A167030 -id:A167030 - cp's OEIS frontend

A167193 a(n) = (1/3)(1 - (-2)^n + 3(-1)^n ) = (-1)^(n+1)*A167030(n).

1, 0, 0, 2, -4, 10, -20, 42, -84, 170, -340, 682, -1364, 2730, -5460, 10922, -21844, 43690, -87380, 174762, -349524, 699050, -1398100, 2796202, -5592404, 11184810, -22369620, 44739242, -89478484, 178956970, -357913940, 715827882, -1431655764, 2863311530, -5726623060, 11453246122, -22906492244, 45812984490

Offset: 0

Paul Curtz, Oct 30 2009

This is the inverse binomial transform of 1, 1, 1, 3, 5, 11,.. (continued as in A001045 and conjectured to be equal to A152046).

Any sequence (like this one) which obeys a(n)= -2a(n-1)+a(n-2)+2a(n-3) also obeys a(n)=5a(n-2)-4a(n-4), proved by telescoping; see A101622.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,1,2).

[( 1-(-1)^n*2^n)/3+(-1)^n: n in [0..40] ]; // Vincenzo Librandi, Aug 06 2011

LinearRecurrence[{-2,1,2},{1,0,0}, 25] (* or *) Table[(1/3)*(1 + 3*(-1)^n - (-2)^n), {n,0,25}] (* G. C. Greubel, Jun 04 2016 *)

a(2n) = (-1)^n* A084240(n). a(2n+1) = A020988(n).
G.f.: ( -1 - 2*x + x^2 ) / ( (x-1)*(1+2*x)*(1+x) ).
a(n) = -a(n-1) + 2*a(n-2) - 2*(-1)^n.
a(n) = -2*a(n-1) + a(n-2) + 2*a(n-3).
E.g.f.: (1/3)*(exp(x) + 3*exp(-x) - exp(-2*x)). - G. C. Greubel, Jun 04 2016

A026644 a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4.

Original entry on oeis.org

1, 2, 4, 10, 20, 42, 84, 170, 340, 682, 1364, 2730, 5460, 10922, 21844, 43690, 87380, 174762, 349524, 699050, 1398100, 2796202, 5592404, 11184810, 22369620, 44739242, 89478484, 178956970, 357913940, 715827882, 1431655764, 2863311530, 5726623060

Offset: 0

Clark Kimberling

Number of moves to solve Chinese rings puzzle.
a(n-1) (with a(0):=0) enumerates all sequences of length m=1,2,...,floor(n/2) with nonzero integer entries n_i satisfying sum |n_i| <= n-m. Rephrasing K. A. Meissner's example p. 6. Example n=4: from length m=1: [1], [2], [3], each in 2 signed versions; from m=2: [1,1] in 2^2 = 4 signed versions. Hence a(3) = a(4-1) = 3*2 + 1*4 = 10.
Also the number of different 3-colorings (out of 4 colors) for the vertices of all triangulated planar polygons on a base with n+1 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Mar 23 2010
For n > 0, also the total distance that the disks travel from the leftmost peg to the rightmost peg in the Tower of Hanoi puzzle, in the unique solution with 2^n-1 moves (see links). - Sela Fried, Dec 17 2023

Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.
Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13th International Puzzle Party, Amsterdam, Aug 20 1993.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
Sela Fried, Economically solving the Tower of Hanoi puzzle.
Nicolas Gastineau and O. Togni, On S-packing edge-colorings of cubic graphs, arXiv preprint arXiv:1711.10906 [cs.DM], 2017.
Lee Hae-hwang, Illustration of initial terms in terms of rosemary plants
Krzysztof A. Meissner, Black hole entropy in Loop Quantum Gravity, arXiv:gr-qc/0407052, 2004.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Row sums of A026637.
For n >= 1, equals twice A000975, also A001045 - 1.
A167030 is an essentially identical sequence.

