A005578 -id:A005578 - cp's OEIS frontend

A263799 T(n,k)=Number of (n+1)X(k+1) 0..1 arrays with each row and column divisible by 3, read as a binary number with top and left being the most significant bits, and rows and columns lexicographically nonincreasing.

2, 2, 2, 3, 2, 3, 3, 3, 3, 3, 4, 3, 7, 3, 4, 4, 4, 7, 7, 4, 4, 5, 4, 14, 7, 14, 4, 5, 5, 5, 14, 17, 17, 14, 5, 5, 6, 5, 25, 18, 61, 18, 25, 5, 6, 6, 6, 25, 56, 130, 130, 56, 25, 6, 6, 7, 6, 41, 66, 494, 616, 494, 66, 41, 6, 7, 7, 7, 41, 218, 1435, 4991, 4991, 1435, 218, 41, 7, 7, 8, 7, 63, 272

Offset: 1

R. H. Hardin, Oct 26 2015

Table starts
.2.2..3...3.....4.......4........5.........5.........6..........6.........7
.2.2..3...3.....4.......4........5.........5.........6..........6.........7
.3.3..7...7....14......14.......25........25........41.........41........63
.3.3..7...7....17......18.......56........66.......218........272.......798
.4.4.14..17....61.....130......494......1435......4917......13962.....41366
.4.4.14..18...130.....616.....4991.....30130....185795....1022105...5241463
.5.5.25..56...494....4991....62904....760671...8468941...90476206.850301770
.5.5.25..66..1435...30130...760671..20141827.445862545.9910247963
.6.6.41.218..4917..185795..8468941.445862545
.6.6.41.272.13962.1022105.90476206

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

R. H. Hardin, Table of n, a(n) for n = 1..144

Column 1 is A005578(n+1).

Empirical for column k:
k=1: a(n) = a(n-1) +a(n-2) -a(n-3)
k=2: a(n) = a(n-1) +a(n-2) -a(n-3)
k=3: a(n) = a(n-1) +3*a(n-2) -3*a(n-3) -3*a(n-4) +3*a(n-5) +a(n-6) -a(n-7)
k=4: [order 19]
k=5: [order 37]
k=6: [order 83]

A052950 Expansion of (2-3x-x^2+x^3)/((1-x)(1+x)(1-2x)).

Original entry on oeis.org

2, 1, 3, 4, 9, 16, 33, 64, 129, 256, 513, 1024, 2049, 4096, 8193, 16384, 32769, 65536, 131073, 262144, 524289, 1048576, 2097153, 4194304, 8388609, 16777216, 33554433, 67108864, 134217729, 268435456, 536870913, 1073741824, 2147483649

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals row sums of triangle A178616 but replacing the 2 with a 1. - Gary W. Adamson, May 30 2010
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
An autosequence of the second kind whose first kind companion is A005578. - Jean-François Alcover, Mar 18 2020

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 9*x^4 + 16*x^5 + 33*x^6 + 64*x^7 + 129*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1009
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A178616. - Gary W. Adamson, May 30 2010

Concatenation([2], List([1..40], n-> (2^n +1 +(-1)^n)/2));  # G. C. Greubel, Oct 21 2019

[2] cat [(2^n +1 +(-1)^n)/2: n in [1..40]]; // G. C. Greubel, Oct 21 2019

spec:= [S,{S=Union(Sequence(Prod(Sequence(Z),Z)), Sequence(Prod(Z,Z)))}, unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
seq(`if`(n=0, 2, (2^n +1 +(-1)^n)/2), n=0..40); # G. C. Greubel, Oct 21 2019

a[n_]:= (2^n +1 +(-1)^n +Boole[n==0])/2; (* Michael Somos, Jun 03 2014 *)
a[n_]:= If[ n<0, (1-n)! SeriesCoefficient[Sinh[x] +Exp[x/2], {x,0,1-n}], n! SeriesCoefficient[Cosh[x](1+Exp[x]), {x,0,n}]]; (* Michael Somos, Jun 04 2014 *)
LinearRecurrence[{2,1,-2}, {2,1,3,4}, 40] (* G. C. Greubel, Oct 21 2019 *)

