A038730 -id:A038730 - cp's OEIS frontend

A038732 T(n,n-3), array T as in A038730.

1, 8, 38, 141, 455, 1350, 3805, 10395, 27875, 73945, 194961, 512303, 1343768, 3521381, 9223435, 24152800, 63239810, 165572615, 433485350, 1134892290, 2971202146, 7778726798, 20364993198, 53316270346, 139583838315

Offset: 3

Clark Kimberling, May 02 2000

nonn

First differences are essentially in A038739. - R. J. Mathar, May 23 2008

G.f.: x^3/[(1-3x+x^2)(1-x)^5].

A038733 T(n,n-4), array T as in A038730.

Original entry on oeis.org

1, 10, 57, 245, 888, 2881, 8679, 24872, 68940, 186953, 499927, 1325204, 3494249, 9184675, 24098536, 63165197, 165471668, 433350754, 1134715190, 2970971916, 7778430788, 20364616458, 53315795326, 139583244540

Offset: 4

Clark Kimberling, May 02 2000

nonn

G.f.: x^4/[(1-3x+x^2)(1-x)^7].

A038734 T(n,n-5), array T as in A038730.

Original entry on oeis.org

1, 12, 80, 393, 1594, 5676, 18437, 56070, 162643, 456169, 1249622, 3368279, 8981185, 23778766, 62674883, 164736197, 432269179, 1133152915, 2968751841, 7775322683, 20360324313, 53309942401, 139575355815, 365424013769

Offset: 5

Clark Kimberling, May 02 2000

nonn

G.f.: x^5/[(1-3x+x^2)(1-x)^9].

A038735 T(n,n-6), array T as in A038730.

Original entry on oeis.org

1, 14, 107, 593, 2673, 10429, 36622, 118885, 363791, 1064866, 3015563, 8334539, 22634700, 60713627, 161467437, 426957444, 1124716630, 2955628731, 7755292673, 20330279298, 53265590236, 139510843575, 365331452729

Offset: 6

Clark Kimberling, May 02 2000

nonn

A026532 Ratios of successive terms are 3, 2, 3, 2, 3, 2, 3, 2, ...

Original entry on oeis.org

1, 3, 6, 18, 36, 108, 216, 648, 1296, 3888, 7776, 23328, 46656, 139968, 279936, 839808, 1679616, 5038848, 10077696, 30233088, 60466176, 181398528, 362797056, 1088391168, 2176782336, 6530347008, 13060694016, 39182082048, 78364164096, 235092492288, 470184984576

Offset: 1

Clark Kimberling

Preface the series with a 1: (1, 1, 3, 6, 18, 36, ...); then the next term in the series = (1, 1, 3, 6, ...) dot (1, 2, 1, 2, ...). Example: 36 = (1, 1, 3, 6, 18) dot (1, 2, 1, 2, 1) = (1 + 2 + 3 + 12 + 18). - Gary W. Adamson, Apr 18 2009

Partial products of A176059. - Reinhard Zumkeller, Apr 04 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..700
Sean A. Irvine, Walks on Graphs.
José L. Ramírez, Bi-periodic incomplete Fibonacci sequences, Annales Mathematicae et Informaticae 42 (2013), 83-92. See the 1st column of Table 1 on p. 85.
Index entries for linear recurrences with constant coefficients, signature (0,6).

Cf. A026519, A026520, A026521, A026522, A026523, A026524, A026525, A026526, A026527, A026528, A026529, A026530, A026531, A026533, A026534, A027262, A027263, A027264, A027265, A027266.
Cf. A026534, A026549, A176059, A208131.
Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

a026532 n = a026532_list !! (n-1)
a026532_list = scanl (*) 1 $ a176059_list
-- Reinhard Zumkeller, Apr 04 2012

[(1/4)*(3-(-1)^n)*6^Floor(n/2) : n in [1..30]]; // Vincenzo Librandi, Jun 08 2011

FoldList[(2 + Boole[EvenQ@ #2]) #1 &, Range@ 28] (* or *)
CoefficientList[Series[x*(1+3x)/(1-6x^2), {x,0,31}], x] (* Michael De Vlieger, Aug 02 2017 *)
LinearRecurrence[{0,6},{1,3},30] (* Harvey P. Dale, Jul 11 2018 *)

a(n)=if(n%2,3,1)*6^(n\2) \\ Charles R Greathouse IV, Jul 02 2013

def a(n): return (3 if n%2 else 1)*6**(n//2)
print([a(n) for n in range(31)]) # Indranil Ghosh, Aug 02 2017

