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A007582
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a(n) = 2^(n-1)*(1+2^n).
(Formerly M2849)
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66
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1, 3, 10, 36, 136, 528, 2080, 8256, 32896, 131328, 524800, 2098176, 8390656, 33558528, 134225920, 536887296, 2147516416, 8590000128, 34359869440, 137439215616, 549756338176, 2199024304128, 8796095119360, 35184376283136
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OFFSET
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0,2
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COMMENTS
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Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_8. Example: a(1)=3 because in the cycle ABCDEFGH we have three walks of length 3 between A and B: ABAB, ABCB and AHAB. - Emeric Deutsch, Apr 01 2004
Smallest number containing in its binary representation two equal non-overlapping subwords of length n: A097295(a(n))=n and A097295(m)<n for m<a(n). - Reinhard Zumkeller, Aug 04 2004
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either x equals y or x does not equal y. - Ross La Haye, Jan 02 2008
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A). This is just a simpler statement of my previous comment for this sequence. - Ross La Haye, Jan 10 2008
a(n+1) written in base 2: 11, 1010, 100100, 10001000, 1000010000, ..., i.e., number 1, n times 0, number 1, n times 0 (A163449(n)). - Jaroslav Krizek, Jul 27 2009
Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times sum_{k=0..n} binomial(n,k)*k^q, then A007582(x)= sum_{k=0..x-1} T(x,k)*2^k. - John M. Campbell, Nov 16 2011
a(n) gives the number of pairs (r, s) such that 0 <= r <= s <= (2^n)-1 that satisfy AND(r, s, XOR(r, s)) = 0. - Ramasamy Chandramouli, Aug 30 2012
Consider the quantum spin-1/2 chain with even number of sites L (physics, condensed matter theory). The spectrum of the Hamiltonian can be classified according to symmetries. If the only symmetry of the spin Hamiltonian is Parity, i.e., reflection with respect to the middle of the chain (see e.g. the transverse-field Ising model with open boundary conditions), then the dimension of the p=+1 parity sector is given by a(n) with n=L/2. - Marin Bukov, Mar 11 2016
a(n) is also the total number of words of length n, over an alphabet of four letters, of which one of them appears an even number of times. See the Lekraj Beedassy, Jul 22 2003, comment on A006516 (4-letter odd case), and the Balakrishnan reference there. For the 1- to 11-letter cases, see the crossrefs. - Wolfdieter Lang, Jul 17 2017
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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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) = 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) = 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) = (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
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MAPLE
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MATHEMATICA
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LinearRecurrence[{6, -8}, {1, 3}, 30] (* Harvey P. Dale, Apr 08 2013 *)
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PROG
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(PARI) a(n)=if(n<0, 0, 2^(n-1)*(1+2^n))
(PARI) a(n)=sum(k=-n\4, n\4, binomial(2*n+1, n+1+4*k))
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CROSSREFS
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The number of words of length n with m letters, one of them appearing an even number of times is for m = 1..11: A000035, A011782, A007051, A007582, A081186, A081187, A081188, A081189, A081190, A060531, A081192. - Wolfdieter Lang, Jul 17 2017
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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