|
| |
|
|
A177049
|
|
Numerator of (3n+1)*(3n+2)/4.
|
|
1
|
|
|
|
1, 5, 14, 55, 91, 68, 95, 253, 325, 203, 248, 595, 703, 410, 473, 1081, 1225, 689, 770, 1711, 1891, 1040, 1139, 2485, 2701, 1463, 1580, 3403, 3655, 1958, 2093, 4465, 4753, 2525, 2678, 5671, 5995, 3164, 3335, 7021, 7381, 3875, 4064, 8515, 8911
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
A trisection of A064038.
|
|
|
LINKS
|
Table of n, a(n) for n=0..44.
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).
|
|
|
FORMULA
|
Conjecture: a(n)= +3*a(n-1) -6*a(n-2) +10*a(n-3) -12*a(n-4) +12*a(n-5) -10*a(n-6) +6*a(n-7) -3*a(n-8) +a(n-9) with g.f. -(x^2+4*x+1)*(x^6-2*x^5+12*x^4-13*x^3+12*x^2-2*x+1) ) / ( (x-1)^3*(x^2+1)^3 ). - R. J. Mathar, Dec 12 2010
The conjecture is correct. - Charles R Greathouse IV, Feb 08 2012
a(n) ~ -27/8*n^2 - 27/8*n. - Ralf Stephan, Jun 16 2014
Sum_{n>=0} 1/a(n) = (4/(3*sqrt(3)) - 1/3)*Pi. - Amiram Eldar, Aug 13 2022
|
|
|
MATHEMATICA
|
Table[Numerator[(3 n + 1) (3 n + 2)/4], {n, 0, 50}] (* Wesley Ivan Hurt, Jun 14 2014 *)
LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 5, 14, 55, 91, 68, 95, 253, 325}, 50] (* Harvey P. Dale, Jan 18 2020 *)
|
|
|
CROSSREFS
|
Cf. A064038, A127922.
Sequence in context: A073541 A268887 A055488 * A127922 A262247 A279511
Adjacent sequences: A177046 A177047 A177048 * A177050 A177051 A177052
|
|
|
KEYWORD
|
nonn,frac,less,easy
|
|
|
AUTHOR
|
Paul Curtz, Dec 09 2010
|
|
|
STATUS
|
approved
|
| |
|
|