[(1/2)*6^((n-2)/2)*(3*(1+(-1)^n) + sqrt(6)*(1-(-1)^n)) for n in (1..30)] # G. C. Greubel, Dec 21 2021

a(n) = T(n, 0) + T(n, 1) + ... + T(n, 2n-2), T given by A026519.
From Benoit Cloitre, Nov 14 2003: (Start)
a(n) = (1/2)*(5+(-1)^n)*a(n-1) for n>1, a(1) = 1.
a(n) = (1/4)*(3-(-1)^n)*6^floor(n/2).  (End)
From Ralf Stephan, Feb 03 2004: (Start)
G.f.: x*(1+3*x)/(1-6*x^2).
a(n+2) = 6*a(n). (End)
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = (1/2)*6^((n-2)/2)*(3*(1+(-1)^n) + sqrt(6)*(1-(-1)^n)). - G. C. Greubel, Dec 21 2021
Sum_{n>=1} 1/a(n) = 8/5. - Amiram Eldar, Feb 13 2023

New definition from Ralf Stephan, Dec 01 2004

Offset changed from 0 to 1 by Vincenzo Librandi, Jun 08 2011

A038792 Rectangular array defined by T(i,1) = T(1,j) = 1 for i >= 1 and j >= 1; T(i,j) = max(T(i-1,j) + T(i-1,j-1), T(i-1,j-1) + T(i,j-1)) for i >= 2, j >= 2, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 8, 8, 5, 1, 1, 6, 12, 13, 12, 6, 1, 1, 7, 17, 21, 21, 17, 7, 1, 1, 8, 23, 33, 34, 33, 23, 8, 1, 1, 9, 30, 50, 55, 55, 50, 30, 9, 1, 1, 10, 38, 73, 88, 89, 88, 73, 38, 10, 1, 1, 11, 47, 103, 138, 144, 144, 138, 103, 47, 11, 1

Offset: 1

Clark Kimberling, May 02 2000

Antidiagonal sums: A029907.
Main diagonal: A001519 (odd-indexed Fibonacci numbers).
Next diagonal: A001906 (even-indexed Fibonacci numbers).

			From _Clark Kimberling_, Jun 20 2011: (Start)
Northwest corner begins at (i,j) = (1,1):
  1,   1,   1,   1,   1,   1,   1,   1, ...
  1,   2,   3,   4,   5,   6,   7,   8, ...
  1,   3,   5,   8,  12,  17,  23,  30, ...
  1,   4,   8,  13,  21,  33,  50,  73, ...
  1,   5,  12,  21,  34,  55,  88, 138, ...
  1,   6,  17,  33,  55,  89, 144, 232, ...
  1,   7,  23,  50,  88, 144, 233, 377, ...
(End)
Antidiagonal triangle begins as:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  3,  1;
  1, 4,  5,  4,  1;
  1, 5,  8,  8,  5,  1;
  1, 6, 12, 13, 12,  6,  1;
  1, 7, 17, 21, 21, 17,  7,  1;
  1, 8, 23, 33, 34, 33, 23,  8, 1;
  1, 9, 30, 50, 55, 55, 50, 30, 9, 1;

G. C. Greubel, Antidiagonals n = 1..50, flattened
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 that contains the incomplete Fibonacci numbers.
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002), 328-338; see Example 4 (appears as a triangular array).
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A001519, A001906, A029907.
Cf. A000045, A038730, A134511, A324242.
Main diagonal gives A001519.

function t(n,k)
  if k eq 0 or n eq 0 then return 1;
  else return Max(t(n-1,k-1) + t(n-1,k), t(n-1,k-1) + t(n,k-1));
  end if; return t;
end function;
T:= func< n,k | t(n-k, k-1) >;
[T(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 05 2022