{a(n)=(2^n+1+(-1)^n+(n==0))/2}; /* Michael Somos, Jun 03 2014 */

[2]+[(2^n +1 +(-1)^n)/2 for n in (1..40)] # G. C. Greubel, Oct 21 2019

G.f.: (2-3*x-x^2+x^3)/((1-x^2)*(1-2*x)).
a(n) = a(n-1) + 2*a(n-2) - 1.
a(n) = 2^(n-1) + Sum_{alpha=RootOf(-1+z^2)} alpha^(-n)/2.
From Paul Barry, Sep 18 2003: (Start)
a(n) = (2^n + 1 + (-1)^n + 0^n)/2.
E.g.f.: cosh(x)*(1+exp(x)). (End)
a(2*n + 1) = 4 * a(2*n - 1) for all n in Z. a(2*n + 2) = 3*a(2*n + 1) + 2*a(2*n) if n>0. - Michael Somos, Jun 04 2014

More terms from James Sellers, Jun 05 2000

A129761 First differences of Fibbinary numbers (A003714).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 22, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 43, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 86, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 22, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 171, 1, 1, 2, 1, 3, 1, 1, 6

Offset: 0

Ralf Stephan, May 14 2007

nonn

Theorem: If the Zeckendorf representation of M ends with exactly k >= 0 zeros, ...10^k, then a(n) = ceiling(2^k/3). Also, if the Zeckendorf representation of n (A014417(n)) is even then a(n) is given by A319952, otherwise a(n) = 1. - Jeffrey Shallit and N. J. A. Sloane, Oct 03 2018

N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (First 2500 terms from Vincenzo Librandi)

Cf. A000045, A000071, A003714, A005578, A035614, A319432.

with(combinat): F:=fibonacci:
A072649:= proc(n) local j; global F; for j from ilog[(1+sqrt(5))/2](n)
       while F(j+1)<=n do od; (j-1); end:
A003714 := proc(n) global F; option remember; if(n < 3) then RETURN(n); else RETURN((2^(A072649(n)-1))+A003714(n-F(1+A072649(n)))); fi; end:
A129761 := n -> A003714(n+1)-A003714(n):
[seq(A129761(n),n=0..120)]; # N. J. A. Sloane, Oct 03 2018, borrowing programs from other sequences

Differences[Select[Range[600], !MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &]] (* Harvey P. Dale, Jul 17 2011 *)

a(n) = A005578(A035614(n)). - Alan Michael Gómez Calderón, Nov 01 2023

a(0)=1 added by N. J. A. Sloane, Oct 02 2018

A131271 Triangular array T(n,k), n>=0, k=1..2^n, read by rows in bracketed pairs such that highest ranked element is bracketed with lowest ranked.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 5, 2, 7, 3, 6, 1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11, 1, 32, 16, 17, 8, 25, 9, 24, 4, 29, 13, 20, 5, 28, 12, 21, 2, 31, 15, 18, 7, 26, 10, 23, 3, 30, 14, 19, 6, 27, 11, 22, 1, 64, 32, 33, 16, 49, 17, 48, 8, 57

Offset: 0

J. Demongeot (Jacques.Demongeot(AT)imag.fr), Jun 24 2007

In a knockout competition with 2^n players, arranging the competition brackets (see Wikipedia) in T(n,k) order, where T(n,k) is the rank of the k-th player, ensures that highest ranked players cannot meet until the later stages of the competition. None of the top 2^p ranked players can meet earlier than the p-th from last round of the competition. At the same time the top ranked players in each match meet the lowest ranked player possible consistent with this rule. The sequence for the top ranked players meeting the highest ranked player possible is A049773. - Colin Hall, Feb 28 2012
Ranks in natural order of 2^n increasing real numbers appearing in limit cycles of interval iterations, or Median Spiral Order. [See the works by Demongeot & Waku]
From Andrey Zabolotskiy, Dec 06 2024 (Start):
For n>0, row n-1 is the permutation relating row n of Kepler's tree of fractions with row n of the left half of Stern-Brocot tree. Specifically, if K_n(k) [resp. SB_n(k)] is the k-th fraction in the n-th row of A294442 [resp. A057432], where 1/2 is in row 1 and k=1..2^(n-1), then K_n(k) = SB_n(T(n-1, k)).
The inverse permutation is row n of A088208.
When 1 is subtracted from each term, rows 3-5 become A240908, A240909, A240910. (End)

			Triangle begins:
1;
1,  2;
1,  4, 2, 3;
1,  8, 4, 5, 2,  7, 3,  6;
1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11;
...

Alois P. Heinz, Table of n, a(n) for n = 0..8190
John M. Campbell, A generalization of Deterministic Finite Automata related to discharging, arXiv:2506.14072 [cs.FL], 2025. See p. 7.
Jacques Demongeot and Jules Waku, Counter-Examples about Lower- and Upper-Bounded Population Growth, Math. Pop. Studies 12 (2005), 199-210.
Jacques Demongeot and Jules Waku, Application of interval iterations to the entrainment problem in respiratory physiology, Phil. Trans. R. Soc. A, 367 (2009), 4717-4739.
Wikipedia, Bracket (tournament)

Cf. A049773, A088208, A294442, A057432.
Cf. A005578 (last elements in rows), A155944 (T(n,2^(n-1)) for n>0).
Cf. A006519, A240908, A240909, A240910.

