A084964 -id:A084964 - cp's OEIS frontend

A109613 Odd numbers repeated.

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 25, 27, 27, 29, 29, 31, 31, 33, 33, 35, 35, 37, 37, 39, 39, 41, 41, 43, 43, 45, 45, 47, 47, 49, 49, 51, 51, 53, 53, 55, 55, 57, 57, 59, 59, 61, 61, 63, 63, 65, 65, 67, 67, 69, 69, 71, 71, 73

Offset: 0

Reinhard Zumkeller, Aug 01 2005

The number of rounds in a round-robin tournament with n competitors. - A. Timothy Royappa, Aug 13 2011
Diagonal sums of number triangle A113126. - Paul Barry, Oct 14 2005
When partitioning a convex n-gon by all the diagonals, the maximum number of sides in resulting polygons is 2*floor(n/2)+1 = a(n-1) (from Moscow Olympiad problem 1950). - Tanya Khovanova, Apr 06 2008
The inverse values of the coefficients in the series expansion of f(x) = (1/2)*(1+x)*log((1+x)/(1-x)) lead to this sequence; cf. A098557. - Johannes W. Meijer, Nov 12 2009
From Reinhard Zumkeller, Dec 05 2009: (Start)
First differences: A010673; partial sums: A000982;
A059329(n) = Sum_{k = 0..n} a(k)*a(n-k);
A167875(n) = Sum_{k = 0..n} a(k)*A005408(n-k);
A171218(n) = Sum_{k = 0..n} a(k)*A005843(n-k);
A008794(n+2) = Sum_{k = 0..n} a(k)*A059841(n-k). (End)
Dimension of the space of weight 2n+4 cusp forms for Gamma_0(5). - Michael Somos, May 29 2013
For n > 4: a(n) = A230584(n) - A230584(n-2). - Reinhard Zumkeller, Feb 10 2015
The arithmetic function v+-(n,2) as defined in A290988. - Robert Price, Aug 22 2017
For n > 0, also the chromatic number of the (n+1)-triangular (Johnson) graph. - Eric W. Weisstein, Nov 17 2017
a(n-1), for n >= 1, is also the upper bound a_{up}(b), where b = 2*n + 1, in the first (top) row of the complete coach system Sigma(b) of Hilton and Pedersen [H-P]. All odd numbers <= a_{up}(b) of the smallest positive restricted residue system of b appear once in the first rows of the c(2*n+1) = A135303(n) coaches. If b is an odd prime a_{up}(b) is the maximum. See a comment in the proof of the quasi-order theorem of H-P, on page 263 ["Furthermore, every possible a_i < b/2 ..."]. For an example see below. - Wolfdieter Lang, Feb 19 2020
Satisfies the nested recurrence a(n) = a(a(n-2)) + 2*a(n-a(n-1)) with a(0) = a(1) = 1. Cf. A004001. - Peter Bala, Aug 30 2022
The binomial transform is 1, 2, 6, 16, 40, 96, 224, 512, 1152, 2560,.. (see A057711). - R. J. Mathar, Feb 25 2023

			G.f. = 1 + x + 3*x^2 + 3*x^3 + 5*x^4 + 5*x^5 + 7*x^6 + 7*x^7 + 9*x^8 + 9*x^9 + ...
Complete coach system for (a composite) b = 2*n + 1 = 33: Sigma(33) ={[1; 5], [5, 7, 13; 2, 1, 2]} (the first two rows are here 1 and 5, 7, 13), a_{up}(33) = a(15) = 15. But 15 is not in the reduced residue system modulo 33, so the maximal (odd) a number is 13. For the prime b = 31, a_{up}(31) = a(14) = 15 appears as maximum of the first rows. - _Wolfdieter Lang_, Feb 19 2020

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, 3rd printing 2012, pp. (260-281).

