login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A277581 Goldbach's problem extended to squares of nonnegative differences of primes: smallest integer >= ((A112823(n) - A234345(n))^2)/n for n >= 2. 1
0, 0, 1, 0, 1, 0, 5, 2, 4, 0, 1, 0, 3, 2, 3, 0, 1, 0, 2, 1, 15, 0, 5, 6, 2, 3, 12, 0, 1, 0, 11, 2, 2, 5, 3, 0, 9, 1, 1, 0, 1, 0, 1, 1, 20, 0, 3, 12, 1, 6, 7, 0, 4, 11, 1, 2, 16, 0, 1, 0, 6, 2, 1, 3, 2, 0, 14, 1, 1, 0, 1, 0, 13, 1, 1, 2, 2, 0, 5, 1, 11, 0, 2, 7, 1, 10, 4, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,7
COMMENTS
Where A112823(n) + A234345(n) = 2n and A112823(n) <= A234345(n) (or nonnegative differences of primes). If n is prime, then a(n) = 0.
Conjecture: 1 <= a(n) <= m for all n, where m is largest value of a(n), i.e., the sequence of records in a(n) {1, 5, 15, 20, ..., m} is finite.
LINKS
EXAMPLE
a(8) = 5 because ((A112823(8) - A234345(8))^2)/8 = ((5 - 11)^2)/8 < 5, where 5(prime) + 11(prime) = 2*8;
a(9) = 2 because ((A112823(9) - A234345(9))^2)/9 = ((7 - 11)^2)/9 < 2, where 7(prime) + 11(prime) = 2*9;
a(10) = 4 because ((A112823(10) - A234345(10))^2)/10 = ((7 - 13)^2/10 < 4, where 7(prime) + 13(prime) = 2*10.
CROSSREFS
Cf. A112823 (2 together with A002374), A234345, A277583 (Goldbach's problem extended to squares of prime gaps >= 2).
Sequence in context: A193797 A350518 A214662 * A307381 A256167 A207528
KEYWORD
nonn
AUTHOR
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified July 29 04:13 EDT 2024. Contains 374727 sequences. (Running on oeis4.)