A226096 Squares with doubled (4*n+2)^2.

1, 4, 4, 9, 16, 25, 36, 36, 49, 64, 81, 100, 100, 121, 144, 169, 196, 196, 225, 256, 289, 324, 324, 361, 400, 441, 484, 484, 529, 576, 625, 676, 676, 729, 784, 841, 900, 900, 961, 1024, 1089, 1156, 1156, 1225, 1296, 1369

Offset: 0

Author

Paul Curtz, May 26 2013

Keywords

• nonn
• easy

Comments

Also nondecreasing ordered values of A226008 (except 0).

Consider A225948/A226008 ordered according to a(n): 0/1, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = b(n)/a(n), and consider the sequence with period 5: 1, 64, 16, 1, 4, ... = t(n); then a(n) = 4*b(n) + t(n).

The recurrences in Formula lines are also valid for b(n).

Note that the fractions b(n)/a(n) of rank 0, 3,4,5, 8,9,10, ... = A047205:

0, 2/9, 3/16, 6/25, 12/49, 15/64, 20/81, ... are all in A226023(n).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2,-2,0,0,0,-1,1).

Crossrefs

Cf. A016826, A226008.

Programs

Mathematica

MapIndexed[ If [Mod[First[#2], 4] == 2, Sequence @@ {#1, #1}, #1] &, Range[40]]^2 (* Jean-François Alcover, May 28 2013 *)

Formula

a(n+5) - a(n) = 8*A090223(n+4).

a(n) = 1 followed by (A090223(n) + 2)^2.

a(n) = 3*a(n-5) -3*a(n-10) +a(n-15).

G.f.: (x^9 + 3*x^8 + 5*x^6 + 7*x^5 + 7*x^4 + 5*x^3 + 3*x + 1)/((1 - x)*(1 - x^5)^2). [Ralf Stephan, May 30 2013]

a(n) = a(n-1) +2*a(n-5) -2*a(n-6) -a(n-10) +a(n-11). [Bruno Berselli, May 30 2013]

a(n) = (24*(16*floor(n/5)^2 + 8*floor(n/5) + 1) - (11 + 24*floor(n/5))*(n - 5*floor(n/5))^4 + 2*(49 + 104*floor(n/5))*(n - 5*floor(n/5))^3 - 23*(11 + 24*floor(n/5))*(n - 5*floor(n/5))^2 + 2*(119 + 280*floor(n/5))*(n - 5*floor(n/5)))/24. - Luce ETIENNE, May 08 2017