A014551 -id:A014551 - cp's OEIS frontend

A274074 a(n) = 6^n+(-1)^n.

2, 5, 37, 215, 1297, 7775, 46657, 279935, 1679617, 10077695, 60466177, 362797055, 2176782337, 13060694015, 78364164097, 470184984575, 2821109907457, 16926659444735, 101559956668417, 609359740010495, 3656158440062977, 21936950640377855, 131621703842267137

Offset: 0

Colin Barker, Jun 09 2016

Michael De Vlieger, Table of n, a(n) for n = 0..1285
Weerayuth Nilsrakoo and Achariya Nilsrakoo, On One-Parameter Generalization of Jacobsthal Numbers, WSEAS Trans. Math. (2025) Vol. 24, 51-61. See p. 3.
Index entries for linear recurrences with constant coefficients, signature (5,6).

Sequences of the type k^n+(-1)^n: A014551 (k=2), A102345 (k=3), A201455 (k=4), A087404 (k=5), this sequence (k=6).

Array[6^# + (-1)^# &, 23, 0] (* or *)
LinearRecurrence[{5, 6}, {2, 5}, 23] (* or *)
CoefficientList[ Series[(5x -2)/(6x^2 + 5x -1), {x, 0, 23}], x] (* Robert G. Wilson v, Jan 01 2017 *)

PARI
```
Vec((2-5*x)/((1+x)*(1-6*x)) + O(x^30))
```

O.g.f.: (2-5*x) / ((1+x)*(1-6*x)).
E.g.f.: exp(-x) + exp(6*x).
a(n) = 5*a(n-1)+6*a(n-2) for n>1.

A284129 Hosoya triangle Jacobsthal Lucas type.

Original entry on oeis.org

1, 5, 5, 7, 25, 7, 17, 35, 35, 17, 31, 85, 49, 85, 31, 65, 155, 119, 119, 155, 65, 127, 325, 217, 289, 217, 325, 127, 257, 635, 455, 527, 527, 455, 635, 257, 511, 1285, 889, 1105, 961, 1105, 889, 1285, 511, 1025, 2555, 1799, 2159, 2015, 2015, 2159, 1799, 2555, 1025, 2047, 5125

Offset: 1

Rigoberto Florez, Mar 20 2017

			Triangle begins:
    1,
    5,    5,
    7,   25,    7,
   17,   35,   35,   17,
   31,   85,   49,   85,   31,
   65,  155,  119,  119,  155,   65,
  127,  325,  217,  289,  217,  325,  127,
  257,  635,  455,  527,  527,  455,  635,  257,
  511, 1285,  889, 1105,  961, 1105,  889, 1285,  511,
  ...

Indranil Ghosh, Rows 1..100, flattened
Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Matrices in the Hosoya triangle, arXiv:1808.05278 [math.CO], 2018.
H. Hosoya, Fibonacci Triangle, The Fibonacci Quarterly, 14;2, 1976, pages 173-178.
Rigoberto Florez, R. Higuita and L. Junes, GCD property of the generalized star of David in the generalized Hosoya triangle, J. Integer Seq., 17 (2014), Article 14.3.6, 17 pp.
Rigoberto Florez and L. Junes, GCD properties in Hosoya's triangle, Fibonacci Quart. 50 (2012), pages 163-174.
Wikipedia, Hosoya's triangle.

Cf. A014551.

a[n_]:= 2^n + (-1)^n; Table[a[k] a[n - k + 1], {n, 10}, {k, n}] // Flatten (* Indranil Ghosh, Mar 30 2017 *)

a(n) = 2^n + (-1)^n;
for(n=1, 10, for(k=1, n, print1(a(k)*a(n - k + 1),", ");); print();); \\ Indranil Ghosh, Mar 30 2017

def a(n): return 2**n + (-1)**n
for n in range(1, 11):
    print([a(k) * a(n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 30 2017

T(n,k) = A014551(k)*A014551(n - k + 1), where n > 0 and 0 < k <= n.

A307688 a(n) = 2a(n-1)-2a(n-2)+a(n-3)+2*a(n-4) with a(0)=a(1)=0, a(2)=2, a(3)=3.