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Chromatic Number
Eric Weisstein's World of Mathematics, Johnson Graph
Eric Weisstein's World of Mathematics, Triangular Graph
Wikipedia, Round-robin tournament
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A063196, A110660, A186421, A186422, A211955, A230584, A290988.
Complement of A052928 with respect to the universe A004526. - Guenther Schrack, Aug 21 2018
First differences of A000982, A061925, A074148, A105343, A116940, and A179207. - Guenther Schrack, Aug 21 2018

a109613 = (+ 1) . (* 2) . (`div` 2)
a109613_list = 1 : 1 : map (+ 2) a109613_list
-- Reinhard Zumkeller, Oct 27 2012, Feb 21 2011

A109613:=n->2*floor(n/2)+1; seq(A109613(k), k=0..100); # Wesley Ivan Hurt, Oct 22 2013

Flatten@ Array[{2# - 1, 2# - 1} &, 37] (* Robert G. Wilson v, Jul 07 2012 *)
(# - Boole[EvenQ[#]] &) /@ Range[80] (* Alonso del Arte, Sep 11 2019 *)
With[{c=2*Range[0,40]+1},Riffle[c,c]] (* Harvey P. Dale, Jan 02 2020 *)

A109613(n)=n>>1<<1+1 \\ Charles R Greathouse IV, Feb 24 2011

def a(n) : return( len( CuspForms( Gamma0( 5), 2*n + 4, prec=1). basis())); # Michael Somos, May 29 2013

((1 to 49) by 2) flatMap { List.fill(2)() } // _Alonso del Arte, Sep 11 2019

a(n) = 2*floor(n/2) + 1.
a(n) = A052928(n) + 1 = 2*A004526(n) + 1.
a(n) = A028242(n) + A110654(n).
a(n) = A052938(n-2) + A084964(n-2) for n > 1. - Reinhard Zumkeller, Aug 27 2005
G.f.: (1 + x + x^2 + x^3)/(1 - x^2)^2. - Paul Barry, Oct 14 2005
a(n) = 2*a(n-2) - a(n-4), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 3. - Philippe Deléham, Nov 03 2008
a(n) = A001477(n) + A059841(n). - Philippe Deléham, Mar 31 2009
a(n) = 2*n - a(n-1), with a(0) = 1. - Vincenzo Librandi, Nov 13 2010
a(n) = R(n, -2), where R(n, x) is the n-th row polynomial of A211955. a(n) = (-1)^n + 2*Sum_{k = 1..n} (-1)^(n - k - 2)*4^(k-1)*binomial(n+k, 2*k). Cf. A084159. - Peter Bala, May 01 2012
a(n) = A182579(n+1, n). - Reinhard Zumkeller, May 06 2012
G.f.: ( 1 + x^2 ) / ( (1 + x)*(x - 1)^2 ). - R. J. Mathar, Jul 12 2016
E.g.f.: x*exp(x) + cosh(x). - Ilya Gutkovskiy, Jul 12 2016
From Guenther Schrack, Sep 10 2018: (Start)
a(-n) = -a(n-1).
a(n) = A047270(n+1) - (2*n + 2).
a(n) = A005408(A004526(n)). (End)
a(n) = A000217(n) / A004526(n+1), n > 0. - Torlach Rush, Nov 10 2023

A152832 a(0) = -2; a(n) = n - a(n-1) for n > 0.

Original entry on oeis.org

-2, 3, -1, 4, 0, 5, 1, 6, 2, 7, 3, 8, 4, 9, 5, 10, 6, 11, 7, 12, 8, 13, 9, 14, 10, 15, 11, 16, 12, 17, 13, 18, 14, 19, 15, 20, 16, 21, 17, 22, 18, 23, 19, 24, 20, 25, 21, 26, 22, 27, 23, 28, 24, 29, 25, 30, 26, 31, 27, 32, 28, 33, 29, 34, 30, 35, 31, 36, 32, 37, 33, 38, 34, 39, 35

Offset: 0

Vladimir Joseph Stephan Orlovsky, Dec 14 2008

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

Cf. A084964

lst={};a=2;Do[a=n-a;AppendTo[lst,a],{n,0,6!}];lst
RecurrenceTable[{a[0]==-2,a[n]==n-a[n-1]},a,{n,80}] (* or *) With[ {nn=40}, Riffle[ Range[-2,nn-2],Range[3,nn+3]]] (* Harvey P. Dale, Jun 02 2019 *)

From R. J. Mathar, Jan 03 2009, Aug 14 2009: (Start)
a(n) = (n+1)/2-(9*(-1)^n+1)/4.
G.f.: -(2*x-1)*(x-2)/((1+x)*(x-1)^2).
a(n) = a(n-1)+a(n-2)-a(n-3). (End)

Definition corrected by N. J. A. Sloane, Jan 11 2009

Formula adapted to offset by R. J. Mathar, Aug 14 2009

A052938 Expansion of (1 + 2x - 2x^2)/( (1+x)*(1-x)^2 ).

