A347637 -id:A347637 - cp's OEIS frontend

A016921 a(n) = 6*n + 1.

1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199, 205, 211, 217, 223, 229, 235, 241, 247, 253, 259, 265, 271, 277, 283, 289, 295, 301, 307, 313, 319, 325, 331

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 22 ).
Also solutions to 2^x + 3^x == 5 (mod 7). - Cino Hilliard, May 10 2003
Except for 1, exponents n > 1 such that x^n - x^2 - 1 is reducible. - N. J. A. Sloane, Jul 19 2005
Let M(n) be the n X n matrix m(i,j) = min(i,j); then the trace of M(n)^(-2) is a(n-1) = 6*n - 5. - Benoit Cloitre, Feb 09 2006
If Y is a 3-subset of an (2n+1)-set X then, for n >= 3, a(n-1) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
All composite terms belong to A269345 as shown in there. - Waldemar Puszkarz, Apr 13 2016
First differences of the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 773", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
For b(n) = A103221(n) one has b(a(n)-1) = b(a(n)+1) = b(a(n)+2) = b(a(n)+3) = b(a(n)+4) = n+1 but b(a(n)) = n. So-called "dips" in A103221. See the Avner and Gross remark on p. 178. - Wolfdieter Lang, Sep 16 2016
A (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. The sequence a(n) is also the minimum number of pebbles such that any assignment of those pebbles on a complete graph with 3 vertices is a next-player winning game in the two player impartial (k+1,k) pebbling game. These games are represented by A347637(3,n). - Joe Miller, Oct 18 2021
Interleaving of A017533 and A017605. - Leo Tavares, Nov 16 2021

			From _Ilya Gutkovskiy_, Apr 15 2016: (Start)
Illustration of initial terms:
                      o
                    o o o
              o     o o o
            o o o   o o o
      o     o o o   o o o
    o o o   o o o   o o o
o   o o o   o o o   o o o
n=0  n=1     n=2     n=3
(End)

Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 178.

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Tanya Khovanova, Recursive Sequences.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N)).
William A. Stein, The modular forms database.
Leo Tavares, Illustration: Hexagonal Lines.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093563 ((6, 1) Pascal, column m=1).
Cf. A000567 (partial sums), A002476 (primes), A005408, A008588, A016813, A016933, A016945, A016957, A017281, A017533, A128470, A158057, A161700, A161705, A161709, A161714.
a(n) = A007310(2*(n+1)); complement of A016969 with respect to A007310.
Cf. A287326 (second column).
Cf. A000217, A003215.
Cf. A001477, A017605.

List([0..60], n-> 6*n+1); # G. C. Greubel, Sep 18 2019

a016921 = (+ 1) . (* 6)
a016921_list = [1, 7 ..]  -- Reinhard Zumkeller, Jan 15 2013

[6*n+1: n in [0..60]]; // G. C. Greubel, Sep 18 2019

a[1]:=1:for n from 2 to 100 do a[n]:=a[n-1]+6 od: seq(a[n], n=1..56); # Zerinvary Lajos, Mar 16 2008

Range[1, 500, 6] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)

a(n)=6*n+1 \\ Charles R Greathouse IV, Mar 22 2016

for n in range(0,10**5):print(6*n+1) # Soumil Mandal, Apr 14 2016

a(n) = 6*n + 1, n >= 0 (see the name).
G.f.: (1+5*x)/(1-x)^2.
A008615(a(n)) = n. - Reinhard Zumkeller, Feb 27 2008
A157176(a(n)) = A013730(n). - Reinhard Zumkeller, Feb 24 2009
a(n) = 4*(3*n-1) - a(n-1) (with a(0)=1). - Vincenzo Librandi, Nov 20 2010
E.g.f.: (1 + 6*x)*exp(x). - G. C. Greubel, Sep 18 2019
a(n) = A003215(n) - 6*A000217(n-1). See Hexagonal Lines illustration. - Leo Tavares, Sep 10 2021
From Leo Tavares, Oct 27 2021: (Start)
a(n) = 6*A001477(n-1) + 7
a(n) = A016813(n) + 2*A001477(n)
a(n) = A017605(n-1) + A008588(n-1)
a(n) = A016933(n) - 1
a(n) = A008588(n) + 1. (End)
Sum_{n>=0} (-1)^n/a(n) = Pi/6 + sqrt(3)*arccoth(sqrt(3))/3. - Amiram Eldar, Dec 10 2021

A361315 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;1,1) pebbling game.

Original entry on oeis.org

31, 26, 19, 17, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41

Offset: 4

Gabrielle Demchak, Eugene Fiorini, Michael J. Herrera, Samuel Murray, Rhaldni Sayaman, Brittany Shelton and Wing Hong Tony Wong, Mar 14 2023

A (3;1,1) move in an impartial two-player pebbling game consists of removing three pebbles from a vertex and adding a pebble to each of two distinct adjacent vertices. The winning player is the one who makes the final allowable move. We start at n = 4 because we have shown that a(3) does not exist while a(2) is clearly undefined.

			For n = 4, a(4) = 31 is the least number of pebbles for which every game 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.

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.

Cf. A340631, A346197, A346401, A347637.

(*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 (3;1,1)-pebbling move; only work for n>=3.*)
pebblemoves[config_] :=  Block[{n, temp}, n = Length[config];   temp = Table[config, {i, n (n - 1) (n - 2)/2}] +     Permutations[Join[{-3, 1, 1}, Table[0, {i, n - 3}]]];   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, 4, 20}]

cp's OEIS Frontend

A016921 a(n) = 6*n + 1.

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