|
Search: id:A099468
|
|
|
| A099468 |
|
Numbers n such that there are no primes < 2n in the sequence m(0)=n, m(k+1)=m(k)+4k. |
|
+0
2
|
|
| 1, 21, 45, 51, 81, 213, 249, 315, 477, 525, 681, 891, 1143, 1221, 1851, 1965, 2415, 5133
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
No others < 10^8. Note that 3 divides all these n > 1. This sequence is conjectured to be complete. Related to a question posed in A036468 by Zhang Ming-Zhi. Let r=2s+1 be an odd number. If n = (s+1)^2+s^2, then the sequence m(0)=n, m(k+1)=m(k)+4k for k=0,1,...s calculates the s+1 distinct sums of two squares (r-i)^2+i^2.
|
|
EXAMPLE
|
45 is here because 45, 49, 57, 69, and 85 are all composite.
|
|
MATHEMATICA
|
lst={}; Do[n=m; found=False; k=0; While[n=n+4k; !found && n<2m, found=PrimeQ[n]; k++ ]; If[ !found, AppendTo[lst, m]], {m, 1, 10000, 2}]; lst
|
|
CROSSREFS
|
Cf. A036468 (number of ways to represent 2n+1 as a+b with a^2+b^2 prime).
Adjacent sequences: A099465 A099466 A099467 this_sequence A099469 A099470 A099471
Sequence in context: A063319 A120071 A003857 this_sequence A063500 A102603 A044098
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
T. D. Noe (noe(AT)sspectra.com), Oct 17 2004
|
|
|
Search completed in 0.002 seconds
|
