A089977 -id:A089977 - cp's OEIS frontend

A297395 T(n,k)=Number of nXk 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 1 neighboring 1.

1, 2, 1, 3, 5, 1, 4, 9, 9, 1, 6, 13, 19, 20, 1, 9, 33, 37, 57, 41, 1, 13, 69, 127, 126, 139, 85, 1, 19, 121, 323, 700, 385, 369, 178, 1, 28, 253, 763, 2569, 3175, 1243, 963, 369, 1, 41, 529, 2121, 7779, 14940, 15541, 3924, 2489, 769, 1, 60, 1013, 5557, 31081, 58901, 99682

Offset: 1

R. H. Hardin, Dec 29 2017

Table starts
.1...2....3.....4.......6........9........13.........19...........28
.1...5....9....13......33.......69.......121........253..........529
.1...9...19....37.....127......323.......763.......2121.........5557
.1..20...57...126.....700.....2569......7779......31081.......117084
.1..41..139...385....3175....14940.....58901.....325922......1616869
.1..85..369..1243...15541....99682....514945....3977868.....27131403
.1.178..963..3924...74736...640562...4279111...46261441....428200086
.1.369.2489.12477..358341..4101278..35870939..540319235...6780786267
.1.769.6523.39625.1729617.26607999.302197213.6362528482.108762242579

			Some solutions for n=5 k=4
..0..0..1..1. .0..1..1..0. .0..1..0..0. .0..0..1..1. .0..0..0..0
..0..0..0..0. .0..0..0..0. .1..0..0..0. .0..0..0..0. .0..0..1..0
..0..1..1..0. .0..0..0..0. .0..0..0..0. .0..0..1..0. .0..0..0..1
..0..0..0..0. .0..0..0..1. .0..0..0..0. .0..0..0..1. .0..1..0..0
..0..0..1..1. .0..0..1..0. .1..1..0..0. .0..0..0..0. .1..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..420

Column 2 is A105309(n+1).
Row 1 is A000930(n+1).
Row 2 is A089977(n+1).

Empirical for column k:
k=1: a(n) = a(n-1)
k=2: a(n) = a(n-1) +2*a(n-2) +a(n-3) -a(n-4)
k=3: a(n) = a(n-1) +4*a(n-2) +2*a(n-3) -4*a(n-4)
k=4: a(n) = a(n-1) +6*a(n-2) +4*a(n-3) -3*a(n-4) -a(n-5) -2*a(n-6) -a(n-7)
k=5: [order 20]
k=6: [order 25]
k=7: [order 55]
Empirical for row n:
n=1: a(n) = a(n-1) +a(n-3)
n=2: a(n) = a(n-1) +4*a(n-3)
n=3: a(n) = a(n-1) +a(n-2) +8*a(n-3) +a(n-4) -2*a(n-6)
n=4: [order 8]
n=5: [order 21]
n=6: [order 31]
n=7: [order 69]

A297595 T(n,k) = Number of n X k 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 1 or 5 neighboring 1s.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 9, 9, 1, 6, 13, 25, 20, 1, 9, 33, 49, 69, 41, 1, 13, 69, 145, 154, 205, 85, 1, 19, 121, 443, 752, 577, 597, 178, 1, 28, 253, 1141, 3145, 3747, 1977, 1701, 369, 1, 41, 529, 3009, 10131, 23066, 18577, 6962, 4949, 769, 1, 60, 1013, 8455, 37929, 103673

Offset: 1

R. H. Hardin, Jan 01 2018

Table starts
.1...2.....3.....4.......6........9........13..........19...........28
.1...5.....9....13......33.......69.......121.........253..........529
.1...9....25....49.....145......443......1141........3009.........8455
.1..20....69...154.....752.....3145.....10131.......37929.......150388
.1..41...205...577....3747....23066....103673......514290......2834897
.1..85...597..1977...18577...163704....975485.....6551844.....50398161
.1.178..1701..6962...93150..1172288...9403199....85828150....919035936
.1.369..4949.24441..464697..8419996..90862063..1120526916..16723808887
.1.769.14389.85803.2320289.60354437.875241087.14592832760.303459238317

