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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
Russell Jay Hendel, Sums of Squares: Methods for Proving Identity Families, arXiv:2103.16756 [math.NT], 2021.
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)).
From G. C. Greubel, Nov 22 2021: (Start)
T(2*n-2, n) = A028557(n-2), n >= 2.
T(4*n-6, n) = 2*A140672(n-2), n >= 2. (End)
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), ", ")))
(Sage) flatten([[(n+3)*(k-n) -4 for n in (1..k-1)] for k in (2..15)]) # G. C. Greubel, Nov 22 2021
CROSSREFS
Sequence in context: A254707 A134417 A116080 * A205296 A143820 A259930
KEYWORD
easy,nonn,tabl
AUTHOR
Russell Jay Hendel, Apr 06 2021
STATUS
approved

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)