Original entry on oeis.org

1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8, 10, 9, 11, 10, 12, 11, 13, 12, 14, 13, 15, 14, 16, 15, 17, 16, 18, 17, 19, 18, 20, 19, 21, 20, 22, 21, 23, 22, 24, 23, 25, 24, 26, 25, 27, 26, 28, 27, 29, 28, 30, 29, 31, 30, 32, 31, 33, 32, 34, 33, 35, 34, 36, 35, 37, 36, 38, 37, 39

Offset: 0

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

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 929
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A028242 (same sequence with 1,0,2 prefix).

Cf. A035106.

List([0..30], n-> (2*n+7-3*(-1)^n)/4); # G. C. Greubel, Oct 18 2019

a052938 n = a052938_list !! n
a052938_list = 1 : 3 : 2 : zipWith (-) [5..] a052938_list
-- Reinhard Zumkeller, Oct 06 2015

[(2*n+7-3*(-1)^n)/4: n in [0..30]]; // G. C. Greubel, Oct 18 2019

R:=PowerSeriesRing(Integers(), 75); Coefficients(R!( (1 + 2*x - 2*x^2)/( (1+x)*(1-x)^2 ))); // Marius A. Burtea, Oct 18 2019

spec := [S,{S=Prod(Union(Sequence(Z),Z,Z),Sequence(Prod(Z,Z)))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
seq((2*n+7-3*(-1)^n)/4, n=0..30); # G. C. Greubel, Oct 18 2019

LinearRecurrence[{1,1,-1},{1,3,2},80] (* Harvey P. Dale, Apr 10 2019 *)

a(n)=([0,1,0; 0,0,1; -1,1,1]^n*[1;3;2])[1,1] \\ Charles R Greathouse IV, Sep 02 2015

[(2*n+7-3*(-1)^n)/4 for n in (0..30)] # G. C. Greubel, Oct 18 2019

G.f.: (1+2*x-2*x^2)/((1+x)*(1-x)^2).
a(n) = -a(n-1) + n + 3, with a(0)=1.
a(n) = (3*(-1)^(n+1) + 2*n + 7)/4.
A112034(n) = 3*2^a(n); a(n) = A109613(n+2) - A084964(n). - Reinhard Zumkeller, Aug 27 2005
a(n) = A035106(n+3) - A035106(n+2). - Reinhard Zumkeller, Oct 06 2015
a(n) = A060762(n+1) - 1. - Filip Zaludek, Nov 19 2016
E.g.f.: ((5+x)*sinh(x) + (2+x)*cosh(x))/2. - G. C. Greubel, Oct 18 2019

More terms from James Sellers, Jun 06 2000

A340631 a(n) is the minimum number of pebbles such that any assignment of those pebbles on a complete graph with n vertices is a next-player winning game in the two-player impartial pebbling game.

Original entry on oeis.org

7, 23, 7, 13, 9, 15, 11, 17, 13, 19, 15, 21, 17, 23, 19, 25, 21, 27, 23, 29, 25, 31, 27, 33, 29, 35, 31, 37, 33, 39, 35, 41, 37, 43, 39, 45, 41, 47, 43, 49, 45, 51, 47, 53, 49, 55, 51, 57, 53, 59, 55, 61, 57, 63, 59, 65, 61, 67, 63, 69, 65, 71, 67, 73, 69, 75

Offset: 3

Eugene Fiorini, Max C. Lind, Andrew Woldar, Wing Hong Tony Wong, Jan 13 2021

A move in an impartial two-player pebbling game consists of removing two pebbles from a vertex and adding one pebble to an adjacent vertex. The winning player is the one who makes the final allowable move. We start at n = 3 because we have shown a(2) does not exist.

			For n=3, a(3)=7 is the least number of pebbles for which every game in a next-player winning game regardless of assignment.