			Some solutions for n=6 k=4
..0..0..0..0. .0..1..0..0. .1..0..0..0. .1..0..0..0. .0..0..1..0
..0..1..1..0. .0..0..1..0. .0..1..0..0. .0..1..0..0. .0..1..0..0
..0..0..0..0. .0..0..0..0. .0..1..0..0. .0..0..0..0. .0..0..0..0
..0..1..0..1. .0..0..1..1. .1..0..0..0. .0..0..0..0. .0..0..0..0
..0..1..1..1. .0..0..0..0. .1..0..0..0. .0..0..0..0. .1..0..0..0
..0..0..0..1. .1..1..0..0. .0..1..0..0. .0..0..0..0. .0..1..0..0

R. H. Hardin, Table of n, a(n) for n = 1..391

Column 2 is A105309(n+1).
Row 1 is A000930(n+1).
Row 2 is A089977(n+1).

Empirical for column k:
k=1: a(n) = a(n-1)
k=2: a(n) = a(n-1) +2*a(n-2) +a(n-3) -a(n-4)
k=3: a(n) = a(n-1) +2*a(n-2) +10*a(n-3) +4*a(n-4) -8*a(n-5) -8*a(n-6)
k=4: [order 9]
k=5: [order 22]
k=6: [order 40]
k=7: [order 83]
Empirical for row n:
n=1: a(n) = a(n-1) +a(n-3)
n=2: a(n) = a(n-1) +4*a(n-3)
n=3: a(n) = a(n-1) +a(n-2) +8*a(n-3) +7*a(n-4) -8*a(n-6) -6*a(n-7)
n=4: [order 12]
n=5: [order 26]
n=6: [order 49]

A183450 T(n,k)=Number of nXk binary arrays with every 1 having exactly two king-move neighbors equal to 1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 9, 9, 1, 1, 13, 18, 13, 1, 1, 33, 30, 30, 33, 1, 1, 69, 107, 72, 107, 69, 1, 1, 121, 265, 283, 283, 265, 121, 1, 1, 253, 553, 831, 2054, 831, 553, 253, 1, 1, 529, 1505, 2399, 9208, 9208, 2399, 1505, 529, 1, 1, 1013, 3852, 7761, 34867, 53608, 34867, 7761

Offset: 1

R. H. Hardin Jan 04 2011

Table starts
.1....1....1.....1.......1........1..........1...........1............1
.1....5....9....13......33.......69........121.........253..........529
.1....9...18....30.....107......265........553........1505.........3852
.1...13...30....72.....283......831.......2399........7761........23840
.1...33..107...283....2054.....9208......34867......176949.......833001
.1...69..265...831....9208....53608.....257733.....1817225.....11414889
.1..121..553..2399...34867...257733....1744887....16192521....132651622
.1..253.1505..7761..176949..1817225...16192521...216103851...2491241396
.1..529.3852.23840..833001.11414889..132651622..2491241396..39184763856
.1.1013.8922.72396.3619285.65073225.1038281076.26807184942.568464533358

			Some solutions for 6X5
..0..1..0..0..0....1..1..0..1..1....1..1..0..0..1....0..0..0..1..1
..1..0..1..0..0....0..1..0..1..0....1..0..0..1..1....0..0..0..1..0
..1..0..1..0..0....0..0..0..0..0....0..0..0..0..0....0..1..0..0..0
..1..0..0..1..0....0..0..1..1..0....0..0..0..1..1....1..0..1..0..0
..0..1..0..1..0....0..1..0..0..1....1..1..0..0..1....1..0..1..0..0
..0..0..1..0..0....0..0..1..1..0....1..0..0..0..0....0..1..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..363

Column 2 is A089977(n+1)

