A023607 -id:A023607 - cp's OEIS frontend

A104796 Triangle read by rows: T(n,k) = (n+1-k)*Fibonacci(n+2-k), for n>=1, 1<=k<=n.

1, 4, 1, 9, 4, 1, 20, 9, 4, 1, 40, 20, 9, 4, 1, 78, 40, 20, 9, 4, 1, 147, 78, 40, 20, 9, 4, 1, 272, 147, 78, 40, 20, 9, 4, 1, 495, 272, 147, 78, 40, 20, 9, 4, 1, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 1584, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 2796, 1584, 890, 495, 272

Offset: 1

Gary W. Adamson, Mar 26 2005

The first column is A023607 (without the leading zero).

			Rows 1,2,3,4,5,6 and columns 1,2,3,4,5,6 of the triangle are:
1;
4, 1;
9, 4, 1;
20, 9, 4, 1;
40, 20, 9, 4, 1;
78, 40, 20, 9, 4, 1;
...
Row 3 for example is 3*F(4), 2*F(3), 1*F(2) = 3*3, 2*2, 1*1 = 9, 4, 1.
Row 4 is 4*F(5), 3*F(4), 2*F(3), 1*F(2) = 4*5, 3*3, 2*2, 1*1 = 20, 9, 4, 1.
Reading the rows backwards gives an initial segment of the terms of A023607 (but without the initial zero).

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Row sums are in A094584.

Cf. A023607, A104762, A104765, A000045.

Table[(n+1-k)Fibonacci[n+2-k],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 24 2020 *)
Module[{nn=20,c},c=LinearRecurrence[{2,1,-2,-1},{1,4,9,20},nn];Table[ Reverse[ Take[c,n]],{n,nn}]]//Flatten (* Harvey P. Dale, Sep 25 2020 *)

Edited by Ralf Stephan, Apr 05 2009

Entry revised by N. J. A. Sloane, Sep 23 2020

A182001 Riordan array ((2*x+1)/(1-x-x^2), x/(1-x-x^2)).

Original entry on oeis.org

1, 3, 1, 4, 4, 1, 7, 9, 5, 1, 11, 20, 15, 6, 1, 18, 40, 40, 22, 7, 1, 29, 78, 95, 68, 30, 8, 1, 47, 147, 213, 185, 105, 39, 9, 1, 76, 272, 455, 466, 320, 152, 49, 10, 1, 123, 495, 940, 1106, 891, 511, 210, 60, 11, 1, 199, 890, 1890, 2512, 2317, 1554, 770, 280, 72, 12, 1

Offset: 0

Philippe Deléham, Apr 05 2012

Subtriangle of the triangle given by (0, 3, -5/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -2/3, 2/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Antidiagonal sums are A001045(n+2).

			Triangle begins :
    1;
    3,   1;
    4,   4,    1;
    7,   9,    5,    1;
   11,  20,   15,    6,    1;
   18,  40,   40,   22,    7,    1;
   29,  78,   95,   68,   30,    8,   1;
   47, 147,  213,  185,  105,   39,   9,   1;
   76, 272,  455,  466,  320,  152,  49,  10, 1;
  123, 495,  940, 1106,  891,  511, 210,  60, 11,  1;
  199, 890, 1890, 2512, 2317, 1554, 770, 280, 72, 12, 1;
(0, 3, -5/3, -1/3, 0, 0, ...) DELTA (1, 0, -2/3, 2/3, 0, 0, ...) begins:
  1;
  0,  1;
  0,  3,  1;
  0,  4,  4,  1;
  0,  7,  9,  5,  1;
  0, 11, 20, 15,  6, 1;
  0, 18, 40, 40, 22, 7, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. Columns : A000032, A023607, A152881

function T(n,k)
  if k lt 0 or k gt n then return 0;
  elif k eq n then return 1;
  elif k eq 0 then return Lucas(n+1);
  else return T(n-1,k) + T(n-1,k-1) + T(n-2,k);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

with(combinat);
T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=n then 1
    elif k=0 then fibonacci(n+2) + fibonacci(n)
    else T(n-1,k) + T(n-1,k-1) + T(n-2,k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Feb 18 2020

