|
| |
|
|
A103274
|
|
Number of ways of writing prime(n) in the form 2*prime(i)+prime(j).
|
|
0
| |
|
|
0, 0, 0, 1, 2, 2, 4, 2, 3, 4, 2, 4, 5, 4, 4, 5, 3, 3, 5, 4, 4, 5, 4, 7, 6, 6, 5, 6, 6, 8, 6, 6, 8, 5, 8, 6, 6, 9, 5, 9, 7, 6, 6, 7, 10, 7, 8, 8, 6, 9, 12, 10, 7, 7, 11, 8, 10, 8, 11, 12, 9, 10, 12, 8, 10, 14, 12, 12, 7, 9, 12, 12, 11, 13, 10, 10, 15, 12, 15, 11, 12, 9, 12, 12, 12, 14, 12, 14, 13
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,5
|
|
|
COMMENTS
| First nonzero entry is for n=4: prime(4)=7=prime(1)+2*prime(3)=2+3*3, hence a(4)=1. Also, a(5)=2 because 11=5+2*3=7+2*2 (two solutions). Note that a(n) is not monotonic. - Zak Seidov (zakseidov(AT)yahoo.com), Jan 21 2006
Marnell conjectures that a(n) > 0 for n > 3. I find no exceptions below 10^9. [From Charles R Greathouse IV (charles.greathouse(AT)case.edu), May 04 2010]
|
|
|
REFERENCES
| Geoffrey R. Marnell, "Ten Prime Conjectures", Journal of Recreational Mathematics 33:3 (2004-2005), pp. 193-196. [From Charles R Greathouse IV (charles.greathouse(AT)case.edu), May 04 2010]
|
|
|
FORMULA
| a(n) = A046926(prime(n)). - David Wasserman (dwasserm(AT)earthlink.net), Oct 08 2005
|
|
|
EXAMPLE
| 11=2*2+7=2*3+5, so a(5)=2
a(100)=13 because p(100)=541=p(i)+2*p(j) for 13 pairs {i, j}: {2, 57}, {17, 53}, {23, 50}, {41, 42}, {49, 37}, {52, 36}, {56, 34}, {69, 25}, {76, 22}, {81, 18}, {91, 12}, {92, 11}, {96, 8}; e.g. 541=prime(96)+2*prime(8)=503+2*19. - Zak Seidov (zakseidov(AT)yahoo.com), Jan 21 2006
|
|
|
CROSSREFS
| Sequence in context: A163371 A061338 A135714 * A046820 A043262 A130860
Adjacent sequences: A103271 A103272 A103273 * A103275 A103276 A103277
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Yasutoshi Kohmoto (zbi74583(AT)boat.zero.ad.jp), Jan 27 2005
|
|
|
EXTENSIONS
| More terms from David Wasserman (dwasserm(AT)earthlink.net), Oct 08 2005
Edited by N. J. A. Sloane (njas(AT)research.att.com) at the suggestion of Andrew Plewe, Jul 14 2007
|
| |
|
|