E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays, Vol. 1, CRC Press, 2001.

Michael De Vlieger, Table of n, a(n) for n = 3..10000
Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, and Tony W. H. Wong, Generalized Impartial Two-player Pebbling Games on K_3 and C_4, J. Int. Seq. (2024) Vol. 27, Issue 5, Art. No. 24.5.8. See p. 3.
Eugene Fiorini, Max Lind, Andrew Woldar, and Tony W. H. Wong, Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, J. Int. Seq., Vol. 24 (2021), Article 21.6.4.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A084964.

(* Given n and m, list all possible assignments. *)
alltuples[n_, m_] := IntegerPartitions[m + n, {n}] - 1;
(* Given an assignment, list all resultant assignments after one pebbling move; only work for n>=3. *)
pebblemoves[config_] := Block[{n, temp}, n = Length[config]; temp = Table[config, {i, n (n - 1)}] + Permutations[Join[{-2, 1}, Table[0, {i, n - 2}]]]; temp = Select[temp, Min[#] >= 0 &]; temp = ReverseSort[DeleteDuplicates[ReverseSort /@ temp]]];
(* Given n and m, list all assignments that are P-games. *)
Plist = {};plist[n_, m_] := Block[{index, tuples}, While[Length[Plist] < n, index = Length[Plist]; AppendTo[Plist, {{Join[{1}, Table[0, {i, index}]]}}]]; Do[AppendTo[Plist[[n]], {}]; tuples = alltuples[n, i]; Do[If[Not[IntersectingQ[pebblemoves[tuples[[j]]], Plist[[n, i - 1]]]], AppendTo[Plist[[n, i]], tuples[[j]]]], {j, Length[tuples]}], {i, Length[Plist[[n]]] + 1, m}]; Plist[[n, m]]];
(* Given n, print out the minimum m such that there are no P-games with m pebbles *)
Do[m = 1; While[plist[n, m] != {}, m++]; Print["n=", n, " m=", m], {n, 3, 20}]

For n>2, a(2n-1)=2n+1, a(2n)=a(2n+5)=2n+7.
G.f.: x^3*(12*x^4-10*x^3-23*x^2+16*x+7)/((x+1)*(x-1)^2).
For n>=5, a(n) = 2*A084964(n+4) - 1 = 2*A084964(n+2) + 1. - Hugo Pfoertner, Jan 14 2021

A217764 Array defined by a(n,k) = floor((k+2)/2)3^n - floor((k+1)/2)2^n, read by antidiagonals.

Original entry on oeis.org

1, 3, 0, 9, 1, 1, 27, 5, 4, 0, 81, 19, 14, 2, 1, 243, 65, 46, 10, 5, 0, 729, 211, 146, 38, 19, 3, 1, 2187, 665, 454, 130, 65, 15, 6, 0, 6561, 2059, 1394, 422, 211, 57, 24, 4, 1, 19683, 6305, 4246, 1330, 665, 195, 84, 20, 7, 0, 59049, 19171, 12866, 4118, 2059, 633, 276, 76, 29, 5, 1

Offset: 0

Ross La Haye, Mar 23 2013

Columns 0,1,2,3 respectively correspond to relations R_3, R_4, R_0, R_1 defined in La Haye paper listed below.

			a(4,4) = 211 because floor((4+2)/2)*3^4 - floor((4+1)/2)*2^4 = 3*3^4 - 2*2^4 = 243 - 32 = 211.

Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.

Cf. a(1,k) = A084964(k+2); a(n,0) = A000244(n); a(n,1) = A001047(n); a(n,2) = A027649(n); a(n,3) = A056182(n); a(n,4) = A001047(n+1); a(n,5) = A210448(n); a(n,6) = A166060(n); a(n,7) = A145563(n); a(n,8) = A102485(n).

a(n,k) = floor((k+2)/2)*3^n - floor((k+1)/2)*2^n. a(n,k) = 5*a(n-1,k) - 6*a(n-2,k); a(0,k) = floor((k+2)/2) - floor((k+1)/2), a(1,k) = floor((k+2)/2)*3 - floor((k+1)/2)*2.

A346197 a(n) is the minimum number of pebbles such that any assignment of those pebbles on K_5 is a next-player winning game in the two-player impartial (n+1,n) pebbling game.