T:= proc(n,k) option remember;
      `if`({n, k} = {1}, 1,
      `if`(irem(k, 2)=1, T(n-1, (k+1)/2), 2^(n-1)+1 -T(n-1, k/2)))
    end:
seq(seq(T(n, k), k=1..2^(n-1)), n=1..7); # Alois P. Heinz, Feb 28 2012, with offset 1

T[0, 1] = 1;
T[n_, k_] := T[n, k] = If[Mod[k, 2] == 1, T[n, (k + 1)/2], 2^n + 1 - T[n, k/2]];
Table[T[n, k], {n, 0, 6}, {k, 2^n}] // Flatten (* Jean-François Alcover, May 31 2018, after Alois P. Heinz *)

T(0,1) = 1, T(n,2k-1) = T(n-1,k), T(n,2k) = 2^n+1 - T(n-1,k).

T(n,1) = 1; for 1 < k <= 2^n, T(n,k) = 1 + (2^n)/m - T(n,k-m), where m = A006519(k-1). - Mathew Englander, Jun 20 2021

Edited (with new name from Colin Hall) by Andrey Zabolotskiy, Dec 06 2024

A241138 T(n,k)=Number of nXk 0..2 arrays with no element equal to the same number of vertical neighbors as horizontal neighbors, with new values 0..2 introduced in row major order.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 5, 5, 2, 3, 17, 24, 17, 3, 6, 56, 152, 152, 56, 6, 11, 195, 935, 1670, 935, 195, 11, 22, 691, 6150, 19619, 19619, 6150, 691, 22, 43, 2476, 40504, 238607, 434444, 238607, 40504, 2476, 43, 86, 8941, 269481, 2939027, 9958520, 9958520, 2939027

Offset: 1

R. H. Hardin, Apr 16 2014

Table starts
..0.....1........1..........2.............3..............6..............11
..1.....2........5.........17............56............195.............691
..1.....5.......24........152...........935...........6150...........40504
..2....17......152.......1670.........19619.........238607.........2939027
..3....56......935......19619........434444........9958520.......230889822
..6...195.....6150.....238607.......9958520......428855442.....18742128588
.11...691....40504....2939027.....230889822....18742128588...1546330012544
.22..2476...269481...36470565....5392206371...825015103540.128518782423157
.43..8941..1797120..454209512..126282435834.36426835430201
.86.32393.12006500.5665536042.2961490680709

			Some solutions for n=4 k=4
..0..1..1..1....0..1..2..0....0..1..0..2....0..1..2..0....0..1..0..1
..0..0..1..0....0..1..2..0....0..1..0..2....0..1..2..0....0..1..0..1
..0..2..2..0....2..2..1..1....2..1..1..1....0..2..1..0....2..2..0..1
..1..1..1..1....0..0..2..2....2..1..0..0....0..2..1..0....0..0..2..2

R. H. Hardin, Table of n, a(n) for n = 1..112

Column 1 is A005578(n-2)

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) for n>4
k=2: a(n) = 4*a(n-1) +2*a(n-2) -11*a(n-3) -6*a(n-4) +8*a(n-5) for n>6
k=3: [order 26] for n>27

A354667 Triangle read by rows: T(n,k) is the number of tilings of an (n+4*k) X 1 board using k (1,1;5)-combs and n-k squares.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 4, 0, 1, 1, 1, 6, 0, 3, 0, 1, 2, 9, 0, 9, 0, 1, 1, 3, 12, 5, 18, 0, 4, 0, 1, 4, 16, 12, 36, 0, 16, 0, 1, 1, 5, 20, 25, 60, 15, 40, 0, 5, 0, 1, 6, 25, 42, 100, 42, 100, 0, 25, 0, 1, 1, 7, 31, 66, 150, 112, 200

Offset: 0

Michael A. Allen, Jun 05 2022

This is the m=2, t=5 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(2*j+r-4*k,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for r=0,1, where f(n,x) is a (1,5)-bonacci polynomial defined by f(n,x)=f(n-1,x)+x*f(n-5,x)+delta(n,0) where f(n<0,x)=0.
T(n+8-4*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 2, 4, 6, or 8.

