A107241 Sum of squares of first n tetranacci numbers (A000288).

1, 2, 3, 4, 20, 69, 238, 863, 3264, 12100, 44861, 166662, 619591, 2301800, 8551800, 31774561, 118060082, 438649107, 1629796276, 6055504952, 22499207241, 83595676570, 310599326171, 1154030334396, 4287794153932, 15931278338957

Offset: 1

Author

Jonathan Vos Post, May 14 2005

Keywords

• easy
• nonn

Comments

Tetranacci numbers are also called Fibonacci 4-step numbers. a(n) is prime for n = 2, 3, 8, 26, ... a(n) is semiprime for n = 4, 6, 11, 13, ... a(10) = 12100 = 94^2 + 3264 = 110^2 = 2^2 * 5^2 * 11^2. For Fibonacci numbers (A000045) F(i) we have Sum_{i=1..n} F(i) = F(n)*F(n+1).

Examples

			a(1) = 1^2 = 1.
a(2) = 1^2 + 1^2 = 1.
a(3) = 1^2 + 1^2 + 1^2 = 3, prime.
a(4) = 1^2 + 1^2 + 1^2 + 1^2 = 4 = 2^2, semiprime.
a(5) = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 = 20.
a(6) = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2 = 69 = 3 * 23, semiprime.
a(8) = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2 + 13^2 + 25^2 = 863, prime.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
• Index entries for linear recurrences with constant coefficients, signature (3,2,2,6,-16,-2,6,-2,2,1,-1).

Crossrefs

Cf. A000288, A107239, A107240, A107241, A107242, A107243, A107244, A107245, A107246, A107247, A107248.

Programs

Maple

T:= proc(n) option remember;
      if n=0 then 0
    elif n<5 then 1
    else add(T(n-j), j=1..4)
      fi; end:
seq( add(T(k)^2, k=1..n), n=1..30); # G. C. Greubel, Dec 18 2019

Mathematica

Accumulate[LinearRecurrence[{1,1,1,1},{1,1,1,1},30]^2] (* Harvey P. Dale, Feb 14 2012 *)
LinearRecurrence[{3,2,2,6,-16,-2,6,-2,2,1,-1}, {1,2,3,4,20,69,238,863,3264, 12100,44861}, 30] (* Ray Chandler, Aug 02 2015 *)
T[n_]:= T[n]= If[n == 0, 0, If[n < 5, 1, Sum[T[n-j], {j,4}]]]; a[n_]:= Sum[T[j]^2, {j,n}]; Table[a[n], {n, 30}] (* G. C. Greubel, Dec 18 2019 *)

Sage

@CachedFunction
def T(n):
    if (n==0): return 0
    elif (n<5): return 1
    else: return sum(T(n-j) for j in (1..4))
def a(n): return sum(T(j)^2 for j in (1..n))
[a(n) for n in (1..30)] # G. C. Greubel, Dec 18 2019

Formula

a(n) = Sum_{i=1..n} A000288(i)^2.

From R. J. Mathar, Aug 11 2009: (Start)

a(n) = 3*a(n-1) + 2*a(n-2) + 2*a(n-3) + 6*a(n-4) - 16*a(n-5) - 2*a(n-6) + 6*a(n-7) - 2*a(n-8) + 2*a(n-9) + a(n-10) - a(n-11).

G.f.: (1 - x - 5*x^2 - 11*x^3 - 8*x^4 - x^5 - x^6 - 7*x^7 + x^8 + 4*x^9)/((1 - x)*(1 - 3*x - 3*x^2 + x^3 + x^4)(1 + x + 2*x^2 + 2*x^3 - 2*x^4 + x^5 - x^6)). (End)