[n eq 0 select 1 else (2^(n+2) -3-(-1)^n)/3 : n in [0..40]]; // G. C. Greubel, Jun 28 2024

f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi;

Join[{1}, Floor[(2^Range[3, 40] - 2)/3]] (* or *) LinearRecurrence[{2,1,-2},{1,2,4,10},40] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
CoefficientList[Series[(1-x^2+2x^3)/((1-x)(1-x-2x^2)),{x,0,1001}],x] (* Vincenzo Librandi, Apr 04 2012 *)

Vec((1-x^2+2*x^3)/(1-x)/(1-x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Apr 04 2012

def A026644(n): return ((4<Chai Wah Wu, Apr 17 2025

[(2^(n+2)-3-(-1)^n)/3 + int(n==0) for n in range(41)] # G. C. Greubel, Jun 28 2024

a(2*k) = 2*a(2*k-1), a(2*k+1) = 2*a(2*k) + 2. - Peter Shor, Apr 11 2002
For n>0: a(n+1) = a(n) + 2*b(n+1) + 4*b(n), where b(k) = A001045(k). - N. J. A. Sloane, May 16 2003
For n>0: if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3. - Richard Hess
a(2*n) = 2*n-1 + Sum_{k=0..2*n-1} a(k), n>0; a(2*n+1) = 2*n+1 + Sum_{k=0..n} a(k). - Lee Hae-hwang, Sep 17 2002; corrected by R. J. Mathar, Oct 21 2008
a(n) = 2*n + 2*Sum_{k=1..n-2} a(k), n>0. - Lee Hae-hwang, Sep 19 2002; corrected by R. J. Mathar, Oct 21 2008
From Paul Barry, Oct 24 2007: (Start)
G.f.: (1 - x^2 + 2*x^3)/((1 - x)*(1 - x - 2*x^2)).
a(n) = J(n+2) - 1 + 0^n, where J(n) = A001045(n) (Jacobsthal numbers).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
a(n) = 0^n + Sum_{k=0..n} (2 - 2*0^(n-k))*J(k+1). (End)
a(n) = A052953(n+1) - 2, n>0. [Moved from A020988, R. J. Mathar, Oct 21 2008]
a(n) = floor(A097074(n+1)/2), n>0. - Gary Detlefs, Dec 19 2010
a(n) = A169969(2*n-1) - 1, n>=2; a(n) = 3*2^(n-1) - 1 - A169969(2*n-7), n>=5 . - Yosu Yurramendi, Jul 05 2016
a(n+3) = 3*2^(n+2) - 2 - a(n), n>=1, a(1)=2, a(2)=4, a(3)=10 . - Yosu Yurramendi, Jul 05 2016
a(n) + A084170(n) = 3*2^n - 2, n>=1. - Yosu Yurramendi, Jul 05 2016
E.g.f: (3 - 4*cosh(x) + 4*cosh(2*x) - 2*sinh(x) + 4*sinh(2*x))/3. - Ilya Gutkovskiy, Jul 05 2016
a(n+3) = 9*2^n + A084170(n), n>=0. - Yosu Yurramendi, Jul 07 2016
a(n) = A000975(n+1) - A000035(n+1), n>0, a(0)=1. - Yuchun Ji, Aug 05 2020

Recurrence in definition line found by Lee Hae-hwang, Apr 03 2002

A128209 Jacobsthal numbers(A001045) + 1.

