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A131569 a(n) = (1/2)*(F(n+2)-1)*(F(n+2)-2) + F(n), where F() are the Fibonacci numbers. 1
1, 2, 8, 24, 71, 198, 541, 1452, 3862, 10208, 26885, 70644, 185369, 485982, 1273420, 3335640, 8735707, 22875050, 59895221, 156819960, 410579786, 1074943872, 2814291433, 7367994504, 19289795761, 50501560538, 132215157296, 346144350552, 906218605007 (list; graph; refs; listen; history; text; internal format)
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.

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

Cf. A000045 (Fibonacci numbers).

Sequence in context: A018045 A050242 A045697 * A290904 A231200 A066973

Adjacent sequences:  A131566 A131567 A131568 * A131570 A131571 A131572

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

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Last modified April 18 07:56 EDT 2019. Contains 322209 sequences. (Running on oeis4.)