OFFSET
1,2
COMMENTS
Consider the infinite array M, containing the positive integers by antidiagonals from lower left to upper right: M(j,k) = (k+j-1)*(k+j)/2-(j-1); j, k >= 1. a(n) is the element in row F(n+1) and column F(n), i.e., a(n) = M(F(n+1),F(n)).
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).
FORMULA
From Colin Barker, Feb 21 2015: (Start)
a(n) = 4*a(n-1)-2*a(n-2)-6*a(n-3)+4*a(n-4)+2*a(n-5)-a(n-6).
G.f.: -x*(x^4-2*x^3-2*x^2+2*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)).
(End)
EXAMPLE
Upper left 6 X 6 submatrix of M is
[ 1 3 6 10 15 21]
[ 2 5 9 14 20 27]
[ 4 8 13 19 26 34]
[ 7 12 18 25 33 42]
[11 17 24 32 41 51]
[16 23 31 40 50 61]
F(0) through F(7) are 0, 1, 1, 2, 3, 5, 8, 13. a(4) = M(F(5),F(4)) = M(5,3) = 24.
MATHEMATICA
LinearRecurrence[{4, -2, -6, 4, 2, -1}, {1, 2, 8, 24, 71, 198}, 30] (* Harvey P. Dale, Aug 08 2021 *)
PROG
(PARI) for(n=1, 27, print1((1/2)*(fibonacci(n+2)-1)*(fibonacci(n+2)-2)+fibonacci(n), ", ")) /* Klaus Brockhaus, Aug 29 2007 */
(Magma) z:=15; m:=Fibonacci(z+1); M:=Matrix(IntegerRing(), m, m, [ [ (k+j-1)*(k+j)/2-(j-1): k in [1..m] ]: j in [1..m] ] ); [ M[Fibonacci(n+1), Fibonacci(n)]: n in [1..z] ]; /* Klaus Brockhaus, Aug 29 2007 */
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Aug 27 2007
EXTENSIONS
Edited and extended by Klaus Brockhaus, Aug 29 2007
STATUS
approved