A220436 a(n) = A127546(n)^2.

4, 36, 196, 1444, 9604, 66564, 454276, 3118756, 21362884, 146458404, 1003749124, 6880038916, 47155859716, 323212716324, 2215328606404, 15184099435684, 104073336269956, 713329336067076, 4889231802533764, 33511293841053604, 229689823620354244, 1574317475335503396, 10790532493690424836

Offset: 0

Author

N. J. A. Sloane, Dec 16 2012

Keywords

• nonn

References

• Giacomo Candido, A Relationship Between the Fourth Powers of the Terms of the Fibonacci Series, Scripta Mathematica, Vol. 17, No. 3-4 (1951), p. 230.
• Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, in: Simon R. Blackburn, Stefanie Gerke and Mark Wildon, eds., Surveys in Combinatorics 2013, Cambridge University Press, 2013, pp. 363-378.

Links

• Claudi Alsina and Roger B. Nelsen, On Candido's Identity, Mathematics Magazine, Vol. 80, No. 3 (2007), pp. 226-228; alternative link.
• Alexander Bogomolny, Candido's Identity, Cut the Knot.
• Brother Alfred Brousseau, Ye Olde Fibonacci Curiosity Shoppe, The Fibonacci Quarterly, Vol. 10, No. 4 (1972), pp. 441-443.
• Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint, arXiv:1206.4864 [math.CO], 2012.
• R. S. Melham, Ye Olde Fibonacci Curiosity Shoppe Revisited, The Fibonacci Quarterly, Vol. 42, No. 2 (2004), pp. 155-160.
• Roger B. Nelsen, Proof without Words: Candido's Identity, Mathematics Magazine, Vol. 78, No. 2 (2005), p. 131.
• Darko Veljan, A note on Candido's identity and Heron's formula, Proceedings of the 1st Croatian Combinatorial Days, Zagreb, September 29-30, 2016 (2016).
• Wikipedia, Candido's identity.

Crossrefs

Cf. A000045, A127546.

Programs

Mathematica

Table[Total[Fibonacci[Range[n, n + 2]]^2]^2, {n, 0, 22}] (* or *)
Table[4 (4 ((-1)^(n + 1) LucasL[2 (n + 1)] + LucasL[4 (n + 1)]) + 9)/25, {n, 0, 22}] (* Michael De Vlieger, Feb 18 2017 *)

Formula

Empirical g.f.: -4*(x^4-4*x^3-11*x^2+4*x+1) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Jul 22 2013

a(n) = 4*(4*((-1)^(n + 1)*Lucas(2*(n + 1)) + Lucas(4*(n + 1))) + 9)/25. - Ehren Metcalfe, Feb 18 2017

a(n) = 2 * (F(n)^4 + F(n+1)^4 + F(n+2)^4), where F(n) is the n-th Fibonacci number (A000045) (Candido, 1951). - Amiram Eldar, Jan 11 2022