a(n) = sum(x=2,n,floor(n/x)) - sum(x=2, n-1, floor( (n-1)/x ) )=1 when prime. - Marius-Paul Dumitrean (marius(AT)neldor.com), Feb 19 2007
A027749(a(n)+1)=n; A027749(a(n)+2)=A020639(n+1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 22 2003
Number of partitions of n into exactly 2 types of part, where one part is 1, e.g. n=7 gives 1111111,111112,11122,1222,11113,133,1114,115 and 16, so a(n)=9. - Jon Perry (perry(AT)globalnet.co.uk), May 26 2004
|
|
REFERENCES
|
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
J. P. Gram, Undersoegelser angaaende maengden af primtal under en given graense, Det Kongelige Danskevidenskabernes Selskabs Skrifter, series 6, vol. 2 (1884), 183-288; see Tab. VII: Vaerdier af Funktionen psi(n) og andre numeriske Funktioner, pp. 281-288, especially p. 281.
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=1..1000
|
|
FORMULA
|
a(n) = -n+sum_{k=1..n} tau(k). - Vladeta Jovovic (vladeta(AT)eunet.rs), Oct 17 2002
G.f.: 1/(1-x)*sum(k>=2, x^k/(1-x^k)). - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 23 2003
a(n)=sum(i=2, n, floor(n/i)) - Jon Perry (perry(AT)globalnet.co.uk), Feb 02 2004
a(n)=sum(i=2, n, ceil((n+1)/2))-n+1 - Jon Perry (perry(AT)globalnet.co.uk), May 26 2004
a(n)=A006218(n)-n. Proof : Floor[(n-k)/k)+1=Floor(n/k) then sum{k=1,n-1}Floor((n-k/k)+(n-1)+1=sum{k=1;n-1)Floor(n/k)+Floor(n/n) =sum{k=1,n}Floor(n/k) i.e a(n)+n=A006218(n) - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
|
|
CROSSREFS
|
Cf. A000005, A006218.
Antidiagonal sums of array A003988.
Cf. A006218.
Sequence in context: A161903 A099628 A137169 this_sequence A102821 A101881 A143989
Adjacent sequences: A002538 A002539 A002540 this_sequence A002542 A002543 A002544
|
|
KEYWORD
|
nonn,easy,nice
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com).
|
|
EXTENSIONS
|
More terms from David W. Wilson (davidwwilson(AT)comcast.net).
|
|
|
Search completed in 0.002 seconds
|
