A084175 Jacobsthal oblong numbers.
0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..500
- N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
- N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
- Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
- Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
- Index entries for linear recurrences with constant coefficients, signature (3,6,-8).
Crossrefs
Programs
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GAP
List([0..30], n-> (2^(2*n+1) -(-2)^n -1)/9); # G. C. Greubel, Sep 21 2019
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Magma
[(2*4^n-(-2)^n-1)/9: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011
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Maple
for n from 1 to 25 do print(round(2^n/3)*round(2^(n+1)/3)) od; # Gary Detlefs, Feb 10 2010
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Mathematica
Table[(2*4^n -(-2)^n -1)/9, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011, modified by G. C. Greubel, Sep 21 2019 *) LinearRecurrence[{3,6,-8}, {0,1,3}, 25] (* Jean-François Alcover, Sep 21 2017 *) -
PARI
a(n)=(2*4^n-(-2)^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015
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Python
def A084175(n): return ((m:=1<
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Sage
[gaussian_binomial(n, 2, -2) for n in range(1, 26)] # Zerinvary Lajos, May 28 2009
Formula
a(n) = (2*4^n - (-2)^n - 1)/9;
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3), a(0)=0, a(1)=1, a(2)=3.
G.f.: x/((1+2*x)*(1-x)*(1-4*x)).
E.g.f.: (2*exp(4*x) - exp(x) - exp(-2*x))/9.
a(n+1) - 4*a(n) = 1, -1, 3, -5, 11, ... = A001045(n+1) signed. - Paul Curtz, May 19 2008
a(n) = round(2^n/3) * round(2^(n+1)/3). - Gary Detlefs, Feb 10 2010
From Peter Bala, Mar 30 2015: (Start)
The shifted o.g.f. A(x) := 1/( (1 + 2*x)*(1 - x)*(1 - 4*x) ) = 1/(1 - 3*x - 6*x^2 + 8*x^3). Hence A(x) == 1/(1 - 3*x + 3*x^2 - x^3) (mod 9) == 1/(1 - x)^3 (mod 9). It follows by Theorem 1 of Heninger et al. that (A(x))^(1/3) = 1 + x + 4*x^2 + 10*x^3 + ... has integral coefficients.
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