Original entry on oeis.org

7, 15, 21, 27, 33, 39, 47, 53, 59, 67, 73, 79, 87, 93, 99, 107, 113, 119, 127, 133, 139

Offset: 1

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, Wing Hong Tony Wong, Jul 09 2021

For n>0, an (n+1,n) pebbling move involves removing n+1 pebbles from a vertex in a simple graph and placing n pebbles on an adjacent vertex.

A two-player impartial (n+1,n) pebbling game involves two players alternating (n+1,n) pebbling moves. The first player unable to make a move loses.

			For n=1, a(1)=7 is the least number of pebbles for which every (2,1) game on K_5 is a next-player winning game regardless of assignment.

E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays, Vol. 1, CRC Press, 2001.

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, and Tony W. H. Wong, Generalized Impartial Two-player Pebbling Games on K_3 and C_4, J. Int. Seq. (2024) Vol. 27, Issue 5, Art. No. 24.5.8. See p. 4.
Eugene Fiorini, Max Lind, Andrew Woldar, and Tony W. H. Wong, Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, Journal of Integer Sequences, (2021) Vol. 24, Issue 6, Art. No. 21.6.4.

Cf. A340631, A084964.

Do[remove = k + 1; add = k;
(*Given n and m, list all possible assignments.*)
alltuples[n_, m_] := IntegerPartitions[m + n, {n}] - 1;
(*Given an assignment, list all resultant assignments after one pebbling move; only work for n>=3.*)
pebblemoves[config_] :=  Block[{n, temp},
    n = Length[config];
    temp = Table[config, {i, n (n - 1)}] +
        Permutations[Join[{-remove, add}, Table[0, {i, n - 2}]]];
    temp = Select[temp, Min[#] >= 0 &];
    temp = ReverseSort[DeleteDuplicates[ReverseSort /@ temp]]];
(*Given n and m, list all assignments that are P-games.*)
Plist = {};
plist[n_, m_] :=  Block[{index, tuples},
    While[Length[Plist] < n, index = Length[Plist];
        AppendTo[Plist, {{Join[{1}, Table[0,{i,index}]]}}]];
    Do[AppendTo[Plist[[n]], {}]; tuples = alltuples[n, i];
        Do[If[Not[IntersectingQ[pebblemoves[tuples[[j]]],
                If[i > (remove - add), Plist[[n, i - (remove - add)]], {}]]],
            AppendTo[Plist[[n, i]], tuples[[j]]]], {j, Length[tuples]}],
    {i, Length[Plist[[n]]] + 1, m}]; Plist[[n, m]]];
Do[m = 1; While[plist[n, m] != {}, m++]; Print[" k=", k, " m=", m], {n, 5, 5}],
{k, 1, 21}]

A097065 Interleave n+1 and n-1.

Original entry on oeis.org

1, -1, 2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 5, 8, 6, 9, 7, 10, 8, 11, 9, 12, 10, 13, 11, 14, 12, 15, 13, 16, 14, 17, 15, 18, 16, 19, 17, 20, 18, 21, 19, 22, 20, 23, 21, 24, 22, 25, 23, 26, 24, 27, 25, 28, 26, 29, 27, 30, 28, 31, 29, 32, 30, 33, 31, 34, 32, 35, 33, 36, 34, 37, 35, 38

Offset: 0

Paul Barry, Jul 22 2004

Pairwise sums are abs(A023443), or n - 1 + 2*0^n. The partial sums of this sequence is A000124, with extra leading 1. Partial sums are A097066. Binomial transform is A097067.

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

Essentially the same as A084964.

Cf. A000124, A023443, A097066, A097067.

import Data.List (transpose)
a097065 n = n' - 2 * m where (n', m) = divMod (n + 2) 2
a097065_list = concat $ transpose [[1 ..], [-1 ..]]

