

A006516


a(n) = 2^(n1)*(2^n  1), n >= 0.
(Formerly M4183)


127



0, 1, 6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528
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OFFSET

0,3


COMMENTS

a(n) is also the number of different lines determined by pair of vertices in an ndimensional hypercube. The number of these lines modulo being parallel is in A003462.  Ola Veshta (olaveshta(AT)mydeja.com), Feb 15 2001
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)mydeja.com), Jun 01 2001
a(n) is the number of nletter words formed using four distinct letters, one of which appears an odd number of times.  Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69.  Wolfdieter Lang, Jul 16 2017]
Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket.  Lekraj Beedassy, May 14 2004
Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8.  Herbert Kociemba, Jul 02 2004
The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...).  Paul Barry, Aug 05 2005
Number of monic irreducible polynomials of degree 2 in GF(2^n)[x].  Max Alekseyev, Jan 23 2006
The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs.  Daniele Parisse, Jul 28 2006
8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology."  Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007
If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n1).  Artur Jasinski, Nov 26 2007
Let P(A) be the power set of an nelement set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y.  Ross La Haye, Jan 02 2008
Wieder calls these "conjoint usual 2combinations." The set of "conjoint strict kcombinations" is the subset of conjoint usual kcombinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n  2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}.  Ross La Haye, Jan 15 2008
a(n) is also the number whose binary representation is A109241(n1), for n>0.  Omar E. Pol, Aug 31 2008
If we define a spoofperfect number as:
A spoofperfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.
And if we define a "strong" spoofperfect number as:
A "strong" spoofperfect number is a spoofperfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.
The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.
Then:
If 2^n1 is odd composite but taken as a spoof prime then 2^(n1)*(2^n  1) is an even spoof perfect number (and moreover "strong" spoofperfect).
For example:
a(8) = 2^(81)*(2^8  1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);
sigma(a(8)) = (2^8  1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoofperfect (and also "strong" spoofperfect since 255 is obtained additively);
a(11) = 2^(111)*(2^11  1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);
sigma(a(11)) = (2^11  1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoofperfect (and also "strong" spoofperfect since 2047 is obtained additively).
I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoofperfect numbers. Maybe some other terminology is used.
An example of an even "weak" spoofperfect number would be:
n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);
sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoofperfect (but is not "strong" spoofperfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).
Euler proved:
If 2^k  1 is a prime number, then 2^(k1)*(2^k  1) is a perfect number and every even perfect number has this form.
The following seems to be true (is there a proof?):
If 2^k  1 is an odd composite number taken as a spoof prime, then 2^(k1)*(2^k  1) is a "strong" spoofperfect number and every even "strong" spoofperfect number has this form?
There is only one known odd spoofperfect number (found by Rene Descartes) but it is a "weak" spoofperfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)
Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...).  Gary W. Adamson, Oct 26 2010
a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n  1 stages.  Omar E. Pol, Nov 20 2010
The number of ndimensional odd theta functions of halfintegral characteristic. (Gunning, p.22)  Michael Somos, Jan 03 2014
a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n.  J. M. Bergot, May 07 2014
Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the kth selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m1))/2m. Reiterated, a(n) is the (m1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy.  Tony Bartoletti, Oct 30 2014
Beginning at a(1) = 1, this sequence is the sum of the first 2^(n1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29.  J. M. Bergot, Dec 07 2014
a(n) is the number of edges in the (2^n  1)dimensional simplex.  Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete plane graph in 2^n points.  Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete parallelotope graph in n dimensions.  Dimitri Boscainos, Oct 05 2015
a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4.  Álvar Ibeas, Nov 26 2015
a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x  1)^(2^n))/2, cf. A201461.  Martin Renner, Jan 14 2017
Let f(x)=x+2*sqrt(x) and g(x)=x2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582.  Luc Rousseau, Dec 06 2018
For n>=1, a(n) is the covering radius of the first order ReedMuller code RM(1,2n).  Christof Beierle, Dec 22 2021
a(n) =


REFERENCES

V. K. Balakrishnan, Theory and problems of Combinatorics, "Schaum's Outline Series", McGrawHill, 1995, p. 69.
Martin Gardner, Mathematical Carnival, "Pascal's Triangle", p. 201, Alfred A. Knopf NY, 1975.
Richard K. Guy, Unsolved problems in number theory, (p 72.) [From Daniel Forgues, Nov 10 2009]
Ross Honsberger, Mathematical Gems, M.A.A., 1973, p. 113.
Clifford A. Pickover, Wonders of Numbers, Chap. 55, Oxford Univ. Press NY 2000.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

William Banks, Ahmet Güloğlu, Wesley Nevans and Filip Saidak, Descartes numbers, Anatomy of integers, 167173, CRM Proc. Lecture Notes, 46, Amer. Math. Soc., Providence, RI, 2008. MathSciNet review (subscription required).


FORMULA

G.f.: x/((1  2*x)*(1  4*x)).
E.g.f. for a(n+1), n>=0: 2*exp(4*x)  exp(2*x).
a(n) = 2^(n1)*StirlingS2(n+1,2), n>=0, with StirlingS2(n,m)=A008277(n,m).
a(n) = StirlingS2(2^n,2^n1) = binomial(2^n,2).  Ross La Haye, Jan 12 2008
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} 4^(nj)*binomial(j,k).  Paul Barry, Aug 05 2005
Row sums of triangle A134346. Also, binomial transform of A048473: (1, 5, 17, 53, 161, ...); double bt of A151821: (1, 4, 8, 16, 32, 64, ...) and triple bt of A010684: (1, 3, 1, 3, 1, 3, ...).  Gary W. Adamson, Oct 21 2007
a(n) = 3*Stirling2(n+1,4) + Stirling2(n+2,3).  Ross La Haye, Jun 01 2008


EXAMPLE

G.f. = x + 6*x^2 + 28*x^3 + 120*x^4 + 496*x^5 + 2016*x^6 + 8128*x^7 + 32640*x^8 + ...


MAPLE

GBC := proc(n, k, q) local i; mul( (q^(ni)1)/(q^(ki)1), i=0..k1); end; # define qary Gaussian binomial coefficient [ n, k ]_q
[ seq(GBC(n+1, 2, 2)GBC(n, 2, 2), n=0..30) ]; # produces A006516


MATHEMATICA

Table[2^(n  1)(2^n  1), {n, 0, 30}] (* or *) LinearRecurrence[{6, 8}, {0, 1}, 30] (* Harvey P. Dale, Jul 15 2011 *)


PROG

(Sage) [lucas_number1(n, 6, 8) for n in range(24)] # Zerinvary Lajos, Apr 22 2009
(Sage) [(4**n  2**n) / 2 for n in range(24)] # Zerinvary Lajos, Jun 05 2009
(PARI) vector(100, n, n; 2^(n1)*(2^n1)) \\ Altug Alkan, Oct 06 2015
(Haskell)
a006516 n = a006516_list !! n
a006516_list = 0 : 1 :
zipWith () (map (* 6) $ tail a006516_list) (map (* 8) a006516_list)
(Python) for n in range(0, 30): print(2**(n1)*(2**n  1), end=', ') # Stefano Spezia, Dec 06 2018
(GAP) List([0..25], n>2^(n1)*(2^n1)); # Muniru A Asiru, Dec 06 2018


CROSSREFS



KEYWORD

nonn,nice,easy


AUTHOR



STATUS

approved