A296128 T(n,k)=Number of nXk 0..1 arrays with each 1 adjacent to 2 or 5 king-move neighboring 1s.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 9, 9, 1, 1, 13, 18, 13, 1, 1, 33, 30, 30, 33, 1, 1, 69, 107, 74, 107, 69, 1, 1, 121, 265, 287, 287, 265, 121, 1, 1, 253, 553, 841, 2098, 841, 553, 253, 1, 1, 529, 1505, 2463, 9344, 9344, 2463, 1505, 529, 1, 1, 1013, 3852, 7953, 35733, 54286, 35733, 7953

Offset: 1

R. H. Hardin, Dec 05 2017

Table starts
.1...1....1.....1......1........1.........1..........1...........1............1
.1...5....9....13.....33.......69.......121........253.........529.........1013
.1...9...18....30....107......265.......553.......1505........3852.........8922
.1..13...30....74....287......841......2463.......7953.......24428........74660
.1..33..107...287...2098.....9344.....35733.....182881......859151......3752449
.1..69..265...841...9344....54286....263489....1867837....11720327.....67180741
.1.121..553..2463..35733...263489...1810441...16855693...138016847...1087543621
.1.253.1505..7953.182881..1867837..16855693..227074899..2614792001..28292082465
.1.529.3852.24428.859151.11720327.138016847.2614792001.41069612676.599261543598

			Some solutions for n=5 k=4
..0..1..1..0. .0..0..1..0. .0..1..1..0. .0..0..0..0. .0..1..1..0
..0..1..0..0. .0..1..1..0. .1..0..0..1. .0..0..0..1. .0..0..1..0
..0..0..0..0. .0..0..0..0. .0..1..0..1. .0..0..1..1. .0..0..0..0
..1..0..0..0. .1..0..0..0. .0..1..0..1. .1..0..0..0. .0..0..0..0
..1..1..0..0. .1..1..0..0. .0..0..1..0. .1..1..0..0. .0..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..287

Column 2 is A089977(n+1).

Column 3 is A183444.

Empirical for column k:
k=1: a(n) = a(n-1)
k=2: a(n) = a(n-1) +4*a(n-3)
k=3: a(n) = 2*a(n-1) -a(n-2) +8*a(n-3) -7*a(n-4) +2*a(n-5) -2*a(n-6)
k=4: [order 9]
k=5: [order 39]

A097334 a(n) = Sum_{k=0..n} C(n-k, floor(k/2))*2^k.

Original entry on oeis.org

1, 3, 3, 7, 19, 31, 59, 135, 259, 495, 1035, 2071, 4051, 8191, 16475, 32679, 65443, 131343, 262059, 523831, 1049203, 2097439, 4192763, 8389575, 16779331, 33550383, 67108683, 134226007, 268427539, 536862271, 1073766299, 2147476455

Offset: 0

Paul Barry, Aug 05 2004

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,4).

Cf. A089977, A097333, A097335.

Table[Sum[Binomial[n-k,Floor[k/2]]2^k,{k,0,n}],{n,0,40}] (* or *) LinearRecurrence[{1,0,4},{1,3,3},40] (* Harvey P. Dale, May 17 2021 *)

G.f. : (1+2x)/((1-2*x)*(1+x+2*x^2)); a(n)=a(n-1)+4a(n-3).

A113726 A Jacobsthal convolution.

Original entry on oeis.org

1, 0, 1, 4, 5, 8, 25, 44, 77, 176, 353, 660, 1365, 2776, 5417, 10876, 21981, 43648, 87153, 175076, 349669, 698280, 1398585, 2797260, 5590381, 11184720, 22373761, 44735284, 89474165, 178969208, 357910345, 715807004, 1431683837, 2863325216

Offset: 0

Paul Barry, Nov 08 2005

Convolution of A001045(n+1) and A001607(n+1).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,4,4).

Cf. A094686, A089977, A001045, A001607.