Original entry on oeis.org

1, 2, 2, 4, 6, 12, 22, 44, 86, 172, 342, 684, 1366, 2732, 5462, 10924, 21846, 43692, 87382, 174764, 349526, 699052, 1398102, 2796204, 5592406, 11184812, 22369622, 44739244, 89478486, 178956972, 357913942, 715827884, 1431655766, 2863311532, 5726623062

Offset: 0

Paul Barry, Feb 19 2007

Row sums of A128208.
Essentially the same as A052953. - R. J. Mathar, Jun 14 2008
Let I=I_n be the n X n identity matrix and P=P_n be the incidence matrix of the cycle (1,2,3,...,n). Then, for n >= 1, a(n+1) is the number of different representations of matrix P^(-1)+I+P by sum of permutation matrices. - Vladimir Shevelev, Apr 12 2010
a(n) is the rank of Fibonacci(n+2) in row n of A049456 (regarded as an irregular triangle read by rows). - N. J. A. Sloane, Nov 23 2016

V. S. Shevelyov (Shevelev), Extension of the Moser class of four-line Latin rectangles, DAN Ukrainy, 3(1992),15-19. [From Vladimir Shevelev, Apr 12 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. H. Lehmer, On Stern's Diatomic Series, Amer. Math. Monthly 36(1) 1929, pp. 59-67. [Annotated and corrected scanned copy]
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A167030, A153643, A049456.

[1+2^n/3-(-1)^n/3: n in [0..40]]; // Vincenzo Librandi, Aug 11 2011

CoefficientList[Series[(1-3*x^2)/(1-2*x-x^2+2*x^3),{x,0,40}],x] (* Vladimir Joseph Stephan Orlovsky, Jan 31 2012 *)

a(n)=2^n\/3+1 \\ Charles R Greathouse IV, Jan 31 2012

def A128209(n): return ((1<Chai Wah Wu, Apr 17 2025

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

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

A153772 a(n) = (2^n + 2*(-1)^n - 6)/3.

Original entry on oeis.org

-1, -2, 0, 0, 4, 8, 20, 40, 84, 168, 340, 680, 1364, 2728, 5460, 10920, 21844, 43688, 87380, 174760, 349524, 699048, 1398100, 2796200, 5592404, 11184808, 22369620, 44739240, 89478484, 178956968, 357913940, 715827880

Offset: 0

Paul Curtz, Jan 01 2009

The array of T(n,k) with T(0,k) = A141325(k) and successive differences T(n,k) = T(n-1,k+1) - T(n-1,k) in further rows is
   1,  1,  1,  1,   3,  5, 9, 13, 21, 33, 55,..
   0,  0,  0,  2,   2,  4, 4,  8, 12, 22,..
   0,  0,  2,  0,   2,  0, 4,  4, 10,...
   0,  2, -2,  2,  -2,  4, 0,  6,..
   2, -4,  4, -4,   6, -4, 6,..
  -6,  8, -8, 10, -10, 10,...
with T(n,n) = A078008(n), T(n,n+1) = -A167030(n), T(n,n+2) = A128209(n), T(n,n+3) = -a(n). All these sequences along the diagonals obey the recurrences a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) and a(n) = 5*a(n-2) - 4*a(n-4).
Conjecture: For n >= 6, a(n) is the third largest natural number whose Collatz orbit has length n+2. - Markus Sigg, Sep 14 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A000975, A005186, A033491.

[2^n/3 +2*(-1)^n/3-2: n in [0..40]]; // Vincenzo Librandi, Aug 07 2011

Table[(2^n + 2*(-1)^n - 6)/3, {n,0,25}] (* or *) LinearRecurrence[{2, 1, -2}, {-1, -2, 0}, 25] (* G. C. Greubel, Aug 27 2016 *)

a(n)=(2^n+2*(-1)^n-6)/3 \\ Charles R Greathouse IV, Aug 28 2016

a(n) = A078008(n) - 2.
a(n) = +2*a(n-1) +a(n-2) -2*a(n-3).
a(n) = a(n-1) + 2*a(n-2) + 4.
G.f.: (1 - 5*x^2) / ( (1-x)*(2*x-1)*(1+x) ).
E.g.f.: (1/3)*(2*exp(-x) - 6*exp(x) + exp(2*x)). - G. C. Greubel, Aug 27 2016
a(n) = 4*A000975(n-3) for n >= 3. - Markus Sigg, Sep 14 2020