Original entry on oeis.org

0, 0, 2, 3, 2, 0, 3, 14, 26, 27, 22, 44, 123, 234, 310, 363, 586, 1224, 2259, 3382, 4642, 7227, 13070, 23092, 36555, 54450, 85022, 143883, 245282, 396720, 616803, 973214, 1600106, 2664027, 4334662, 6887804, 10970523, 17828154, 29272390, 47634603, 76493626

Offset: 0

Jean-François Alcover and Paul Curtz, Apr 22 2019

This is an autosequence of the second kind, the companion to A192395.
The array D(n, k) of successive differences begins:
0,   0,   2,   3,   2,   0,   3,  14,  26,  27, ...
0,   2,   1,  -1,  -2,   3,  11,  12,   1,  -5, ...
2,  -1,  -2,  -1,   5,   8,   1, -11,  -6,  27, ...
-3, -1,   1,   6,   3,  -7, -12,   5,  33,  30, ...
2,   2,   5,  -3, -10,  -5,  17,  28,  -3, -55, ...
0,   3,  -8,  -7,   5,  22,  11, -31, -52,  13, ...
...
The main diagonal (0,2,-2,6,-10,22,...) is essentially the same as A151575.
It can be seen that abs(D(n, 1)) = D(1, n).
The diagonal starting from the third 0 is -(-1)^n*11*A001045(n), inverse binomial transform of 11*A001045(n).

Colin Barker, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (2,-2,1,2).

Cf. A151575, A192395.

Cf. A001045 (first and fifth upper diagonals), A014551 (second upper diagonal), A115102 (third), A155980 (fourth).

a[0] = a[1] = 0; a[2] = 2; a[3] = 3; a[n_] := a[n] = 2*a[n-1] - 2*a[n-2] + a[n-3] + 2*a[n-4]; Table[a[n], {n, 0, 40}]
LinearRecurrence[{2,-2,1,2},{0,0,2,3},50] (* Harvey P. Dale, Oct 01 2021 *)

concat([0,0], Vec(x^2*(2 - x) / ((1 - x - x^2)*(1 - x + 2*x^2)) + O(x^40))) \\ Colin Barker, Apr 22 2019

G.f.: x^2*(2 - x) / ((1 - x - x^2)*(1 - x + 2*x^2)). - Colin Barker, Apr 22 2019

A321373 Array T(n,k) read by antidiagonals where the first row is (-1)^k*A140966(k) and each subsequent row is obtained by adding A001045(k) to the preceding one.

Original entry on oeis.org

2, 2, -1, 2, 0, 3, 2, 1, 4, 1, 2, 2, 5, 4, 7, 2, 3, 6, 7, 12, 9, 2, 4, 7, 10, 17, 20, 23, 2, 5, 8, 13, 22, 31, 44, 41, 2, 6, 9, 16, 27, 42, 65, 84, 87, 2, 7, 10, 19, 32, 53, 86, 127, 172, 169, 2, 8, 11, 22, 37, 64, 107, 170, 257, 340, 343

Offset: 0

Paul Curtz, Nov 08 2018

Array:
2, -1,  3,  1,  7,  9,  23,  41,  87, ... = (-1)^n*A140966(n)
2,  0,  4,  4, 12, 20,  44,  84, 172, ... = abs(A084247(n+1))
2,  1,  5,  7, 17, 31,  65, 127, 257, ... = A014551(n)
2,  2,  6, 10, 22, 42,  86, 170, 342, ... = A078008(n+2) = A014113(n+1)
2,  3,  7, 13, 27, 53, 107, 213, 427, ... = A048573(n)
2,  4,  8, 16, 32, 64, 128, 256, 512, ... = A000079(n+1)
2,  5,  9, 19, 37, 75, 149, 299, 597, ... = A062092(n)
2,  6, 10, 22, 42, 86, 170, 342, 682, ... = A078008(n+3) = A014113(n+2).
T(n+1,k) = (-1)^k*A140966(k) + (n+1)*A001045(k).
Every row T(n+1,k) has the signature (1,2).
T(0,k) = 2, -2, 2, -2, ... = (-1)^n*2.
T(n+1,k) - T(0,k) = (n+1)*A001045(n).
5*A001045(n) is not in the OEIS.

			Triangle a(n):
  2;
  2, -1;
  2,  0,  3;
  2,  1,  4,  1;
  2,  2,  5,  4,  7;
  2,  3,  6,  7, 12,  9;
  2,  4,  7, 10, 17, 20, 23;
  etc.
Row sums: 2, 1, 5, 8, 20, 39, 83, 166, 338, 677, 1361, 2724, ... = b(n+2).
With b(0) = 2 and b(1) = 0, b(n) = b(n-1) + 2*b(n-2)  + n - 4, n > 1.
b(n) = A001045(n) - A097065(n-1).
b(n) = b(n-2) + A000225(n-2).