G := x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)); G := convert(series(G, x=0, 11),polynom):
for i from 1 to 10 do series(coeff(G,x,i),y=0,11) od; # Mark van Hoeij, Nov 09 2011
# second Maple program:
G:= x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)):
T:= (i, j)-> coeff(series(coeff(series(G, y, j+1), y, j), x, i+1), x, i):
seq(seq(T(i, 1+d-i), i=1..d), d=1..12); # Alois P. Heinz, Sep 02 2019
# third Maple program:
T:= proc(i,j) option remember; `if`(i=1 or j=1, 1,
      max(T(i-1,j) + T(i-1,j-1), T(i-1,j-1) + T(i,j-1)))
    end:
seq(seq(T(i, 1+d-i), i=1..d), d=1..12); # Alois P. Heinz, Sep 02 2019

f[i_, 0]:= 1; f[0, i_]:= 1
f[i_, j_]:= f[i,j]= Max[f[i-1,j] +f[i-1,j-1], f[i-1,j-1] +f[i,j-1]];
T[i_, j_]:= f[i-j, j-1];
TableForm[Table[f[i, j], {i,0,7}, {j,0,7}]]
Table[T[i, j], {i,10}, {j,i}]//Flatten (* modified by G. C. Greubel, Apr 05 2022 *)

def t(n,k):
    if (k==0 or n==0): return 1
    else: return max(t(n-1,k-1) + t(n-1,k), t(n-1,k-1) + t(n,k-1))
def A038792(n,k): return t(n-k, k-1)
flatten([[A038792(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 05 2022

G.f.: x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)). - Mark van Hoeij, Nov 09 2011
From Petros Hadjicostas, Sep 02 2019: (Start)
Following Dil and Mezo (2008), define the incomplete Fibonacci numbers by F(n,k) = Sum_{s = 0..k} binomial(n-1-s, s) for n >= 1 and 0 <= k <= floor((n-1)/2).
Then T(i, j) = F(i+j-1, min(i-1, j-1)) for i,j >= 1.
(End)

New description from Benoit Cloitre, Aug 05 2003

Updated from pre-2003 triangular format to present rectangular, from Clark Kimberling, Jun 20 2011

A134511 abs(A049310) * A128174 provided both arrays are read with offset (n,k) = (0,0).

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 3, 0, 1, 5, 0, 4, 0, 1, 0, 8, 0, 5, 0, 1, 13, 0, 12, 0, 6, 0, 1, 0, 21, 0, 17, 0, 7, 0, 1, 34, 0, 33, 0, 23, 0, 8, 0, 1, 0, 55, 0, 50, 0, 30, 0, 9, 0, 1, 89, 0, 88, 0, 73, 0, 38, 0, 10, 0, 1, 0, 144, 0, 138, 0, 103, 0, 47, 0, 11, 0, 1, 233, 0, 232, 0, 211, 0, 141, 0, 57, 0, 12, 0, 1

Offset: 0

Gary W. Adamson, Oct 28 2007

A112552(unsigned) = A128174 * A049310.
Row sums = A134512: (1, 1, 3, 4, 10, 14, 32, 46, 99, 145, ...).
From Petros Hadjicostas, Sep 03 2019: (Start)
To prove Alois P. Heinz's claim (see the Formula section and his Maple program below) we note that, for n >= 0 and 0 <= k <= n, T(n, n-k) = Sum_{r = 0 .. infinity} abs(A049310(n,r)) * A128174(r,n-k) = Sum_{r = n-k..n} abs(A049310(n,r)) * A128174(r,n-k). But A049310(n,r) = 0 when n + r is odd and A128174(r,n-k) = 1 iff r + n - k is even. Thus, when k is odd, T(n, n-k) = 0.
Assume now k is even. Then T(n, n-k) = Sum_{r = n-k..n and n+r even} abs(A049310(n,r)) =  Sum_{r = n-k..n and n+r even} binomial((n+r)/2, r). Letting m = n-r (which is even), we see that the summation ranges from m = 0 to k over even numbers. Thus, let s = m/2, and so T(n, n-k) = Sum_{s = 0 .. k/2} binomial(n-s, n-2*s) = Sum_{s = 0 .. k/2} binomial(n-s, s) = F(n+1, k/2), where F(.,.) is the incomplete Fibonacci number from the references (see also the Formula section below).
(End)

			First few rows of the triangle T(n,k):
   1;
   0,  1;
   2,  0,  1;
   0,  3,  0,  1;
   5,  0,  4,  0,  1;
   0,  8,  0,  5,  0,  1;
  13,  0, 12,  0,  6,  0,  1;
   0, 21,  0, 17,  0,  7,  0,  1;
  34,  0, 33,  0, 23,  0,  8,  0,  1;
   0, 55,  0, 50,  0, 30,  0,  9,  0,  1;
  ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 141, flattened)
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 (p. 39) that contains the incomplete Fibonacci numbers.
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A038730, A038792, A049310, A128174, A134512.