With[{m = 10}, CoefficientList[CoefficientList[Series[(1+2*x)/(1-x-y*x-x^2), {x, 0, m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 18 2020 *)
T[n_, k_]:= T[n, k]= If[k<0||k>n, 0, If[k==n, 1, If[k==0, LucasL[n+1], T[n-1, k] + T[n-1, k-1] + T[n-2, k] ]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

G.f.: (1+2*x)/(1-x-y*x-x^2).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = 3, T(n,k) = 0 if k<0 or if k>n.
Sum_{k=0..nn} T(n,k)*x^k = A000034(n), A000032(n+1), A048654(n), A108300(n), A048875(n) for x = -1, 0, 1, 2, 3 respectively.

a(29) corrected by and a(55)-a(65) from Georg Fischer, Feb 18 2020

A093835 n*Jacobsthal(n).

Original entry on oeis.org

0, 1, 2, 9, 20, 55, 126, 301, 680, 1539, 3410, 7513, 16380, 35503, 76454, 163845, 349520, 742747, 1572858, 3320497, 6990500, 14680071, 30758222, 64312669, 134217720, 279620275, 581610146, 1207959561, 2505397580, 5189752159

Offset: 0

Paul Barry, Apr 20 2004

Convolution of Jacobsthal numbers A001045 and modified Jacobsthal-Lucas numbers (in A014551, change the leading 2 to a 1).

Index entries for linear recurrences with constant coefficients, signature (2,3,-4,-4).

Cf. A023607.

LinearRecurrence[{2,3,-4,-4},{0,1,2,9},30] (* Harvey P. Dale, Jun 17 2017 *)

G.f.: x(1+2x^2)/(1-x-2x^2)^2; a(n)=n2^n/3-n(-1)^n/3.

a(n) = n*A001045(n). - R. J. Mathar, May 07 2019

A123194 a(n) = (n+1)*Fibonacci(n+2) + 3.

Original entry on oeis.org

4, 7, 12, 23, 43, 81, 150, 275, 498, 893, 1587, 2799, 4904, 8543, 14808, 25555, 43931, 75261, 128538, 218923, 371934, 630457, 1066467, 1800603, 3034828, 5106871, 8580900, 14398415, 24129163, 40388073, 67527582, 112786499, 188195274, 313733813, 522562323

Offset: 0

N. J. A. Sloane, Oct 04 2006

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Philip Matchett Wood, A bijective proof of f_{n+4}+f_1+2f_2+...+nf_n=(n+1)f_{n+2}+3, Integers 6 (2006), A2, 4 pp.
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,1,1).

Cf. A023607.

[(n+1)*Fibonacci(n+2) + 3: n in [0..40]]; // Vincenzo Librandi, Feb 25 2017

Table[(n + 1) Fibonacci[n+2] + 3, {n, 0, 40}] (* Vincenzo Librandi, Feb 25 2017 *)
LinearRecurrence[{3,-1,-3,1,1},{4,7,12,23,43},40] (* Harvey P. Dale, Jan 12 2018 *)

a(n) = (n+1)*fibonacci(n+2) + 3; \\ Michel Marcus, Feb 25 2017

Vec((4 - 5*x - 5*x^2 + 6*x^3 + 3*x^4)/((1 - x)*(1 - x - x^2)^2) + O(x^50)) \\ Colin Barker, Feb 25 2017

G.f.: (4 - 5*x - 5*x^2 + 6*x^3 + 3*x^4)/((1 - x)*(1 - x - x^2)^2). - Ilya Gutkovskiy, Feb 24 2017
From Colin Barker, Feb 25 2017: (Start)
a(n) = 3 - 2^(-1-n)*((1-sqrt(5))^n*(-5+3*sqrt(5)) - (1+sqrt(5))^n*(5+3*sqrt(5)))/5*(1+n).
a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + a(n-4) + a(n-5) for n>4.
(End)

A215108 a(n) = A215082(2*n).

Original entry on oeis.org

0, 1, 4, 8, 17, 35, 66, 124, 229, 414, 742, 1318, 2320, 4059, 7062, 12224, 21071, 36185, 61930, 105678, 179847, 305326, 517212, 874380, 1475472, 2485573, 4180648, 7021544, 11777117, 19728911, 33011202

Offset: 0

Philippe Deléham, Aug 03 2012

Cf. A000045, A023607

a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) -a(n-6), a(0) = 0, a(1) = 1, a(2) = 4, a(3) = 8, a(4) = 17, a(5) = 35 .