Cf. A000079, A001045, A014113, A014551, A048573, A062092, A078008, A084247, A092297, A097073, A140360, A140966.

Cf. A000225, A097065.

T[_, 0] = 2;
T[0, k_] := (2^k + 5(-1)^k)/3;
T[n_ /; n>0, k_ /; k>0] := T[n, k] = T[n-1, k] + (2^k + (-1)^(k+1))/3;
T[, ] = 0;
Table[T[n-k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 10 2018 *)

A345380 Number of Jacobsthal-Lucas numbers m <= n.

Original entry on oeis.org

0, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 0

Ovidiu Bagdasar, Jun 16 2021

			a(0)=0 since the least term in A014551 is 1.
a(1)=1 since A014551(0) = 2 is followed in that sequence by 1.
a(k)=2 for 2 <= k <= 4 since the first terms of A014551 are {2, 1, 5}.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36. See Section 5.6, pp. 35, Table 7.

Cf. A014551, A108852 (Fibonacci), A130245 (Lucas), A130253.

Block[{a = 1, b = -2, nn = 105, u, v = {}}, u = {2, a}; Do[AppendTo[u, Total[{-b, a} u[[-2 ;; -1]]]]; AppendTo[v, Count[u, _?(# <= i &)]], {i, nn}]; {Boole[First[u] <= 0]}~Join~v]  (* or *)
{0}~Join~Accumulate@ ReplacePart[ConstantArray[0, Last[#]], Map[# -> 1 &, #]] &@ LinearRecurrence[{1, 2}, {2, 1}, 8] (* Michael De Vlieger, Jun 16 2021 *)

A369404 a(n) = 32^n + 5(-1)^n.

Original entry on oeis.org

8, 1, 17, 19, 53, 91, 197, 379, 773, 1531, 3077, 6139, 12293, 24571, 49157, 98299, 196613, 393211, 786437, 1572859, 3145733, 6291451, 12582917, 25165819, 50331653, 100663291, 201326597, 402653179, 805306373, 1610612731, 3221225477, 6442450939, 12884901893

Offset: 0

Philippe Deléham, Jan 22 2024

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

Cf. A000079, A001045, A014551, A033999.

LinearRecurrence[{1,2},{8,1},33] (* James C. McMahon, Jan 31 2024 *)

def A369404(n): return (3<Chai Wah Wu, Feb 25 2024

a(n) = 3*A000079(n) + 5*A033999(n).
a(n) = 4*A014551(n) - 3*A001045(n).
a(n) = a(n-1) + 2*a(n-2).
G.f.: (8 - 7*x)/((1 + x)*(1 - 2*x)).

A137374 Triangular array of the coefficients of the Jacobsthal-Lucas polynomials ordered by descending powers, read by rows.

Original entry on oeis.org

2, 1, 4, 1, 6, 1, 8, 8, 1, 20, 10, 1, 16, 36, 12, 1, 56, 56, 14, 1, 32, 128, 80, 16, 1, 144, 240, 108, 18, 1, 64, 400, 400, 140, 20, 1, 352, 880, 616, 176, 22, 1, 128, 1152, 1680, 896, 216, 24, 1, 832, 2912, 2912, 1248, 260, 26, 1, 256, 3136, 6272, 4704, 1680, 308, 28, 1

Offset: 0

Roger L. Bagula, Apr 09 2008

The even rows which start with 4, 8, 16 ... appear to be the absolute values of the Riordan array A128414. - Georg Fischer, Feb 25 2020

			The triangle starts:
    2;
    1;
    4,   1;
    6,   1;
    8,   8,   1;
   20,  10,   1;
   16,  36,  12,   1;
   56,  56,  14,   1;
   32, 128,  80,  16,  1;
  144, 240, 108,  18,  1;
   64, 400, 400, 140, 20, 1;
  352, 880, 616, 176, 22, 1;
  ...

Eric Weisstein's World of Mathematics, Jacobsthal-Lucas Polynomial.

Row sums give A014551.

