A361732 -id:A361732 - cp's OEIS frontend

A367211 Triangular array read by rows: T(n, k) = binomial(n, k) * A000129(n - k) for 0 <= k < n.

1, 2, 2, 5, 6, 3, 12, 20, 12, 4, 29, 60, 50, 20, 5, 70, 174, 180, 100, 30, 6, 169, 490, 609, 420, 175, 42, 7, 408, 1352, 1960, 1624, 840, 280, 56, 8, 985, 3672, 6084, 5880, 3654, 1512, 420, 72, 9, 2378, 9850, 18360, 20280, 14700, 7308, 2520, 600, 90, 10

Offset: 1

Clark Kimberling, Nov 13 2023

T(n, k) are the coefficients of the polynomials p(1, x) = 1, p(2, x) = 2 + 2*x, p(n, x) = u*p(n-1, x) + v*p(n-2, x) for n >= 3, where u = p(2, x), v = 1 - 2*x - x^2.

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.

			First nine rows:
  [n\k] 0     1     2     3     4     5    6   7  8
  [1]   1;
  [2]   2     2;
  [3]   5     6    3;
  [4]  12    20    12     4;
  [5]  29    60    50    20     5;
  [6]  70   174   180   100    30     6;
  [7] 169   490   609   420   175    42   7;
  [8] 408  1352  1960  1624   840   280   56   8;
  [9] 985  3672  6084  5880  3654  1512  420  72  9;
.
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.

Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, Characterization of the strong divisibility property for generalized Fibonacci polynomials, Integers, 18 (2018), Paper No. A14.

Cf. A000129 (column 1, Pell numbers), 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.

P := proc(n) option remember; ifelse(n <= 1, n, 2*P(n - 1) + P(n - 2)) end:
T := (n, k) -> P(n - k) * binomial(n, k):
for n from 1 to 9 do [n], seq(T(n, k), k = 0..n-1) od;
# (after Werner Schulte)  Peter Luschny, Nov 24 2023

p[1, x_] := 1; p[2, x_] := 2 + 2 x; u[x_] := p[2, x]; v[x_] := 1 - 2 x - x^2;
p[n_, x_] := Expand[u[x]*p[n - 1, x] + v[x]*p[n - 2, x]]
Grid[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
Flatten[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
(* Or: *)
T[n_, k_] := Module[{P},
  P[m_] := P[m] = If[m <= 1, m, 2*P[m - 1] + P[m - 2]];
  P[n - k] * Binomial[n, k] ];
Table[T[n, k], {n, 1, 9}, {k, 0, n - 1}]  (* Peter Luschny, Mar 07 2025 *)

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 - 2*x - x^2.
p(n, x) = k*(b^n - c^n), where k = sqrt(1/8), b = x + 1 - sqrt(2), c = x + 1 + sqrt(2).
From Werner Schulte, Nov 24 2023 and Nov 25 2023: (Start)
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.
T(n, k) = T(n-1, k-1) * n / k for 0 < k < n and T(n, 0) = A000129(n) for n > 0.
T(n, k) = A000129(n-k) * binomial(n, k) for 0 <= k < n.
G.f.: t / (1 - (2+2*x) * t - (1-2*x-x^2) * t^2). (End)

New name using a formula of Werner Schulte by Peter Luschny, Mar 07 2025

A026937 a(n) = Sum_{k=0..n} (k+1)*T(n, n-k), where T is given by A008288.

Original entry on oeis.org

1, 3, 10, 30, 87, 245, 676, 1836, 4925, 13079, 34446, 90090, 234227, 605865, 1560200, 4002072, 10230201, 26069995, 66251090, 167941494, 424753615, 1072057117, 2700704172, 6791746500, 17052595573, 42752015487, 107035180630, 267634562754, 668407232235, 1667467065425

Offset: 0

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Ricardo Gómez Aíza, Trees with flowers: A catalog of integer partition and integer composition trees with their asymptotic analysis, arXiv:2402.16111 [math.CO], 2024. See p. 23.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-4,-1).

Cf. A000129, A001333, A006645, A008288, A364553.

I:=[1, 3, 10, 30]; [n le 4 select I[n] else 4*Self(n-1)-2*Self(n-2)-4*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 20 2012

with (combinat):seq(add(fibonacci(n,2),k=0..n)/2,n=1..27); # Zerinvary Lajos, May 25 2008

CoefficientList[Series[(1-x)/(1-2x-x^2)^2,{x,0,40}],x]  (* Harvey P. Dale, Mar 22 2011 *)
LinearRecurrence[{4,-2,-4,-1},{1,3,10,30},40] (* Vincenzo Librandi, Jun 20 2012 *)
Table[(1/2)*(n+2)*Fibonacci[n+1, 2], {n, 0, 40}] (* G. C. Greubel, May 25 2021 *)

my(x='x+O('x^40)); Vec((1-x)/(1-2*x-x^2)^2) \\ Altug Alkan, Sep 20 2018

a(n) = my(w=quadgen(8)); (n/8)*((2+w)*(1+w)^n - (w-2)*(1-w)^n); \\ Michel Marcus, Jul 31 2023

[(1/2)*(n+2)*lucas_number1(n+1,2,-1) for n in (0..40)] # G. C. Greubel, May 25 2021

G.f.: (1-x)/(1 - 2*x - x^2)^2.
a(n) = Sum_{k=0..n+1} A000129(k)*A001333(n+1-k). - Graeme McRae, Aug 03 2006 and Michel Marcus, Aug 01 2023
a(n) = A006645(n+2) - A006645(n+1). - R. J. Mathar, Jan 27 2011
a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 20 2012
a(n) = ((n+2)/2)*A000129(n+1). - G. C. Greubel, May 25 2021
a(n) = ((n+2)/8)*((sqrt(2) + 2)*(1 + sqrt(2))^n - (sqrt(2) - 2)*(1 - sqrt(2))^n). - Peter Luschny, Jul 31 2023
a(n) = A361732(n+2)/2. - R. J. Mathar, Jun 30 2025

cp's OEIS Frontend

A367211 Triangular array read by rows: T(n, k) = binomial(n, k) * A000129(n - k) for 0 <= k < n.

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