G.f.: x*(x+1)*(2x+1)/((x^2+x+1)*(x^2+x-1)^2)

A179023 a(n) = n(F(n+2) - 1) where F(n) is defined by A000045.

Original entry on oeis.org

0, 1, 4, 12, 28, 60, 120, 231, 432, 792, 1430, 2552, 4512, 7917, 13804, 23940, 41328, 71060, 121752, 207955, 354200, 601776, 1020074, 1725552, 2913408, 4910425, 8263060, 13884156, 23297092, 39041772, 65349240, 109261887, 182492352

Offset: 0

K.V.Iyer, Jun 25 2010

Let the 'Fibonacci weighted stars' T(i)'s be defined as: T(1) is an edge with one vertex as a distinguished vertex; the weight on the edge is taken to be F(1); for n>1, T(n) is formed by taking a copy of T(n-1) and attaching an edge to its distinguished vertex; the weight on the new edge is taken to be F(n). The sum of the weighted distances over all pairs of vertices in T(n) is this sequence.

Index entries for linear recurrences with constant coefficients, signature (4, -4, -2, 4, 0, -1).

f[n_] := n(Fibonacci[n + 2] - 1); Array[f, 33, 0] (* Robert G. Wilson v, Jun 27 2010 *)

a(0)=0, a(1)=1 and for n>1, a(n) = a(n-1) + F(n+1) +nF(n) -1.

a(n)= +4*a(n-1) -4*a(n-2) -2*a(n-3) +4*a(n-4) -a(n-6). = -n + A023607(n+1) - A000045(n+2). G.f.: -x*(-1+2*x^3) / ( (x-1)^2*(x^2+x-1)^2 ). - R. J. Mathar, Sep 15 2010

More terms from Robert G. Wilson v, Jun 27 2010

A181974 Triangle T(n,k), read by rows, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -3, 2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 4, 2, 1, 5, 7, 5, 4, 1, 8, 11, 10, 9, 3, 1, 13, 18, 20, 20, 9, 5, 1, 21, 29, 38, 40, 22, 15, 4, 1, 34, 47, 71, 78, 51, 40, 14, 6, 1, 55, 76, 130, 147, 111, 95, 40, 22, 5, 1, 89, 123, 235, 272, 233, 213, 105, 68, 20, 7, 1

Offset: 0

Philippe Deléham, Apr 06 2012

			Triangle begins :
1
1, 1
2, 3, 1
3, 4, 2, 1
5, 7, 5, 4, 1
8, 11, 10, 9, 3, 1
13, 18, 20, 20, 9, 5, 1
21, 29, 38, 40, 22, 15, 4, 1
34, 47, 71, 78, 51, 40, 14, 6, 1
55, 76, 130, 147, 111, 95, 40, 22, 5, 1
89, 123, 235, 272, 233, 213, 105, 68, 20, 7, 1
144, 199, 420, 495, 474, 455, 256, 185, 65, 30, 6, 1

Cf. Columns : A000045, A000032, A001629, A023607, A001628, A152881, A001872, A001873

G.f.: (1+y*x+2*y*x^2)/(1-x-x^2-y^2*x^2).
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-2), T(0,0) = T(1,0) = T(1,1) = T(2,2) = 1, T(2,0) = 2, T(2,1) = 3 and T(n,k) = 0 if k<0 or if k>n.
T(n + 2k, 2k) = A037027(n + k, k).
T(n + 2k +1, 2k + 1) = A182001(n + k, k).
T(n,0) = Fibonacci(n+1).

A215109 a(n) = A215082(2*n+1).

Original entry on oeis.org

1, 3, 5, 12, 23, 43, 81, 148, 266, 476, 842, 1478, 2581, 4481, 7743, 13328, 22857, 39073, 66605, 113242, 192084, 325128, 549252, 926220, 1559353, 2621295, 4400249, 7376868, 12352043, 20659159, 34516377

Offset: 0

Philippe Deléham, Aug 03 2012

Index entries for linear recurrences with constant coefficients, signature (1,2,1,-2,-3,-1).

