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%I #23 Sep 08 2022 08:45:31
%S 1,2,8,24,71,198,541,1452,3862,10208,26885,70644,185369,485982,
%T 1273420,3335640,8735707,22875050,59895221,156819960,410579786,
%U 1074943872,2814291433,7367994504,19289795761,50501560538,132215157296,346144350552,906218605007
%N a(n) = (1/2)*(F(n+2)-1)*(F(n+2)-2) + F(n), where F() are the Fibonacci numbers.
%C 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)).
%H Colin Barker, <a href="/A131569/b131569.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-6,4,2,-1).
%F From _Colin Barker_, Feb 21 2015: (Start)
%F 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).
%F 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)).
%F (End)
%e Upper left 6 X 6 submatrix of M is
%e [ 1 3 6 10 15 21]
%e [ 2 5 9 14 20 27]
%e [ 4 8 13 19 26 34]
%e [ 7 12 18 25 33 42]
%e [11 17 24 32 41 51]
%e [16 23 31 40 50 61]
%e 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.
%t LinearRecurrence[{4,-2,-6,4,2,-1},{1,2,8,24,71,198},30] (* _Harvey P. Dale_, Aug 08 2021 *)
%o (PARI) for(n=1, 27, print1((1/2)*(fibonacci(n+2)-1)*(fibonacci(n+2)-2)+fibonacci(n), ",")) /* _Klaus Brockhaus_, Aug 29 2007 */
%o (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 */
%Y Cf. A000045 (Fibonacci numbers).
%K easy,nonn
%O 1,2
%A Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Aug 27 2007
%E Edited and extended by _Klaus Brockhaus_, Aug 29 2007
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