[(2*n-1)/4 + 5*(-1)^n/4 : n in [0..100]]; // Wesley Ivan Hurt, Jan 10 2017

A097065:=n->(2*n-1)/4 + 5*(-1)^n/4: seq(A097065(n), n=0..150); # Wesley Ivan Hurt, Jan 10 2017

Table[(2n - 1)/4 + 5(-1)^n/4, {n, 0, 75}] (* Or *) Flatten[ Table[{n + 1, n - 1}, {n, 0, 37}]] (* Or *) CoefficientList[Series[(1 - 2x + 2x^2)/((1 + x)(1 - x)^2), {x, 0, 75}], x] (* Robert G. Wilson v, Jul 24 2004 *)

a(n)=n\2+1-n%2*2 \\ Charles R Greathouse IV, Sep 02 2015

G.f.: (1 - 2*x + 2*x^2)/((1 + x)*(1 - x)^2).
a(n) = (2*n - 1)/4 + 5*(-1)^n/4.
a(n) = floor((n+2)/2) - 2 * (n mod 2). - Reinhard Zumkeller, Apr 06 2015
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2. - Wesley Ivan Hurt, Jan 10 2017
E.g.f.: ((2 + x)*cosh(x) - (3 - x)*sinh(x))/2. - Stefano Spezia, Jul 01 2023

A112032 Denominator of 3/4 + 1/4 - 3/8 - 1/8 + 3/16 + 1/16 - 3/32 - 1/32 + 3/64 ...

Original entry on oeis.org

4, 1, 8, 2, 16, 4, 32, 8, 64, 16, 128, 32, 256, 64, 512, 128, 1024, 256, 2048, 512, 4096, 1024, 8192, 2048, 16384, 4096, 32768, 8192, 65536, 16384, 131072, 32768, 262144, 65536, 524288, 131072, 1048576, 262144, 2097152, 524288, 4194304, 1048576

Offset: 0

Reinhard Zumkeller, Aug 27 2005

Denominator of partial sums of A112030(n)/A016116(n+4), numerators = A112031;
A112031(n)/a(n) - 2/3 = (-1)^floor(n/2) / A112033(n);
lim_{n->infinity} A112031(n)/a(n) = 2/3.

G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 4, Sect. 1, Problem 148.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A016116, A112030, A112031, A112033.

[2^(Floor(n/2) + 1 + (-1)^n): n in [0..50]]; // Vincenzo Librandi, Aug 17 2011

LinearRecurrence[{0,2},{4,1},50] (* following conjecture in Formula field above *) (* Harvey P. Dale, Dec 21 2014 *)

m=50; v=concat([4,1], vector(m-2)); for(n=3, m, v[n]=2*v[n-2]); v \\ G. C. Greubel, Nov 08 2018

a(n) = 2^(floor(n/2) + 1 + (-1)^n) = 2^A084964(n).
Conjectures from Colin Barker, Apr 05 2013: (Start)
a(n) = 2*a(n-2).
G.f.: (x+4) / (1-2*x^2). (End)

a(21) corrected by Vincenzo Librandi, Aug 17 2011

A112033 a(n) = 3 * 2^(floor(n/2) + 1 + (-1)^n).

Original entry on oeis.org

12, 3, 24, 6, 48, 12, 96, 24, 192, 48, 384, 96, 768, 192, 1536, 384, 3072, 768, 6144, 1536, 12288, 3072, 24576, 6144, 49152, 12288, 98304, 24576, 196608, 49152, 393216, 98304, 786432, 196608, 1572864, 393216, 3145728, 786432, 6291456, 1572864

Offset: 0

Reinhard Zumkeller, Aug 27 2005

George Pólya and Gábor Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chapter 4, Sect. 1, Problem 148.

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A084964, A112030, A112031, A112032.

A112033:=n->3*2^(floor(n/2) + 1 + (-1)^n); seq(A112033(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013

Table[3*2^(Floor[n/2] + 1 + (-1)^n), {n,0,50}] (* Wesley Ivan Hurt, Nov 01 2013 *)

a(n) = 3 * 2^(n\2 + 1 + (-1)^n); \\ Michel Marcus, Nov 02 2013

def A112033(n): return 3*(1<<(n>>1)+(int(not n&1)<<1)) # Chai Wah Wu, Jan 17 2023

a(n) = 1 / abs(A112031(n)/A112032(n) - 2/3). (previous name)
a(n) = 3*2^A084964(n) = 3*A112032(n).
From Ralf Stephan, Jul 16 2013: (Start)
Recurrence: a(n) = 2a(n-2), a(0)=12, a(1)=3.
G.f.: (6*x+24)/(1-2*x^2). (End)
From Amiram Eldar, May 11 2025: (Start)
Sum_{n>=0} 1/a(n) = 5/6.
Sum_{n>=0} (-1)^n/a(n) = -1/2. (End)

A346401 a(n) is the minimum number of pebbles such that any assignment of those pebbles on a complete graph with n vertices is a next-player winning game in the two-player impartial (3, 2) pebbling game.

