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A016921 a(n) = 6*n + 1. 125
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
REFERENCES
Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 178.
LINKS
Tanya Khovanova, Recursive Sequences.
William A. Stein, The modular forms database.
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
EXAMPLE
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)
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 *)
PROG
(Haskell)
a016921 = (+ 1) . (* 6)
a016921_list = [1, 7 ..] -- Reinhard Zumkeller, Jan 15 2013
(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
(Magma) [6*n+1: n in [0..60]]; // G. C. Greubel, Sep 18 2019
(GAP) List([0..60], n-> 6*n+1); # G. C. Greubel, Sep 18 2019
CROSSREFS
Cf. A093563 ((6, 1) Pascal, column m=1).
a(n) = A007310(2*(n+1)); complement of A016969 with respect to A007310.
Cf. A287326 (second column).
Sequence in context: A059335 A070419 A080199 * A331425 A260682 A354936
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified March 18 22:34 EDT 2024. Contains 370951 sequences. (Running on oeis4.)