LinearRecurrence[{0,1,4,4},{1,0,1,4},40] (* Harvey P. Dale, Apr 30 2025 *)

G.f.: 1/((1-x-2*x^2)*(1+x+2*x^2)).
a(n) = a(n-2) + 4*a(n-3) + 4*a(n-4).
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*2^k*(1+(-1)^(n-k))/2.
a(n) = 2^n/3 + (-1)^n/6 + A001607(n+1)/2. - R. J. Mathar, Aug 23 2011
a(n) = sum(A128099(n, n-2*k), k=0..floor(n/2)). - Johannes W. Meijer, Aug 28 2013

A122946 a(0)=a(1)=0, a(2)=2; for n >= 3, a(n) = a(n-1) + 4*a(n-3).

Original entry on oeis.org

0, 0, 2, 2, 2, 10, 18, 26, 66, 138, 242, 506, 1058, 2026, 4050, 8282, 16386, 32586, 65714, 131258, 261602, 524458, 1049490, 2095898, 4193730, 8391690, 16775282, 33550202, 67116962, 134218090, 268418898, 536886746, 1073759106, 2147434698, 4294981682, 8590018106

Offset: 0

Benoit Cloitre, Oct 24 2006

See lemma 5.2 of Reznick's preprint.
Conjecture: count of even Markov numbers in generation n (with generations 0, 1 and 2 labeled as {5}, {13, 29} and {34, 194, 433, 169}. (Checked up to generation 20.) - Wouter Meeussen, Jan 16 2024
Wouter Meeussen's conjecture is true. Proof: label the Markov tree with Markov triples according to the scheme described at A368546. Mod 2, the triples are: row 0: (1,1,0); row 1: (1,1,1), (1,1,0); row 2: (1,0,1), (1,0,1), (1,1,1), (1,1,0); row 3: (1,1,0), (0,1,1), (1,1,0), (0,1,1), (1,0,1), (1,0,1), (1,1,1), (1,1,0); etc. Note that the Markov number labels of the tree (the center numbers of the triples) in rows 0 and 1 include no even numbers, while those in row 2 include two even numbers. Observing that the second triple in row 1 and the first four triples in row 3 are the same or the reverse of the root triple, and noting that every vertex in row 3 and beyond is in a subtree with one of these triples as root, the recurrence follows. - William P. Orrick, Mar 05 2024

A.H.M. Smeets, Table of n, a(n) for n = 0..300
Bruce Reznick, Regularity properties of the Stern enumeration of the rationals, arXiv:math/0610601 [math.NT], 2006.
Index entries for linear recurrences with constant coefficients, signature (1,0,4).

Cf. A002487, A002559, A089977, A110512, A368546.

LinearRecurrence[{1,0,4},{0,0,2},36] (* James C. McMahon, Jan 16 2024 *)

a0=a1=0;a2=2;for(n=3,50,a3=a2+4*a0;a0=a1;a1=a2;a2=a3;print1(a3,","))

a(n) = (1/7)*2^(-2 + n/2)*(7*2^(n/2) - 7*cos(n*(Pi - arctan(sqrt(7)))) + 5*sqrt(7)*sin(n*(-Pi + arctan(sqrt(7))))). - Zak Seidov, Oct 26 2006
G.f.: 2*x^2 / ((1-2*x)*(2*x^2+x+1)). - Colin Barker, Jun 20 2013
a(n) = 2 * A089977(n-2) for n >= 2. - Alois P. Heinz, Jan 16 2024
From A.H.M. Smeets, Jan 16 2024: (Start)
Limit_{n -> oo} a(n)/a(n-1) = 2.
a(n) = 2^(n-2) + A110512(n-2), for n >= 2. (End)

Entries checked by Zak Seidov, Oct 26 2006

A193515 T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 6, 1, 1, 6, 9, 10, 13, 9, 1, 1, 7, 11, 13, 22, 23, 13, 1, 1, 8, 13, 16, 33, 43, 37, 19, 1, 1, 9, 15, 19, 46, 69, 73, 63, 28, 1, 1, 10, 17, 22, 61, 101, 121, 139, 109, 41, 1, 1, 11, 19, 25, 78, 139, 181, 253, 268, 183, 60, 1, 1, 12