A(4n,2n) gives: A038736.

N:= 20: # for the first N rows
T128174:= Matrix(N,N,(i,j) -> `if`(j<=i, (i-j+1) mod 2, 0)):
T049310:= Matrix(N,N):
for i from 1 to N do
     P:= orthopoly[U](i-1,x/2);
     for j from 1 to i do
       T049310[i,j]:= abs(coeff(P,x,j-1))
     od
od:
A:= T049310 . T128174:
for i from 1 to N do
convert(A[i,1..i],list)
od;  # Robert Israel, Mar 02 2018
# second Maple program:
T:= (n, k)-> `if`((n+k)::odd, 0, add(binomial(n-s, s), s=0..(n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..12); # Alois P. Heinz, Sep 02 2019

T[n_, k_] := If[OddQ[n+k], 0, Sum[Binomial[n-s, s], {s, 0, (n-k)/2}]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Dec 31 2021, after Alois P. Heinz *)

abs(A049310) * A128174 as infinite lower triangular matrices assuming both of them have offset (n,k) = (0,0).
From Petros Hadjicostas, Sep 03 2019: (Start)
Let F(m,r) = Sum_{j = 0..r} binomial(m-1-j, j) be the incomplete Fibonacci numbers from the references (defined for m >= 1 and 0 <= r <= floor((m-1)/2)).
As Alois P. Heinz observed, for n >= 0 and 0 <= k <= n, T(n, n-k) = F(n+1, k/2) when k is even, and = 0 otherwise (see his Maple program below).
(End)

Edited by Robert Israel, Mar 02 2018

A038731 Number of columns in all directed column-convex polyominoes of area n+1.

Original entry on oeis.org

1, 3, 10, 32, 99, 299, 887, 2595, 7508, 21526, 61251, 173173, 486925, 1362627, 3797374, 10543724, 29180067, 80521055, 221610563, 608468451, 1667040776, 4558234018, 12441155715, 33900136297, 92230468249, 250570010499, 679844574322, 1842280003640

Offset: 0

Clark Kimberling, May 02 2000

Apply Riordan array (1/(1-x), x/(1-x)^2) to n+1. - Paul Barry, Oct 13 2009

Binomial transform of (A001629 shifted left twice). - R. J. Mathar, Feb 06 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Elena Barcucci, Renzo Pinzani, and Renzo Sprugnoli , Directed column-convex polyominoes by recurrence relations, Lecture Notes in Computer Science, No. 668, Springer, Berlin (1993), pp. 282-298.
Éva Czabarka, Rigoberto Flórez, Leandro Junes, and José L. Ramírez, Enumeration of peaks and valleys on non-decreasing Dyck paths, Disc. Math. 341(10) (2018), 2789-2807; see Theorem 2 on p. 2791.
Rigoberto Flórez, Leandro Junes, Luisa M. Montoya, and José L. Ramírez, Counting Subwords in Non-Decreasing Dyck Paths, Journal of Integer Sequences, Vol. 28 (2025), Article 25.1.6. See pp. 5, 15, 17, 19.
Jesus Salas and Alan D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys. 135 (2009) 279-373, arXiv:0711.1738. Mentions this sequence. - _N. J. A. Sloane_, Mar 14 2014
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Row-sums of array T as in A038730.
First differences of A030267.
Row sums of A318942(n+1).
Cf. A000045.

a038731 n = a038731_list !! n
a038731_list = c [1] $ tail a000045_list where
   c us vs'@(v:vs) = (sum $ zipWith (*) us vs') : c (v:us) vs
-- Reinhard Zumkeller, Oct 31 2013

I:=[1, 3, 10, 32]; [n le 4 select I[n] else 6*Self(n-1)-11*Self(n-2)+6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Feb 04 2012

Table[Sum[Binomial[n, k]*CoefficientList[Series[1/(1 - x - x^2)^2, {x, 0, k}], x][[-1]], {k, 0, n}], {n, 0, 27}] (* Arkadiusz Wesolowski, Feb 03 2012 *)
LinearRecurrence[{6, -11, 6, -1}, {1, 3, 10, 32}, 30] (* Vincenzo Librandi, Feb 04 2012 *)

