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A343125 Triangle T(k, n) = (n+3)*(k-n) - 4, k >= 2, 1 <= n <= k-1, read by rows. 1
0, 4, 1, 8, 6, 2, 12, 11, 8, 3, 16, 16, 14, 10, 4, 20, 21, 20, 17, 12, 5, 24, 26, 26, 24, 20, 14, 6, 28, 31, 32, 31, 28, 23, 16, 7, 32, 36, 38, 38, 36, 32, 26, 18, 8, 36, 41, 44, 45, 44, 41, 36, 29, 20, 9, 40, 46, 50, 52, 52, 50, 46, 40, 32, 22, 10 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

2,2

COMMENTS

T(k, n) is even if k is odd.

T(k, n) = T(k, n+1) for n = k/2 - 2 if k >= 6 is even.

T(k, n) = T(k, n+2) for n = (k-1)/2 - 2 if k >= 7 is odd.

For fixed n, T(k, n) is linear in k.

The T(k, j) contribute coefficients to a closed formula for the sum of the first n+1 squares of the k-generalized Fibonacci numbers, F(k, j) = A092921(k, j). See A343138 for sums of squares of F(k, j). See the Formula section for closed formula. Although other sequences occur in coefficients in the closed formula for sums of squares, they are linear in nature. All coefficient sequences are mentioned in the arXiv link. The closed formula generalizes results of Schumacher (see References) for the cases k=3 and k=4 with a uniform proof method (see arXiv link).

REFERENCES

Raphael Schumacher, How to Sum the Squares of the Tetranacci Numbers and the Fibonacci m-step Numbers, Fibonacci Quarterly, 57, (2019), 168-175.

Raphael Schumacher, Explicit Formulas for Sums Involving the Squares of the First n Tribonacci Numbers, Fibonacci Quarterly, 58 (2020), 194-202.

LINKS

Table of n, a(n) for n=2..67.

Russell Jay Hendel, Sums of Squares: Methods for Proving Identity Families, arXiv:2103.16756 [math.NT], 2021

Proof Wiki, Sum of Sequence of Squares of Fibonacci Numbers

FORMULA

Let F(k, n) = A092921(k, n), the k-generalized Fibonacci numbers. Let A(k, n) = A343138(k, n) = Sum_{m=0..n} F(k, m)^2, the sum of the first m+1 k-generalized Fibonacci numbers. Then, for k >= 2, a closed formula for A(k, n) is:

A(k, n) = (1/(2*k-2)) * (Sum_{j=0..k-2, m=j+1..k-1} 2*(j+1)*(m-k+1) * F(k, n+j) * F(k, n+m)) - (k-2)*F(k, n)^2 - Sum_{j=1..k}(T(k, j) * F(k, n+j)^2) + (k-2)).

EXAMPLE

Triangle T(k, n) begins:

   k \ n|  1  2  3  4  5  6  7  8  9  10 11

  ------+----------------------------------

   2    |  0

   3    |  4  1

   4    |  8  6  2

   5    | 12 11  8  3

   6    | 16 16 14 10  4

   7    | 20 21 20 17 12  5

   8    | 24 26 26 24 20 14  6

   9    | 28 31 32 31 28 23 16  7

  10    | 32 36 38 38 36 32 26 18  8

  11    | 36 41 44 45 44 41 36 29 20  9

  12    | 40 46 50 52 52 50 46 40 32 22 10

.

The following are the closed formulas for k = 3, 4 for A(k, n) = Sum_{m=0..n} F(k, m)^2, with F(k, n) = A092921(k, n), the k-generalized Fibonacci numbers, and A(k, n) = A343138(k, n), the sum of squares of F(k, n). These formulas are derived from the closed formula in the formula section. Of course further simplifications are possible. For k = 2, T(2, 1) = 0 so illustrations start with k = 3.

k | Formula

--+--------------------------------------------------------

3 | Sum_{m=0..n} F(3,m)^2 = (1/4)*(2*F(3,n)*F(3,n+2) + 4*F(3,n+1)*F(3,n+2) - (k - 2)*F(3,n)^2 - T(3,1)*F(3,n+1)^2 - T(3,2)*F(3,n+2)^2 + 1).

4 | Sum_{m=0..n} F(3,m)^2 = (1/6)*(-2*F(4,n)*F(4,n+1) + 2*F(4,n)*F(4,n+3) + 4*F(4,n+1)*F(4,n+3) + 6*F(4,n+2)*F(4,n+3) - (k-2)*F(4,n)^2 - T(4,1)*F(4,n+1)^2 - T(4, 2)*F(4,n+2)^2 - T(4,3)*F(4,n+3)^2 + 2).

MAPLE

T := (k, n) -> (n + 3)*(k - n) - 4:

seq(print(seq(T(k, n), n=1..k-1)), k = 2..12); # Peter Luschny, Apr 02 2021

MATHEMATICA

Table[(n + 3) (k - n) - 4, {k, 2, 12}, {n, k - 1}] // Flatten (* Michael De Vlieger, Apr 06 2021 *)

PROG

(PARI) T(k, n)=(n + 3)*(k - n) - 4

for(k = 2, 12, for(n = 1, k - 1, print1(T(k, n), ", ")))

CROSSREFS

Cf. A092921, A085697, A343138.

Sequence in context: A254707 A134417 A116080 * A205296 A143820 A259930

Adjacent sequences:  A343122 A343123 A343124 * A343126 A343127 A343128

KEYWORD

easy,nonn,tabl

AUTHOR

Russell Jay Hendel, Apr 06 2021

STATUS

approved

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Last modified July 24 01:41 EDT 2021. Contains 346269 sequences. (Running on oeis4.)