			Triangle begins:
  1;
  1,   0;
  1,   0,   1;
  1,   0,   2,   0;
  1,   0,   4,   0,   1;
  1,   1,   6,   0,   3,   0;
  1,   2,   9,   0,   9,   0,   1;
  1,   3,  12,   5,  18,   0,   4,   0;
  1,   4,  16,  12,  36,   0,  16,   0,   1;
  1,   5,  20,  25,  60,  15,  40,   0,   5,   0;
  1,   6,  25,  42, 100,  42, 100,   0,  25,   0,   1;
  1,   7,  31,  66, 150, 112, 200,  35,  75,   0,   6,   0;
...

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.

Row sums are A005578.
Sums over k of T(n-4*k,k) are A224811.
Other members of the family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354668 (m=3,t=3).
Other triangles related to tiling using combs: A059259, A123521, A157897, A335964.

Mathematica
```
T[n_,k_]:=If[k<0 || n
				
```

T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + 2*T(n-2,k-2) + T(n-3,k-1) - T(n-3,k-2) - 2*T(n-3,k-3) - T(n-4,k-1) + T(n-4,k-2) + T(n-4,k-3) - T(n-4,k-4) + T(n-5,k-1) - 2*T(n-5,k-3) + T(n-5,k-5) + delta(n,0)*delta(k,0) - delta(n,1)*delta(k,1) - delta(n,2)*delta(k,2) - delta(n,3)*(delta(k,1) - delta(k,3)) with T(n,k<0) = T(n

T(n,0) = 1.

T(n,n) = delta(n mod 2,0).

T(n,1) = n-4 for n>3.

T(2*j+r,2*j-1) = 0 for j>0, r=-1,0,1,2.

T(n,2*j) = C(n/2,j)^2 for j>0 and n even and 2*j <= n <= 2*j+8.

T(n,2*j) = C((n-1)/2,j)*C((n+1)/2,j) for j>0 and n odd and 2*j < n < 2*j+8.

T(2*j+3*p,2*j-p) = C(j+3,4)^p for j>0 and p=0,1,2.

G.f. of row sums: (1-x-x^2)/(1-2*x-x^2+2*x^3).

G.f. of sums of T(n-4*k,k) over k: (1-x^5-x^7-x^10+x^15)/(1-x-x^5+x^6-x^7+x^8-x^9-2*x^10+x^11-x^12+2*x^15-x^16+2*x^17+x^20-x^25).

T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=4*k+1 if k>=0.

A085423 a(n) = floor(log_2(3n)).

Original entry on oeis.org

1, 2, 3, 3, 3, 4, 4, 4, 4, 4, 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, 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, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Ralf Stephan, Jun 30 2003

Length of the symmetric signed digit expansion of n with q=2 (i.e., the length of the representation of n in the (-1,0,1)_2 number system).

n occurs A001045(n) times. - Amiram Eldar, Feb 18 2024

C. Heuberger and H. Prodinger, On minimal expansions in redundant number systems: Algorithms and quantitative analysis, Computing 66(2001), 377-393.

Cf. A005578, A001045.

Cf. A000523, A008585, A002162.

a085423 = a000523 . a008585  -- Reinhard Zumkeller, Mar 16 2013

Floor[Log[2,3*Range[100]]] (* Harvey P. Dale, Oct 15 2016 *)

a(n) = A000523(A008585(n)). - Reinhard Zumkeller, Mar 16 2013

Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) (A002162). - Amiram Eldar, Feb 18 2024

A072176 Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..floor(n/2)} A071922(n-k,k).

Original entry on oeis.org

1, 1, 2, 3, 5, 9, 16, 30, 56, 106, 201, 382, 727, 1384, 2636, 5021, 9565, 18222, 34715, 66137, 126001, 240052, 457338, 871304, 1659978, 3162533, 6025150, 11478911, 21869232, 41664520, 79377833, 151227961, 288114394, 548905795

Offset: 1

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002

Based on the observation that F_{n+1} = Sum_{k} binomial (n-k,k). In both cases the sum is extended to 0<=2k<=n.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1,1).

Cf. A071922, A005578.

a:=[1,1,2,3,5];; for n in [6..40] do a[n]:=2*a[n-1]+a[n-2] -2*a[n-3]-a[n-4]+a[n-5]; od; a; # G. C. Greubel, Aug 26 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)) )); // G. C. Greubel, Aug 26 2019

seq(coeff(series(x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)), x, n+1), x, n), n = 1..40); # G. C. Greubel, Aug 26 2019

Rest@CoefficientList[ Series[x(1-x-x^2)/((1-x)(1-x-2x^2+x^4)), {x, 0, 40}], x] (* or *) LinearRecurrence[{2,1,-2,-1,1}, {1,1,2,3,5}, 40] (* Harvey P. Dale, Jun 23 2011 *)

my(x='x+O('x^40)); Vec(x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4))) \\ G. C. Greubel, Aug 26 2019

def A072176_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)) ).list()
a=A072176_list(40); a[1:] # G. C. Greubel, Aug 26 2019

G.f.: x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)).

a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=5, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Jun 23 2011

A073097 Let x(n) denote the number of 4's among the n first elements of the continued fraction for sum k>=0 1/2^(2^k) (A007400), y(n) the number of 6's and z(n) the number of 2's. Then a(n)=x(n)-y(n)-z(n)-1.

