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

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Examples

			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)
		

References

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

Links

• 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).

Crossrefs

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.

Programs

GAP

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

Haskell

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

Magma

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

Maple

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

Mathematica

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

PARI

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

Python

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

Formula

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