A293961 Number T(n,k) of linear chord diagrams having n chords and maximal chord length k (or k=0 if n=0); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 4, 10, 0, 1, 10, 24, 70, 0, 1, 20, 82, 212, 630, 0, 1, 42, 300, 798, 2324, 6930, 0, 1, 84, 894, 3800, 10078, 30188, 90090, 0, 1, 170, 2744, 18186, 51804, 150046, 452724, 1351350, 0, 1, 340, 8594, 71624, 313006, 851692, 2545390, 7695828, 22972950

Offset: 0

Alois P. Heinz, Oct 20 2017

All terms in columns k > 1 are even.

			Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,   2;
  0, 1,   4,   10;
  0, 1,  10,   24,    70;
  0, 1,  20,   82,   212,   630;
  0, 1,  42,  300,   798,  2324,   6930;
  0, 1,  84,  894,  3800, 10078,  30188,  90090;
  0, 1, 170, 2744, 18186, 51804, 150046, 452724, 1351350;
  ...

Alois P. Heinz, Rows n = 0..20, flattened

Columns k=0-2 give: A000007, A057427, A167030(n+1).
Row sums give A001147.
Main diagonal gives A293962.
T(2n,n) gives A293963.
Cf. A293881, A293960.

A(n,k) = A293960(n,k) - A293960(n,k-1) for k>0, A(n,0) = A000007(n).

A257198 Number of permutations of length n having exactly one descent such that the first element of the permutation is an odd number.

Original entry on oeis.org

0, 0, 2, 6, 16, 36, 78, 162, 332, 672, 1354, 2718, 5448, 10908, 21830, 43674, 87364, 174744, 349506, 699030, 1398080, 2796180, 5592382, 11184786, 22369596, 44739216, 89478458, 178956942, 357913912, 715827852, 1431655734, 2863311498, 5726623028

Offset: 1

Ran Pan, Apr 18 2015

			a(3)=2: (1 3 2, 3 1 2).
a(4)=6: (1 2 4 3, 1 3 2 4, 1 4 2 3, 1 3 4 2, 3 1 2 4, 3 4 1 2).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).

Cf. A178420, A000295, A000975, A167030 (first differences).

[2*Floor((2*2^n-3*n-1)/6): n in [1..40]]; // Vincenzo Librandi, Apr 18 2015

Table[2 Floor[(2 2^n - 3 n - 1) / 6], {n, 50}] (* Vincenzo Librandi, Apr 18 2015 *)

concat([0,0], Vec(-2*x^3/((x-1)^2*(x+1)*(2*x-1)) + O(x^100))) \\ Colin Barker, Apr 19 2015

a(n)=(2<Charles R Greathouse IV, Apr 21 2015

a(n) = 2*floor((2*2^n-3*n-1)/6).
a(n) = 2*A178420(n-1).
a(n) = A000295(n)-A000975(n-1).
From Colin Barker, Apr 19 2015: (Start)
a(n) = (-3-(-1)^n+2^(2+n)-6*n)/6.
a(n) = 3*a(n-1)-a(n-2)-3*a(n-3)+2*a(n-4).
G.f.: -2*x^3 / ((x-1)^2*(x+1)*(2*x-1)).
(End)

cp's OEIS Frontend

A167193 a(n) = (1/3)*(1 - (-2)^n + 3*(-1)^n ) = (-1)^(n+1)*A167030(n).

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A026644 a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4.

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A128209 Jacobsthal numbers(A001045) + 1.

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A153772 a(n) = (2^n + 2*(-1)^n - 6)/3.

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A293961 Number T(n,k) of linear chord diagrams having n chords and maximal chord length k (or k=0 if n=0); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A257198 Number of permutations of length n having exactly one descent such that the first element of the permutation is an odd number.

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A167193 a(n) = (1/3)(1 - (-2)^n + 3(-1)^n ) = (-1)^(n+1)*A167030(n).