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%I #27 Nov 25 2023 23:54:53
%S 1,2,2,5,6,3,12,20,12,4,29,60,50,20,5,70,174,180,100,30,6,169,490,609,
%T 420,175,42,7,408,1352,1960,1624,840,280,56,8,985,3672,6084,5880,3654,
%U 1512,420,72,9,2378,9850,18360,20280,14700,7308,2520,600,90,10
%N Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 2 + 2x, p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where u = p(2,x), v = 1 - 2x - x^2.
%C Because (p(n,x)) is a strong divisibility sequence, for each integer k, the sequence (p(n,k)) is a strong divisibility sequence of integers.
%H Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, <a href="http://math.colgate.edu/~integers/s14/s14.Abstract.html">Characterization of the strong divisibility property for generalized Fibonacci polynomials</a>, Integers, 18 (2018), Paper No. A14.
%F p(n, x) = u*p(n-1, x) + v*p(n-2, x) for n >= 3, where p(1, x) = 1, p(2, x) = 2 + 2 x, u = p(2, x), and v = 1 - 2x - x^2.
%F p(n, x) = k*(b^n - c^n), where k = sqrt(1/8), b = x + 1 - sqrt(2), c = x + 1 + sqrt(2).
%F From _Werner Schulte_, Nov 24 2023 and Nov 25 2023: (Start)
%F The row polynomials p(n, x) = Sum_{k=0..n-1} T(n, k) * x^k satisfy the equation p'(n, x) = n * p(n-1, x) where p' is the first derivative of p.
%F T(n, k) = T(n-1, k-1) * n / k for 0 < k < n and T(n, 0) = A000129(n) for n > 0.
%F T(n, k) = A000129(n-k) * binomial(n, k) for 0 <= k < n.
%F G.f.: t / (1 - (2+2*x) * t - (1-2*x-x^2) * t^2). (End)
%e First nine rows:
%e [n\k] 0 1 2 3 4 5 6 7 8
%e [1] 1;
%e [2] 2 2;
%e [3] 5 6 3;
%e [4] 12 20 12 4;
%e [5] 29 60 50 20 5;
%e [6] 70 174 180 100 30 6;
%e [7] 169 490 609 420 175 42 7;
%e [8] 408 1352 1960 1624 840 280 56 8;
%e [9] 985 3672 6084 5880 3654 1512 420 72 9;
%e .
%e Row 4 represents the polynomial p(4,x) = 12 + 20 x + 12 x^2 + 4 x^3, so that (T(4,k)) = (12, 20, 12, 4), k = 0..3.
%p P := proc(n) option remember; ifelse(n <= 1, n, 2*P(n - 1) + P(n - 2)) end:
%p T := (n, k) -> P(n - k) * binomial(n, k):
%p for n from 1 to 9 do [n], seq(T(n, k), k = 0..n-1) od;
%p # (after _Werner Schulte_) _Peter Luschny_, Nov 24 2023
%t p[1, x_] := 1; p[2, x_] := 2 + 2 x; u[x_] := p[2, x]; v[x_] := 1 - 2 x - x^2;
%t p[n_, x_] := Expand[u[x]*p[n - 1, x] + v[x]*p[n - 2, x]]
%t Grid[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
%t Flatten[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
%Y Cf. A000129 (column 1), A361732 (column 2), A000027 (T(n,n-1)), A007070 (row sums, p(n,1)), A077957 (alternating row sums, p(n,-1)), A081179 (p(n,2), A077985 (p(n,-2), A081180 (p(n,3)), A007070 (p(n,-3)), A081182 (p(n,4)), A094440, A367208, A367209, A367210.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Nov 13 2023
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