|
Search: id:A075888
|
|
|
| A075888 |
|
Difference of successive primes squared divided by 24, (prime(n+1)^2-prime(n)^2)/24, n >= 3. |
|
+0
4
|
|
| 1, 3, 2, 5, 3, 7, 13, 5, 17, 13, 7, 15, 25, 28, 10, 32, 23, 12, 38, 27, 43, 62, 33, 17, 35, 18, 37, 140, 43, 67, 23, 120, 25, 77, 80, 55, 85, 88, 30, 155, 32, 65, 33, 205, 217, 75, 38, 77, 118, 40, 205, 127, 130, 133, 45, 137, 93, 47, 240, 350, 103, 52, 105, 378, 167, 285
(list; graph; listen)
|
|
|
OFFSET
|
3,2
|
|
|
COMMENT
|
For n>=3, prime(n+1)^2-prime(n)^2 is always divisible by 24.
It follows from the previous comment that for n>=3, prime(n)= sqrt(5^2 + k*24) where integer k>= 0 Then it follows from above that for n>=3, ((prime(n))^2 - 1)/24 always gives integral values - see A024702 [From Alexander R. Povolotsky (pevnev(AT)juno.com), Sep 20 2008]
|
|
FORMULA
|
a(n)=(prime(n+1)^2-prime(n)^2)/24, n >= 3.
|
|
EXAMPLE
|
a(4)=3 because (p(5)^2-p(4)^2)/24=(11^2-7^2)/24=3.
|
|
PROGRAM
|
(PARI) j=[]; for(n=3, 300, if(((floor((((prime(n+1))^2)-((prime(n))^2))/24))==(ceil(((((prime(n+1))^2)-((pr\ ime(n))^2))/24)))), j=concat(j, ((((prime(n+1))^2) - ((prime(n))^2))/24)), j=concat(j, -1))); j [From Alexander R. Povolotsky (pevnev(AT)juno.com), Sep 08 2008]
|
|
CROSSREFS
|
Cf. A024702 [From Alexander R. Povolotsky (pevnev(AT)juno.com), Sep 20 2008]
Sequence in context: A111618 A107128 A086670 this_sequence A075889 A045766 A132817
Adjacent sequences: A075885 A075886 A075887 this_sequence A075889 A075890 A075891
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Zak Seidov (zakseidov(AT)yahoo.com), Oct 17 2002
|
|
|
Search completed in 0.002 seconds
|