Cf. A000045, A023607, A215082, A215108.

a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) - a(n-6), a(0) = 1, a(1) = 3, a(2) = 5, a(3) = 12, a(4) = 23, a(5) = 43.

G.f.: (2*x+1)/((x^2+x+1)*(x^2+x-1)^2).

A378943 Numbers obtained from the tribonacci triangle formed by the number of connection points in the paths obtained by Pell walk on the square grid.

Original entry on oeis.org

2, 3, 7, 13, 25, 46, 86, 158, 292, 537, 989, 1819, 3347, 6156, 11324, 20828, 38310, 70463, 129603, 238377, 438445, 806426, 1483250, 2728122, 5017800, 9229173, 16975097, 31222071, 57426343, 105623512, 194271928, 357321784, 657217226, 1208810939, 2223349951, 4089378117

Offset: 1

Irem Eski, Arzu Caglar, Cansu Aytar, and Emre Kaya, Dec 11 2024

The study was inspired by the stepping system discussed in the study titled Pell Walks by Thomas Koshy (2014). In the study, the points formed in this walk are called connected points. There is a direct connection between the Pell-Lucas number sequence and the tribonacci number sequence.

The connection points obtained from Thomas Koshy (2014) were examined on the Pascal triangle. The intermediate sequences of this A number sequence were examined. The counting numbers were extracted from the sequence. The resulting new number sequence is 1,1,4,9,20,40,79,150,283,523... This is called the D number sequence. It is similar to A023607, which is a convolution of Fibonacci and Lucas numbers.

			For n = 4 a(4) = a(3) + a(2) + a(1) + 1 = 2 + 3 + 7 + 1 = 13.
For n = 5 a(5) = a(4) + a(3) + a(2) + 2 = 3 + 7 + 13 + 2 = 25.

T. Koshy, Pell and Pell-Lucas Numbers with Applications, Springer Science-Business Media New York, 2014, 227-253.

E. Namlı, Tribonacci ve lucas sayılarının üreteç fonksiyonları yardımıyla oluşan bazı özdeşlikler, Master's thesis, Bursa Uludag University (Turkey), 2020, 2-16.
Index entries for linear recurrences with constant coefficients, signature (1,2,0,-1,-1).

Cf. A000045, A000073, A001333, A000129.

LinearRecurrence[{1, 2, 0, -1, -1}, {2, 3, 7, 13, 25}, 36] (* Hugo Pfoertner, Jan 09 2025 *)

def a(n, memo={}):
    if n == 1:
        return 2
    elif n == 2:
        return 3
    elif n == 3:
        return 7
    if n in memo:
        return memo[n]
    if n % 2 == 1:
        memo[n] = a(n - 3, memo) + a(n - 2, memo) + a(n - 1, memo) + 2
    else:
        memo[n] = a(n - 3, memo) + a(n - 2, memo) + a(n - 1, memo) + 1
    return memo[n]
sequence = [a(n) for n in range(1, n)]
print(sequence)

a(1)=2, a(2)=3, a(3)=7, and a(n+3) = a(n)+a(n+1)+a(n+2)+2 if n is odd, a(n+3) = a(n)+a(n+1)+ a(n+2)+1 if n is even.

Edited - N. J. A. Sloane, Jan 23 2025

cp's OEIS Frontend

A104796 Triangle read by rows: T(n,k) = (n+1-k)*Fibonacci(n+2-k), for n>=1, 1<=k<=n.

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A182001 Riordan array ((2*x+1)/(1-x-x^2), x/(1-x-x^2)).

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A093835 n*Jacobsthal(n).

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A123194 a(n) = (n+1)*Fibonacci(n+2) + 3.

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A215108 a(n) = A215082(2*n).

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A179023 a(n) = n(F(n+2) - 1) where F(n) is defined by A000045.

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A181974 Triangle T(n,k), read by rows, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -3, 2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

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A215109 a(n) = A215082(2*n+1).

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A378943 Numbers obtained from the tribonacci triangle formed by the number of connection points in the paths obtained by Pell walk on the square grid.

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