Offset: 1

R. H. Hardin, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011

Table starts:
..1...1...1...1...1....1....1....1....1....1....1....1.....1.....1.....1.....1
..1...1...1...1...1....1....1....1....1....1....1....1.....1.....1.....1.....1
..2...3...4...5...6....7....8....9...10...11...12...13....14....15....16....17
..3...5...7...9..11...13...15...17...19...21...23...25....27....29....31....33
..4...7..10..13..16...19...22...25...28...31...34...37....40....43....46....49
..6..13..22..33..46...61...78...97..118..141..166..193...222...253...286...321
..9..23..43..69.101..139..183..233..289..351..419..493...573...659...751...849
.13..37..73.121.181..253..337..433..541..661..793..937..1093..1261..1441..1633
.19..63.139.253.411..619..883.1209.1603.2071.2619.3253..3979..4803..5731..6769
.28.109.268.529.916.1453.2164.3073.4204.5581.7228.9169.11428.14029.16996.20353

			Some solutions for n=7 k=3; colors=1,2,3 and empty=0
..3....0....0....2....0....1....3....0....0....0....1....0....3....1....0....0
..3....0....0....2....2....1....3....2....1....0....1....3....3....1....0....0
..3....1....0....2....2....1....3....2....1....2....1....3....3....1....0....3
..1....1....3....0....2....0....0....2....1....2....3....3....0....2....0....3
..1....1....3....0....0....2....2....2....1....2....3....2....1....2....1....3
..1....0....3....0....0....2....2....2....1....0....3....2....1....2....1....0
..0....0....0....0....0....2....2....2....1....0....0....2....1....0....1....0

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A000930,
Column 2 is A003229(n-1),
Column 3 is A084386,
Column 4 is A089977,
Column 10 is A178205,
Row 6 is A028872(n+2),
Row 7 is A144390(n+1),
Row 8 is A003154(n+1).

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<3 or k=0, 1, k*T(n-3, k) +T(n-1, k)))
    end:
seq(seq(T(n, d+1-n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011

nmax = 13; t[?Negative, ] = 0; t[n_, k_] /; (n < 3 || k == 0) = 1; t[n_, k_] := t[n, k] = k*t[n-3, k] + t[n-1, k]; Flatten[ Table[ t[n-k+1, k], {n , 1, nmax}, {k, n, 1, -1}]](* Jean-François Alcover, Nov 28 2011, after Maple *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. So T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) = sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the polynomial k x^z + x - 1.
T(n,k) = sum {s=0..[n/3]} (binomial(n-2*s,s)*k^s).
For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n-(z-1)*s,s)*k^s). - R. H. Hardin, Jul 31 2011

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

A071911 Numbers m such that Stern's diatomic A002487(m) is divisible by 3.

Original entry on oeis.org

0, 5, 7, 10, 14, 20, 28, 33, 35, 40, 45, 47, 49, 51, 56, 61, 63, 66, 70, 73, 75, 80, 85, 87, 90, 94, 98, 102, 105, 107, 112, 117, 119, 122, 126, 132, 140, 146, 150, 153, 155, 160, 165, 167, 170, 174, 180, 188, 196, 204, 210, 214, 217, 219, 224, 229, 231, 234, 238, 244, 252