Cf. A034807.

b:= proc(n) option remember;
      `if`(n<2, 2-n, b(n-1)+2*expand(x*b(n-2)))
    end:
T:= n-> (p-> (d-> seq(coeff(p, x, d-i), i=0..d))(degree(p)))(b(n)):
seq(T(n), n=0..20);  # Alois P. Heinz, Feb 25 2020

f[0] = 2; f[1] = 1; f[n_] := 2 x f[n - 2] + f[n - 1];
Table[Reverse[CoefficientList[f[n], x]], {n, 0, 14}] // Flatten (* Jinyuan Wang, Feb 25 2020 *)

Let p(n, x) = 2*x*p(n-2, x) + p(n-1, x) with p(0, x) = 2 and p(1, x) = 1. The coefficients of the polynomial p(n, x), listed in reverse order, give row n. - Jinyuan Wang, Feb 25 2020

Offset set to 0 by Peter Luschny, Feb 25 2020

A317793 a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2.

Original entry on oeis.org

1, 15, 22, 177, 406, 2445, 7162, 36177, 121486, 554325, 2009602, 8656377, 32761366, 136617405, 529712842, 2169039777, 8525430046, 34553579685, 136858084882, 551499730377, 2193794127526, 8811785649165, 35137304693722, 140878711512177, 562526325893806

Offset: 1

Jinyuan Wang, Aug 07 2018

This sequence is an extension of A014551; the sequences A014551(n) = 2^n + (-1)^n, a(n) = 4^n + (-3)^n + 2^n + (-1)^n and b(n) = 6^n + (-5)^n + 4^n + (-3)^n + 2^n + (-1)^n, ... can be considered to be of the same type.

For k>0, a(4k-2)/5, a(2k)/3 and a(2k+1)/2 are integers.

Index entries for linear recurrences with constant coefficients, signature (2,13,-14,-24).

Cf. A014551, A087451.

[(4^n+(-3)^n+2^n+(-1)^n)/2: n in [1..30]]; // Vincenzo Librandi, Aug 08 2018

CoefficientList[ Series[(-48x^3 - 21x^2 + 13x + 1)/(24x^4 + 14x^3 - 13x^2 - 2x + 1), {x, 0, 25}], x]  (* or *)LinearRecurrence[{2, 13, -14, -24}, {1, 15, 22, 177}, 26] (* Robert G. Wilson v, Aug 07 2018 *)

Vec(x*(1 + 13*x - 21*x^2 - 48*x^3) / ((1 + x)*(1 - 2*x)*(1 + 3*x)*(1 - 4*x)) + O(x^40)) \\ Colin Barker, Aug 07 2018

a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2 for n > 0.
From Colin Barker, Aug 07 2018: (Start)
G.f.: x*(1 + 13*x - 21*x^2 - 48*x^3) / ((1 + x)*(1 - 2*x)*(1 + 3*x)*(1 - 4*x)).
a(n) = 2*a(n-1) + 13*a(n-2) - 14*a(n-3) - 24*a(n-4) for n>4.
(End)
E.g.f.: (cosh(3*x/2) + cosh(7*x/2))*(cosh(x/2) + sinh(x/2)) - 2. - Stefano Spezia, Mar 20 2022

More terms from Colin Barker, Aug 07 2018

A338198 Triangle read by rows, T(n,k) = ((k+1)2^(n-k)-(k-2)(-1)^(n-k))/3 for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 2, 3, 2, 1, 6, 5, 4, 3, 1, 10, 11, 8, 5, 4, 1, 22, 21, 16, 11, 6, 5, 1, 42, 43, 32, 21, 14, 7, 6, 1, 86, 85, 64, 43, 26, 17, 8, 7, 1, 170, 171, 128, 85, 54, 31, 20, 9, 8, 1, 342, 341, 256, 171, 106, 65, 36, 23, 10, 9, 1, 682, 683, 512, 341, 214, 127, 76, 41, 26, 11, 10, 1

Offset: 0

Werner Schulte, Oct 15 2020

This triangle is related to the Jacobsthal numbers (A001045).

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :    0     1     2    3    4    5    6   7   8   9
======================================================
  0 :    1
  1 :    0     1
  2 :    2     1     1
  3 :    2     3     2    1
  4 :    6     5     4    3    1
  5 :   10    11     8    5    4    1
  6 :   22    21    16   11    6    5    1
  7 :   42    43    32   21   14    7    6   1
  8 :   86    85    64   43   26   17    8   7   1
  9 :  170   171   128   85   54   31   20   9   8   1
etc.

For columns k = 0 to 8 see A078008, A001045, A000079, A001045, A084214, A014551, A083595, A083582, A259713 respectively.