Original entry on oeis.org

13, 21, 15, 21, 17, 25, 21, 29, 25, 33, 29, 37, 33, 41, 37, 45, 41, 49, 45, 53, 49, 57

Offset: 3

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, Wing Hong Tony Wong, Jul 15 2021

A (3,2) pebbling move involves removing 3 pebbles from a vertex in a simple graph and placing 2 pebbles on an adjacent vertex.

A two-player impartial (3,2) pebbling game involves two players alternating (3,2) pebbling moves. The first player unable to make a move loses.

			For n=6, a(6)=21 is the least number of pebbles for which every (3,2) game on K_6 is a next-player winning game regardless of assignment.
For n=7, a(7)=17 is the least number of pebbles for which every (3,2) game on K_7 is a next-player winning game regardless of assignment.

E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays, Vol. 1, CRC Press, 2001.

E. Fiorini, M. Lind, A. Woldar, and T. W. H. Wong,Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, Journal of Integer Sequences, 24(6), 2021.

Cf. A340631, A084964, A346197.

remove = 3; add = 2;
(*Given n and m,list all possible assignments.*)
alltuples[n_, m_] := IntegerPartitions[m + n, {n}] - 1;
(*Given an assignment,list all resultant assignments after one pebbling move; only work for n>=3.*)
pebblemoves[config_] := Block[{n, temp},
    n = Length[config];
    temp = Table[config, {i, n (n - 1)}] +
        Permutations[Join[{-remove, add}, Table[0, {i, n - 2}]]];
    temp = Select[temp, Min[#] >= 0 &];
    temp = ReverseSort[DeleteDuplicates[ReverseSort /@ temp]]];
(*Given n and m,list all assignments that are P-games.*)
Plist = {};
plist[n_, m_] :=  Block[{index, tuples},
    While[Length[Plist] < n, index = Length[Plist];
        AppendTo[Plist, {{Join[{1}, Table[0, {i, index}]]}}]];
    Do[AppendTo[Plist[[n]], {}]; tuples = alltuples[n, i];
        Do[If[Not[IntersectingQ[pebblemoves[tuples[[j]]],
                If[i > (remove - add), Plist[[n, i - (remove - add)]], {}]]],
            AppendTo[Plist[[n, i]], tuples[[j]]]], {j, Length[tuples]}],
    {i, Length[Plist[[n]]] + 1, m}]; Plist[[n, m]]];
Do[m = 1; While[plist[n, m] != {}, m++]; Print[" n=", n, " m=", m], {n, 3, 24}]

a(n) = 2n+3 when n >= 7 is odd (conjectured).

a(n) = 2n+9 when n >= 6 is even (conjectured).

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A109613 Odd numbers repeated.

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A152832 a(0) = -2; a(n) = n - a(n-1) for n > 0.

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A052938 Expansion of (1 + 2*x - 2*x^2)/( (1+x)*(1-x)^2 ).

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A340631 a(n) is the minimum number of pebbles such that any assignment of those pebbles on a complete graph with n vertices is a next-player winning game in the two-player impartial pebbling game.

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A217764 Array defined by a(n,k) = floor((k+2)/2)*3^n - floor((k+1)/2)*2^n, read by antidiagonals.

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A346197 a(n) is the minimum number of pebbles such that any assignment of those pebbles on K_5 is a next-player winning game in the two-player impartial (n+1,n) pebbling game.

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A097065 Interleave n+1 and n-1.

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A052938 Expansion of (1 + 2x - 2x^2)/( (1+x)*(1-x)^2 ).

A217764 Array defined by a(n,k) = floor((k+2)/2)3^n - floor((k+1)/2)2^n, read by antidiagonals.