A343138 Array A(k, n) read by descending antidiagonals: A(k, n) = Sum_{m=0..n} F(k, m)^2, where F are the k-generalized Fibonacci numbers A092921.
0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 6, 2, 1, 0, 1, 5, 15, 6, 2, 1, 0, 1, 6, 40, 22, 6, 2, 1, 0, 1, 7, 104, 71, 22, 6, 2, 1, 0, 1, 8, 273, 240, 86, 22, 6, 2, 1, 0, 1, 9, 714, 816, 311, 86, 22, 6, 2, 1, 0, 1, 10, 1870, 2752, 1152, 342, 86, 22, 6, 2, 1, 0
Offset: 0
Examples
Array starts: n = 0 1 2 3 4 5 6 7 8 9 10 ------------------------------------------------------------ [k=0] 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... [A057427] [k=1] 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... [A001477] [k=2] 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, ... [A001654] [k=3] 0, 1, 2, 6, 22, 71, 240, 816, 2752, 9313, 31514, ... [A107239] [k=4] 0, 1, 2, 6, 22, 86, 311, 1152, 4288, 15952, 59216, ... [k=5] 0, 1, 2, 6, 22, 86, 342, 1303, 5024, 19424, 75120, ... [k=6] 0, 1, 2, 6, 22, 86, 342, 1366, 5335, 20960, 82464, ... [k=7] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21591, 85600, ... [k=8] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 86871, ... [k=9] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 87382, ... [...] [ oo] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 87382, ... [A047849] Note that the first parameter in A(k, n) refers to rows, and the second parameter refers to columns, as always. The usual naming convention for the indices is not adhered to because the row sequences are the sums of the squares of the k-bonacci numbers.
References
- Greg Dresden and Yichen Wang, Sums and convolutions of k-bonacci and k-Lucas numbers, draft 2021.
Links
- Russell Jay Hendel, Sums of Squares: Methods for Proving Identity Families, arXiv:2103.16756 [math.NT], 2021
Programs
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Maple
F := (k, n) -> (F(k, n) := `if`(n<2, n, add(F(k, n-j), j = 1..min(k, n)))): A := (k, n) -> add(F(k, m)^2, m = 0..n): seq(seq(A(k, n-k), k=0..n), n = 0..11); # The following two functions implement Russell Jay Hendel's formula for k >= 2: T := (k, n) -> (n + 3)*(k - n) - 4: H := (k, n) -> (2*add(j*add((m-k+1)*F(k, n+j)*F(k, n+m), m = j+1..k), j = 1..k-1) - add(T(k, j-1)*F(k, n+j)^2, j = 1..k) + (k - 2))/(2*k - 2): seq(lprint([k], seq(H(k, n), n = 0..11)), k=2..9); # Peter Luschny, Apr 07 2021
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Mathematica
A343138[k_, len_] := Take[Accumulate[LinearRecurrence[PadLeft[{1}, k, 1], PadLeft[{1}, k], len + k]^2], -len - 2]; A343138[0, len_] := Table[Boole[n != 0], {n, 0, len}]; A343138[1, len_] := Table[n, {n, 0, len}]; (* Table: *) Table[A343138[k, 12], {k, 0, 9}] (* Sequence / descending antidiagonals: *) Table[Table[Take[A343138[j, 12], {k + 1 - j, k + 1 - j}], {j, 0, k}], {k, 0, 10}] // Flatten (* Georg Fischer, Apr 08 2021 *)
Formula
Russell Jay Hendel gives the following representation, valid for k >= 2:
A(n, k) = Sum_{m=0..n} F(k, m)^2 = (1/(2*k-2)) * (2*Sum_{j=1..k-1}(j*Sum_{m=j+1..k} (m-k+1) * F(k, n+j) * F(k, n+m)) - Sum_{j=1..k}(A343125(k, j-1) * F(k, n+j)^2) + (k - 2)). - Peter Luschny, Apr 07 2021