5*a(n) = (2n+1)*F(2n+2) - (n-4)*F(2n+1), where the F(n)'s are the Fibonacci numbers, F(0)=0, F(1)=1.
a(n) = Sum_{k=1..n+1} k*binomial(n+k-1, 2k-2). - Emeric Deutsch, Jun 11 2003
From Paul Barry, Oct 13 2009: (Start)
G.f.: (1-x)^3/(1-3x+x^2)^2.
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*(k+1). (End)
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - R. J. Mathar, Feb 06 2010
a(n) = Sum_{k=0..n} (F(2k)+0^k)*F(2n-2k+1). - Paul Barry, Jun 23 2010
E.g.f.: exp(3*x/2)*(5*(5 + 4*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(7 + 10*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Entry improved by comments from Emeric Deutsch, Jun 14 2001

A137176 Hyperlucas number array T(r,n) = L(n)^(r), read by ascending antidiagonals (r >= 0, n >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 4, 4, 0, 1, 5, 8, 7, 0, 1, 6, 13, 15, 11, 0, 1, 7, 19, 28, 26, 18, 0, 1, 8, 26, 47, 54, 44, 29, 0, 1, 9, 34, 73, 101, 98, 73, 47, 0, 1, 10, 43, 107, 174, 199, 171, 120, 76, 0, 1, 11, 53, 150, 281, 373, 370, 291, 196, 123

Offset: 0

Jonathan Vos Post, Apr 04 2008

In Theorem 17, Dil and Mezo (2008) connect the hyperlucas numbers (this array) with the incomplete Lucas numbers (A324242). - Petros Hadjicostas, Sep 03 2019

			The array T(r,n) = L(n)^(r) begins:
.....|n=0|n=1|.n=2|.n=3|.n=4.|.n=5.|..n=6.|.n=7..|..n=8..|..n=9..|.n=10..|.in.OEIS
r=0..|.0.|.1.|..3.|..4.|...7.|..11.|...18.|...29.|....47.|....76.|...123.|.A000204
r=1..|.0.|.1.|..4.|..8.|..15.|..26.|...44.|...73.|...120.|...196.|...319.|.A027961
r=2..|.0.|.1.|..5.|.13.|..28.|..54.|...98.|..171.|...291.|...487.|...806.|.A023537
r=3..|.0.|.1.|..6.|.19.|..47.|.101.|..199.|..370.|...661.|..1148.|..1954.|.A027963
r=4..|.0.|.1.|..7.|.26.|..73.|.174.|..373.|..743.|..1404.|..2552.|..4506.|.A027964
r=5..|.0.|.1.|..8.|.34.|.107.|.281.|..654.|.1397.|..2801.|..5353.|..9859.|.A053298
r=6..|.0.|.1.|..9.|.43.|.150.|.431.|.1085.|.2482.|..5283.|.10636.|.20495.|.new
r=7..|.0.|.1.|.10.|.53.|.203.|.634.|.1719.|.4201.|..9484.|.20120.|.40615.|.new
r=8..|.0.|.1.|.11.|.64.|.267.|.901.|.2620.|.6821.|.16305.|.36425.|.77040.|.new
r=9..|.0.|.1.|.12.|.76.|.343.|1244.|.3864.|10685.|.26990.|.63415.|140455.|.new
For example, T(4,5) = L(5)^(4) = L(0)^(3) + L(1)^(3) + L(2)^(3) + L(3)^(3) + L(4)^(3) + L(5)^(3) = 0 + 1 + 6 + 19 + 47 + 101 = 174. - _Petros Hadjicostas_, Sep 03 2019

Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, arXiv:0803.4388 [math.NT], 2008.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, Applied Mathematics and Computation 206(2) (2008), 942-951.
Eric W. Weisstein, Steenrod Algebra.

Cf. A000204, A027961, A023537, A027963, A027964, A053298, A123736, A136431.

Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

L:= proc(r, n) option remember; `if`(n=0, 0, `if`(r=0,
      `if`(n<3, 2*n-1, L(0, n-2)+L(0, n-1)), L(r-1, n)+L(r, n-1)))
    end:
seq(seq(L(d-n, n), n=0..d), d=0..12);  # Alois P. Heinz, Sep 03 2019

L[r_, n_] := L[r, n] = If[n == 0, 0, If[r == 0, If[n < 3, 2n-1, L[0, n-2] + L[0, n-1]], L[r-1, n] + L[r, n-1]]];
Table[L[d-n, n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

T(r,n) = L(n)^(r) = Apply partial sum operator r times to Lucas numbers A000204.
From Petros Hadjicostas, Sep 03 2019: (Start)
T(r, n) = L(n)^(r) = Sum_{k = 0..n} L(k)^(r-1) for r >= 1, with T(0,n) = L(n)^(0) = L(n) = A000204(n), T(r,0) = L(0)^(r) = 0, and T(r,1) = L(1)^(r) = 1. (See Definition 13 in Dil and Mezo (2008).)
G.f. for row r: Sum_{n >= 0} L(n)^(r)*t^n = t * (1+2*t)/((1-t-t^2) * (1-t)^r). (Corrected from Proposition 14 in Dil and Mezo (2008).)
(End)

A324242 Incomplete Lucas numbers: irregular triangular array L(n,k) = Sum_{j = 0..k} (n/(n-j)) * binomial(n-j, j), read by rows, with n >= 1 and 0 <= k <= floor(n/2).

Original entry on oeis.org

1, 1, 3, 1, 4, 1, 5, 7, 1, 6, 11, 1, 7, 16, 18, 1, 8, 22, 29, 1, 9, 29, 45, 47, 1, 10, 37, 67, 76, 1, 11, 46, 96, 121, 123, 1, 12, 56, 133, 188, 199, 1, 13, 67, 179, 284, 320, 322, 1, 14, 79, 235, 417, 508, 521, 1, 15, 92, 302, 596, 792, 841, 843, 1, 16, 106, 381, 831, 1209, 1349, 1364

Offset: 1

Petros Hadjicostas, Sep 02 2019

For additional properties of the incomplete Lucas numbers and special cases not listed here, see Filipponi (1996, pp. 45-53).

			Triangle L(n,k) (with rows n >= 1 and columns k >= 0) begins as follows:
  1;
  1,  3;
  1,  4;
  1,  5,  7;
  1,  6, 11;
  1,  7, 16,  18;
  1,  8, 22,  29;
  1,  9, 29,  45,  47;
  1, 10, 37,  67,  76;
  1, 11, 46,  96, 121, 123;
  1, 12, 56, 133, 188, 199;
  ...
Row sums are 1, 4, 5, 13, 18, 42, 60, 131, 191, 398, 589, 1186, 1775, 3482, 5257, 10103, 15360, ...

A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Lucas numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 2 (p. 46) that contains the incomplete Lucas numbers.
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A038730, A038792, and A134511 for various versions of the incomplete Fibonacci numbers.

Cf. A000032, A000045, A000124, A000204, A137176.

Flatten[Table[Sum[(n/(n-j))*Binomial[n-j, j],{j,0,k}],{n,1,15},{k,0,Floor[n/2]}]] (* Stefano Spezia, Sep 03 2019 *)

L(n,k) = F(n-1, k-1) + F(n+1, k) for n >= 1 and 0 <= k <= floor(n/2), where F(n,k) = Sum_{j = 0..k} binomial(n-1-j, j) are the incomplete Fibonacci numbers (defined for n >= 1 and 0 <= k <= floor((n-1)/2)).
L(n+2, k+1) = L(n+1, k+1) + L(n,k) for n >= 1 and 0 <= k <= floor((n-1)/2).
L(n,k) = F(n+2,k) - F(n-2, k-2) for n >= 3 and 2 <= k <= floor((n+1)/2).
Special cases: L(n,0) = 1 (n >= 1), L(n,1) = n+1 (n >= 2), L(n,2) = (n^2-n+2)/2 = A000124(n-1) (n >= 4), and L(n, floor(n/2)) = A000204(n) (n >= 1).
Sum of row n = (3 + (-1)^n)*A000204(n)/4 + n*A000045(n)/2.
G.f. for column k >= 1: t^(2*k)*((A000204(2*k) + t*A000204(2*k-1))*(1-t)^(k+1) - t^2*(2-t))/((1-t)^(k+1) * (1-t-t^2)).

cp's OEIS Frontend

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