A007582 a(n) = 2^(n-1)*(1+2^n).
1, 3, 10, 36, 136, 528, 2080, 8256, 32896, 131328, 524800, 2098176, 8390656, 33558528, 134225920, 536887296, 2147516416, 8590000128, 34359869440, 137439215616, 549756338176, 2199024304128, 8796095119360, 35184376283136, 140737496743936, 562949970198528
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n=0..200
- M. Archibald, A. Blecher, A. Knopfmacher, M. E. Mays, Inversions and Parity in Compositions of Integers, J. Int. Seq., Vol. 23 (2020), Article 20.4.1.
- S. Hong and J. H. Kwak, Regular fourfold coverings with respect to the identity automorphism, J. Graph Theory, 17 (1993), 621-627.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 168
- Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
- Mircea Merca, A Note on Cosine Power Sums, J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
- Index entries for linear recurrences with constant coefficients, signature (6,-8)
Programs
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Magma
[Binomial(2^n + 1, 2) : n in [0..30]]; // Wesley Ivan Hurt, Jul 03 2020
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Maple
seq(binomial(-2^n, 2), n=0..23); # Zerinvary Lajos, Feb 22 2008
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Mathematica
Table[ Binomial[2^n + 1, 2], {n, 0, 23}] (* Robert G. Wilson v, Jul 30 2004 *) LinearRecurrence[{6,-8},{1,3},30] (* Harvey P. Dale, Apr 08 2013 *)
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Maxima
A007582(n):=2^(n-1)*(1+2^n)$ makelist(A007582(n),n,0,30); /* Martin Ettl, Nov 15 2012 */
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PARI
a(n)=if(n<0,0,2^(n-1)*(1+2^n))
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PARI
a(n)=sum(k=-n\4,n\4,binomial(2*n+1,n+1+4*k))
Formula
G.f.: (1-3*x)/((1-2*x)*(1-4*x)). C(1+2^n, 2) where C(n, 2) is n-th triangular number A000217.
E.g.f.: exp(3*x)*cosh(x). - Paul Barry, Apr 07 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*3^(n-2*k). - Paul Barry, May 08 2003
a(n+1) = 4*a(n) - 2^n; see also A049775. a(n) = 2^(n-1)*A000051(n). - Philippe Deléham, Feb 20 2004
a(n) = 6*a(n-1) - 8*a(n-2). - Emeric Deutsch, Apr 01 2004
Row sums of triangle A134308. - Gary W. Adamson, Oct 19 2007
a(n) = StirlingS2(2^n + 1,2^n) = 1 + 2*StirlingS2(n+1,2) + 3*StirlingS2(n+1,3) + 3*StirlingS2(n+1,4) = StirlingS2(n+2,2) + 3(StirlingS2(n+1,3) + StirlingS2(n+1,4)). - Ross La Haye, Mar 01 2008
a(n) = StirlingS2(2^n + 1,2^n) = 1 + 2*StirlingS2(n+1,2) + 3*StirlingS2(n+1,3) + 3*StirlingS2(n+1,4) = StirlingS2(n+2,2) + 3(StirlingS2(n+1,3) + StirlingS2(n+1,4)). - Ross La Haye, Apr 02 2008
a(n) = A028403(n+1) / 4. - Jaroslav Krizek, Jul 27 2009
a(n) = Sum_{k=-floor(n/4)..floor(n/4)} binomial(2*n,n+4*k)/2. - Mircea Merca, Jan 28 2012
G.f.: Q(0)/2 where Q(k) = 1 + 2^k/(1 - 2*x/(2*x + 2^k/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 10 2013
a(n) = Sum_{k=1..2^n} k. - Joerg Arndt, Sep 01 2013
a(n) = (1/3) * Sum_{k=2^n..2^(n+1)} k. - J. M. Bergot, Jan 26 2015
a(n+1) = 2*a(n) + 4^n. - Yuchun Ji, Mar 10 2017
Comments
. - Sean A. Irvine, Nov 06 2024