Original entry on oeis.org

-1, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1

Offset: 0

Benoit Cloitre, Aug 18 2002

sign

The positive sequence has a(n) = mod(A000120(A047849(n)),2) = mod(A000120(A078008(2n)),2) - Paul Barry, Jan 13 2005

Cosh(1) in 'reflected factorial' base is 1.10101010101010101010101010101010101010101010... - see A091337 for Sinh(1) (from Robert G. Wilson v, May 04 2005)

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A000120, A005578, A007400, A008619, A047849, A078008, A091337.

up_to = 65537;
A007400(n) = if(n<3, [0, 1, 4][n+1], if(n%8==1, A007400((n+1)/2), if(n%8==2, A007400((n+2)/2), [2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4][(n%16)+1]))); \\ From A007400
A073097list(up_to) = { my(v=vector(up_to), x4=0, y6=0, z2=0, k); for(n=1, up_to, k=A007400(n); if(2==k,z2++,if(4==k,x4++,if(6==k,y6++))); v[n] = (x4-y6-z2-1)); (v); };
v073097 = A073097list(up_to);
A073097(n) = if(!n,-1,v073097[n]); \\ Antti Karttunen, Jan 12 2019

It seems that a(2k+1) = 0 for k>=1.

The positive sequence (assuming the pattern continues) has g.f. (1+x-x^2)/((1-x)(1-x^2)), with a(n)=(1-(1)^n)/2+0^n = mod((1+A001045(n+1))/2, 2) = mod(A005578, 2). The partial sums are A008619(n+1). - Paul Barry, Apr 28 2004

A099754 a(n) = (3^n +1)/2 + 2^n.

Original entry on oeis.org

2, 4, 9, 22, 57, 154, 429, 1222, 3537, 10354, 30549, 90622, 269817, 805354, 2407869, 7207222, 21588897, 64701154, 193972389, 581655022, 1744440777, 5232273754, 15694724109, 47079978022, 141231545457, 423677859154, 1271000023029

Offset: 0

Miklos Kristof, Nov 11 2004

Let b(0)=1, b(n) = A005578(n-1) = {1,1,2,3,6,11,22,43,86,171,342, ...} then a(n) = Sum_{k=0..n+1} C(n+1,k)*b(k).

Binomial transform of A135351. - R. J. Mathar, Aug 05 2009

			a(6) = (3^6+1)/2 + 2^6 = 365+64 = 429.
a(6) = 1 + 7*1 + 21*1 + 35*2 + 35*3 + 21*6 + 7*11 + 1*22 = 429.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Yilmaz Simsek, New families of special numbers for computing negative order Euler numbers, arXiv:1604.05601 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A005578.

List([0..30], n-> (3^n +2^(n+1) +1)/2); # G. C. Greubel, Sep 03 2019

[(3^n +2^(n+1) +1)/2: n in [0..30]]; // G. C. Greubel, Sep 03 2019

seq((3^n +2^(n+1) +1)/2, n=0..30); # G. C. Greubel, Sep 03 2019

Table[(3^n +2^(n+1) +1)/2, {n,0,30}] (* G. C. Greubel, Sep 03 2019 *)
LinearRecurrence[{6,-11,6},{2,4,9},30] (* Harvey P. Dale, May 23 2021 *)

a(n) = (3^n+1)/2 + 2^n; \\ Michel Marcus, Aug 15 2013

[(3^n +2^(n+1) +1)/2 for n in (0..30)] # G. C. Greubel, Sep 03 2019

a(n) = (3^n + 2^(n+1) + 1)/2.
G.f.: (2-8*x+7*x^2)/((1-x)*(1-2*x)*(1-3*x)). - Jaume Oliver Lafont, Mar 06 2009
a(n) = A007051(n) + A000079(n). - Michel Marcus, Aug 15 2013
E.g.f.: (exp(x) + 2*exp(2*x) + exp(3*x))/2. - G. C. Greubel, Sep 03 2019

Corrected and extended by T. D. Noe, Nov 07 2006

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