Table[((k + 1)*2^(n - k) - (k - 2)*(-1)^(n - k))/3, {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 15 2020 *)

T(n,n) = 1 for n >= 0; T(n,n-1) = n-1 for n > 0.
T(n,k) = T(n-1,k) + 2 * T(n-2,k) for 0 <= k <= n-2.
T(n,k) = 2 * T(n-1,k) - (k-2) * (-1)^(n-k) for 0 <= k < n.
T(n,k) = T(n+1-k,1) + (k-1) * T(n-k,1) for 0 <= k < n.
T(n+1,k) * T(n-1,k) - T(n,k+1) * T(n,k-1) = T(n-k,1)^2 for 0 < k < n.
Row sums are A083579(n+1) for n >= 0.
G.f. of column k >= 0: (1+(k-1)*t) * t^k / (1-t-2*t^2).
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = (1 - (1+x)*t + 2*x*t^2) / ((1 - x*t)^2 * (1 - t - 2*t^2)).
Conjecture: Let M(n,k) be the matrix inverse of T(n,k), seen as a matrix. Then M(i,j) = 0 if j < 0 or j > i, M(n,n) = 1 for n >= 0, M(n,n-1) = 1-n for n > 0, and M(n,k) = (-1)^(n-k) * (k^2-2) * (n-2)! / k! for 0 <= k <= n-2.

A344109 a(n) = (52^n + 7(-1)^n)/3.

Original entry on oeis.org

4, 1, 9, 11, 29, 51, 109, 211, 429, 851, 1709, 3411, 6829, 13651, 27309, 54611, 109229, 218451, 436909, 873811, 1747629, 3495251, 6990509, 13981011, 27962029, 55924051, 111848109, 223696211, 447392429, 894784851, 1789569709, 3579139411, 7158278829, 14316557651, 28633115309, 57266230611, 114532461229, 229064922451

Offset: 0

Paul Curtz, May 09 2021

Élis Gardel da Costa Mesquita, Eudes Antonio Costa, Paula M. M. C. Catarino, and Francisco R. V. Alves, Jacobsthal-Mulatu Numbers, Latin Amer. J. Math. (2025) Vol. 4, No. 1, 23-45. See pp. 24, 26, 43.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A001045, A020714, A110286, A154879, A321421.

Cf. A014551, A156550, A084214.

LinearRecurrence[{1,2}, {4,1}, 28] (* Amiram Eldar, May 10 2021 *)

a(n+1) = 5*2^n - a(n) for n >= 0, with a(0) = 4.
a(n+2) = 5*2^n + a(n) for n >= 0, with a(0) = 4, a(1) = 1.
a(n+3) = 15*2^n - a(n) for n >= 0, with a(0) = 4, a(1) = 1, a(2) = 9.
a(n) = A001045(n+2) + A154879(n).
a(2*n+1) = A321421(n).
a(n) = a(n-1) + 2*a(n-2) for n >= 2. - Pontus von Brömssen, May 09 2021
G.f.: (4 - 3*x)/(1 - x - 2*x^2). - Stefano Spezia, May 10 2021
a(n) = 2*A014551(n) - A001045(n).
a(n) = abs(A156550(n)) - (-1)^n.
a(n+3) = a(n) + 7*A084214(n+1) for n >= 0, with a(0) = 4.
a(n) = 5*A001045(n+1) - A084214(n+1) for n >= 0.
a(n) = A084214(n+1) + 3*(-1)^n for n >= 0.

cp's OEIS Frontend

A274074 a(n) = 6^n+(-1)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A284129 Hosoya triangle Jacobsthal Lucas type.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A307688 a(n) = 2*a(n-1)-2*a(n-2)+a(n-3)+2*a(n-4) with a(0)=a(1)=0, a(2)=2, a(3)=3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A321373 Array T(n,k) read by antidiagonals where the first row is (-1)^k*A140966(k) and each subsequent row is obtained by adding A001045(k) to the preceding one.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A345380 Number of Jacobsthal-Lucas numbers m <= n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A369404 a(n) = 3*2^n + 5*(-1)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

A137374 Triangular array of the coefficients of the Jacobsthal-Lucas polynomials ordered by descending powers, read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A317793 a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2.

Views

Author

Keywords

Comments

A307688 a(n) = 2a(n-1)-2a(n-2)+a(n-3)+2*a(n-4) with a(0)=a(1)=0, a(2)=2, a(3)=3.

A369404 a(n) = 32^n + 5(-1)^n.

A338198 Triangle read by rows, T(n,k) = ((k+1)2^(n-k)-(k-2)(-1)^(n-k))/3 for 0 <= k <= n.

A344109 a(n) = (52^n + 7(-1)^n)/3.