OFFSET
0,2
COMMENTS
Sum of divisors of 4^n. - Paul Barry, Oct 13 2005
If x = a(n), y = A000079(n+1) and z = A087289(n), then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
It seems that a(n) divides A001676(3+4n). Several other entries apparently have this sequence embedded in them, e.g., A014551, A168604, A213243, A213246-8, and A279872. - Tom Copeland, Dec 27 2016
To elaborate on Librandi's comment from 2014: all these numbers, even if prime in Z, are sure not to be prime in Z[sqrt(2)], since a(n) can at least be factored as ((2^(2n + 1) - 1) - (2^(2n) - 1)*sqrt(2))((2^(2n + 1) - 1) + (2^(2n) - 1)*sqrt(2)). For example, 7 = (3 - sqrt(2))(3 + sqrt(2)), 31 = (7 - 3*sqrt(2))(7 + 3*sqrt(2)), 127 = (15 - 7*sqrt(2))(15 + 7*sqrt(2)). - Alonso del Arte, Oct 17 2017
Largest odd factors of A147590. - César Aguilera, Jan 07 2020
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Roudy El Haddad, Recurrent Sums and Partition Identities, arXiv:2101.09089 [math.NT], 2021.
Roudy El Haddad, A generalization of multiple zeta value. Part 1: Recurrent sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata with even rule numbers, Fig 11.
Robert Schneider, Partition zeta functions, Research in Number Theory, 2(1):9, 2016.
Eric Weisstein's World of Mathematics, Rule 220
Index entries for linear recurrences with constant coefficients, signature (5,-4).
FORMULA
G.f.: (1+2*x)/((1-x)*(1-4*x)).
E.g.f.: 2*exp(4*x)-exp(x).
With a leading zero, this is a(n) = (4^n - 2 + 0^n)/2, the binomial transform of A080925. - Paul Barry, May 19 2003
From Benoit Cloitre, Jun 18 2004: (Start)
a(n) = (-16^n/2)*B(2n, 1/4)/B(2n) where B(n, x) is the n-th Bernoulli polynomial and B(k) = B(k, 0) is the k-th Bernoulli number.
a(n) = 5*a(n-1) - 4*a(n-2).
a(n) = (-4^n/2)*B(2*n, 1/2)/B(2*n). (End)
a(n) = Stirling2(2*(n+1), 2). - Zerinvary Lajos, Dec 06 2006
a(n) = 4*a(n-1) + 3 with n > 0, a(0) = 1. - Vincenzo Librandi, Dec 30 2010
a(n) = 6*A002450(n) + 1. - Roderick MacPhee, Jul 06 2012
a(n) = Sum_{i = 0..n} binomial(2n+2, 2i). - Wesley Ivan Hurt, Mar 14 2015
a(n) = (1/4^n) * Sum_{k = 0..n} binomial(2*n+1,2*k)*9^k. - Peter Bala, Feb 06 2019
a(n) = numerator(zeta_star({2}_(n + 1))/zeta(2*n + 2)) where zeta_star is the multiple zeta star values and ({2}_n) represents (2, ..., 2) where the multiplicity of 2 is n. - Roudy El Haddad, Feb 22 2022
MAPLE
seq(2*4^n-1, n = 0..22); # Peter Luschny, Aug 17 2011
MATHEMATICA
2 * 4^Range[0, 31] - 1 (* Alonso del Arte, Oct 17 2017 *)
PROG
(Magma) [2*4^n-1 : n in [0..30]]; // Wesley Ivan Hurt, Mar 14 2015
(PARI) a(n)=2*4^n-1 \\ Charles R Greathouse IV, Sep 24 2015
(Haskell)
a083420 = subtract 1 . (* 2) . (4 ^) -- Reinhard Zumkeller, Dec 22 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Apr 29 2003
STATUS
approved