

A020988


(2/3)*(4^n1).


39



0, 2, 10, 42, 170, 682, 2730, 10922, 43690, 174762, 699050, 2796202, 11184810, 44739242, 178956970, 715827882, 2863311530, 11453246122, 45812984490, 183251937962, 733007751850, 2932031007402, 11728124029610, 46912496118442
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OFFSET

0,2


COMMENTS

Numbers whose binary representations is 10, n times (see A163662(n) for n >= 1).  Alexandre Wajnberg, May 31 2005
Numbers whose base 4 representation consists entirely of 2's; twice base 4 repunits.  Franklin T. AdamsWatters, Mar 29 2006
Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n)=1) a(n)=A060590(2n).  Henry Bottomley, Apr 05 2001
a(n)=number of derangements of [2n+3] with runs consisting of consecutive integers. E.g. a(1)=10 because the derangements of {1,2,3,4,5} with runs consisting of consecutive integers are 51234, 45123, 34512, 23451, 54123, 53412, 45231, 34521, 54231, 53421 (the bars delimit the runs).  Emeric Deutsch, May 26 2003
For n>0 also smallest numbers having in binary representation exactly n+1 maximal groups of consecutive zeros: A087120(n)=a(n1), see A087116.  Reinhard Zumkeller, Aug 14 2003
Number of walks of length 2n+3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0)=2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED.  Emeric Deutsch, Apr 01 2004
Comment from Paul Barry, May 18, 2003. (Start):
row sums of triangle using cumulative sums of oddindexed rows of Pascal's triangle (start with zeros for completeness):
. . . . 0 . 0
. . . . 1 . 1
. . . 1 4 . 4 1
. . 1 6 14 14 6 1
.1 8 27 49 49 27 8 1 (End).
a(n) gives the position of the nth zero in A173732, i.e. A173732(a(n))=0 for all n and this gives all the zeros in A173732. [From Howard A. Landman, Mar 14 2010]
Smallest number having alternating bit sum n. Cf. A065359. For n=0,1,..., the last digit of a(n) is 0,2,0,2,... .  W. Bomfim, Jan 22, 2011
Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages.  Omar E. Pol, Mar 15 2012
For n>0 also partial sums of the odd powers of 2 (A004171).  K. G. Stier, Nov 04 201


REFERENCES

J. Brillhart and P. Morton, A case study in mathematical research: the GolayRudinShapiro sequence, Amer. Math. Monthly, 103 (1996) 854869.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Index entries for sequences related to linear recurrences with constant coefficients


FORMULA

a(n) = 4a(n1) + 2, a(0)=0.
E.g.f. : (2/3)(exp(4x)exp(x)).  Paul Barry, May 18 2003
a(n) = A007583(n+1)1 = A039301(n+2)2 = A083584(n)+1.  Ralf Stephan, Jun 14 2003
G.f.: 2x/((1x)(14x)). [From R. J. Mathar, Sep 17 2008]
a(n) = a(n1) + 2^(2n1), a(0) = 0.  W. Bomfim, Jan 22, 2011
a(n) = A193652(2*n). [Reinhard Zumkeller, Aug 08 2011]
a(n) = 5*a(n1)  4*a(n2) (n>1), a(0)=0, a(1)=2.  L. Edson Jeffery, Mar 02 2012
a(n) = (2/3)*A024036(n).  Omar E. Pol, Mar 15 2012


MATHEMATICA

Table[ FromDigits[ Flatten[ Table[{1, 0}, {i, n}]], 2], {n, 0, 23}] (from Robert G. Wilson v, Jun 01 2005)
(2(4^Range[0, 30]1))/3 (* or *) LinearRecurrence[{5, 4}, {0, 2}, 30] (* Harvey P. Dale, Sep 25 2013 *)


PROG

(PARI) an=0; print1(an, ", "); for(n=1, 23, an+=2^(2*n1); print1(an, ", "))  W. Bomfim, Jan 22, 2011
(MAGMA) [(2/3)*(4^n1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011


CROSSREFS

a(n) = A026644(2n)
a(n) = 2*A002450(n). These two sequences are both subsets of A000975.
a(n) = A007583(n)1 = A039301(n+1)2 = A083584(n1)+1.
Cf. A020989.
Sequence in context: A181052 A024483 A084180 * A177238 A084480 A099553
Adjacent sequences: A020985 A020986 A020987 * A020989 A020990 A020991


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane.


EXTENSIONS

Edited by N. J. A. Sloane, Sep 06 2006


STATUS

approved



