|
|
A107118
|
|
Numbers that are both centered triangular numbers (A005448) and centered hexagonal numbers (A003215).
|
|
1
|
|
|
1, 19, 631, 21421, 727669, 24719311, 839728891, 28526062969, 969046412041, 32919051946411, 1118278719765919, 37988557420094821, 1290492673563457981, 43838762343737476519, 1489227427013510743651, 50589893756115627807601, 1718567160280917834714769
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The centered hexagonal numbers are given by 3*p^2 - 3*p + 1 while the centered triangular numbers are given by (3*r^2 + 3*r + 2)/2. A natural number is both of the above numbers if and only if there exist numbers p and r such that 2*(2p-1)^2 = (2*r+1)^2+1. The Diophantine equation X^2 = 2*Y^2 - 1 has the following solutions: X is given by 1, 7, 41, 239, ..., i.e., A002315, and Y is given by A001653. The first equation gives r with 0, 3, 20, 119, 6906, i.e., A001652, and p with 1, 3, 15, 85, 493, ..., i.e., A011900.
|
|
LINKS
|
|
|
FORMULA
|
a(n+2) = 34*a(n+1) - a(n) - 14.
a(n+1) = 17*a(n) - 7 + sqrt(288*a(n)^2 - 252*a(n) + 45).
G.f.: h(z)=(z*(1-16*z+z^2))/((1-z)*(1-34*z+z^2)).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Colin Barker, Jan 02 2015
a(n) = (14+(9+6*sqrt(2))*(17+12*sqrt(2))^(-n)+(9-6*sqrt(2))*(17+12*sqrt(2))^n)/32. - Colin Barker, Mar 02 2016
|
|
MATHEMATICA
|
a[n_] := 17*n - 7 + Sqrt[288*n^2 - 252*n + 45]; NestList[a, 1, 20] (* Stefan Steinerberger, Sep 18 2007 *)
LinearRecurrence[{35, -35, 1}, {1, 19, 631}, 30] (* Harvey P. Dale, Jan 16 2016 *)
|
|
PROG
|
(PARI) Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015
|
|
CROSSREFS
|
Cf. A003215 (Centered hexagonal numbers), A005448 (Centered triangular numbers).
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|