|
|
A076049
|
|
Numbers k such that the sum of the k-th triangular number and (k+2)-nd triangular number is a triangular number.
|
|
2
|
|
|
0, 3, 8, 25, 54, 153, 322, 899, 1884, 5247, 10988, 30589, 64050, 178293, 373318, 1039175, 2175864, 6056763, 12681872, 35301409, 73915374, 205751697, 430810378, 1199208779, 2510946900, 6989500983, 14634871028, 40737797125
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
T(a(n)) + T(a(n)+2) = A069017(n+1) where T(k) = k*(k+1)/2.
|
|
LINKS
|
|
|
FORMULA
|
Let b(n) = A001109(n). Then we have a pair of recursion formulas:
a(2n+2) = 2*a(2n+1) - a(2n) + 2*b(n+1);
a(2n+3) = 2*a(2n+2) - a(2n+1) + 2*b(n+2).
G.f.: x*(3 + 5*x - x^2 - x^3)/((1-x)*(1 - 6*x^2 + x^4)).
a(n) = -3 + (1/8)*(-1^n)((7 + 5*sqrt(2))*(-1 - sqrt(2))^n + (7 - 5*sqrt(2))*(-1 + sqrt(2))^n - (1 + sqrt(2))^n - (1 - sqrt(2))^n).
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Bruce Corrigan (scentman(AT)myfamily.com), Oct 29 2002
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|