Offset: 0

N. J. A. Sloane, Jun 13 2002

nonn

From Kevin Ryde, Jan 09 2021: (Start)
Dijkstra gives bit pattern {0}1{?0{1}0|?1{0}1}?1{0} for the terms of this sequence, where | is alternative, {x} is zero or more of x, and ? is a single 0 or 1.  Dijkstra formed this from a finite state automaton (states as pairs of Stern diatomic values mod 3 = A071412).  The minimized automaton is as follows.  State A is the start and the sole accepting state.
                 +---+
        1 +----> | B | <-----+ 0
          |      +---+       |
  0 +-- +===+      |       +---+ --+ 1
    +-> | A |      |0,1    | D | <-+
  start +===+      v       +---+
          ^      +---+       ^
        1 +----- | C | ------+ 0
                 +---+
a(n) can be calculated from n by a usual unranking in this automaton, using the number of strings of a given length k accepted from each state.  Reznick's A122946(k) is the number of strings accepted starting from B.  A089977(k-1) is the number accepted starting from C.
Dijkstra shows that for any m, a bit reversal (A030101) or an internal bit complement (A122155) of m are no change to the resulting Diatomic(m) value.  So here if m is a term then so are A030101(m) and A122155(m).  In the bit pattern, a reversal or complement between (but not including) the outermost 1's is no change.
(End)

Edsger W. Dijkstra, An exercise for Dr. R. M. Burstall, 1976. Reprinted in Edsger W. Dijkstra, Selected Writings on Computing, Springer-Verlag, 1982, pages 215-216.
Edsger W. Dijkstra, More about the function ``fusc'', 1976. Reprinted in Edsger W. Dijkstra, Selected Writings on Computing, Springer-Verlag, 1982, pages 230-232.
Bruce Reznick, Regularity Properties of the Stern Enumeration of the Rationals, Journal of Integer Sequences, volume 11, 2008, article 08.4.1. Also arXiv:math/0610601 [math.NT] 2006, and author's copy. Section 5 theorem 18 onward.

Cf. A071412 (diatomic mod 3), A122946 (count by bit length), A089977 (half that).

{ my(M=Mod('x, 'x^2+'x+2),
            f=[2,1, 0,-'x-1, -2,1, 0,'x-1],
            table=[1,3, 5,5, 7,1, 3,7]);
a(n) = n<<=2; my(k=if(n,logint(n,2)+1), p=M^k, s=1);
  while(k>=0,
    my(t = n + (3<>2; }  \\ Kevin Ryde, Jan 09 2021

def aupto(nn):
  ok = [1] + [0 for i in range(nn)]
  for m in range(nn+1):
    if ok[m]:  # from formula
      for i in [2*m, 8*m-5, 8*m+5, 8*m-7, 8*m+7]:
        if 0 <= i <= nn: ok[i] = 1
  return [m for m in range(nn+1) if ok[m]]
print(aupto(252)) # Michael S. Branicky, Jan 09 2021

from itertools import count, islice
from functools import reduce
def inA071911(n): return not (n and sum(reduce(lambda x,y:(x[0],(x[0]+x[1])%3) if int(y) else ((x[0]+x[1])%3,x[1]),bin(n)[-1:2:-1],(1,0)))%3)
def A071911_gen(startvalue=0): # generator of terms >= startvalue
    return filter(inA071911, count(max(startvalue,0)))
A071911_list = list(islice(A071911_gen(),20)) # Chai Wah Wu, May 18 2023

If m is in the sequence, then 2*m, 8*m +- 5, and 8*m +- 7 (when nonnegative) are in the sequence. Starting from m=0, this rule generates the sequence. [Reznick section 5 theorem 18] - Kevin Ryde, Jan 09 2021

cp's OEIS Frontend

A297395 T(n,k)=Number of nXk 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 1 neighboring 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A297595 T(n,k) = Number of n X k 0..1 arrays with every 1 horizontally, diagonally or antidiagonally adjacent to 1 or 5 neighboring 1s.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A183450 T(n,k)=Number of nXk binary arrays with every 1 having exactly two king-move neighbors equal to 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A296128 T(n,k)=Number of nXk 0..1 arrays with each 1 adjacent to 2 or 5 king-move neighboring 1s.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A097334 a(n) = Sum_{k=0..n} C(n-k, floor(k/2))*2^k.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A113726 A Jacobsthal convolution.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A122946 a(0)=a(1)=0, a(2)=2; for n >= 3, a(n) = a(n-1) + 4*a(n-3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A193515 T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.