|
Search: id:A133312
|
|
|
| A133312 |
|
a(n) is the first pentagonal number which is nontrivially the sum of two pentagonal number of the type P(p)+P(p+n) (we always have P(k)=P(0)+P(k)). |
|
+0 1
|
|
| 70, 1926, 6305, 92, 22632, 34580, 49051, 66045, 85562, 1426, 925, 159251, 188860, 220992, 255647, 292825, 852, 2625, 7107, 466767, 516560, 568876, 623715, 681077, 5192, 803370, 7957, 935755, 1005732, 1078232, 22265, 8626, 1310870, 3577
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
The sequence is globally increasing with, at first sight, "unpredictible holes" ; the regular values satisfy a(n)=(2523/2)*n^2+(1189/2)*n+70 that is approximatively (35.5176013*n+ 8.3690808)^2. in fact we have to solve (6*k-1)^2=2*(6*p+3*n-1)^2+18*n^2-1 (1) with given n that is X^2=2Y^2+18*n^2-1 where X=6*k-1 and Y=6*p+3*n-1. p=20*n+5 and k=29*n+7 give allways a solution of the equation (1) but it is not sure that it is the best i.e the first.
|
|
EXAMPLE
|
a(0)=70 because the first interesting relation is P(70)=P(35)+P(35) i.e 70=2*35
a(1)=1926 because the first non obvious relation is P(36)=1926=P(25)+P(26).
|
|
MAPLE
|
for n from 1 to n ; a:=proc(k) if type (sqrt(18*k^2-6*k+1-9*n^2)/6-(3*n-1)/6, integer)=true then k*(3*k-1)/2 else fi end : seq (a(k), k=n..100000) od;
|
|
CROSSREFS
|
Sequence in context: A076430 A006296 A047835 this_sequence A093757 A163022 A006437
Adjacent sequences: A133309 A133310 A133311 this_sequence A133313 A133314 A133315
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Richard Choulet (richardchoulet(AT)yahoo.fr), Dec 20 2007
|
|
|
Search completed in 0.005 seconds
|