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A000302
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Powers of 4: a(n) = 4^n.
(Formerly M3518 N1428)
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538
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1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656
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OFFSET
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0,2
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COMMENTS
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Same as Pisot sequences E(1, 4), L(1, 4), P(1, 4), T(1, 4). Essentially same as Pisot sequences E(4, 16), L(4, 16), P(4, 16), T(4, 16). See A008776 for definitions of Pisot sequences.
The convolution square root of this sequence is A000984, the central binomial coefficients: C(2n,n). - T. D. Noe, Jun 11 2002
With P(n) being the number of integer partitions of n, p(i) as the number of parts of the i-th partition of n, d(i) as the number of different parts of the i-th partition of n, m(i, j) the multiplicity of the j-th part of the i-th partition of n, one has a(n) = Sum_{i = 1..P(n)} p(i)!/(Product_{j = 1..d(i)} m(i, j)!) * 2^(n-1). - Thomas Wieder, May 18 2005
Equals the Catalan sequence: (1, 1, 2, 5, 14, ...), convolved with A032443: (1, 3, 11, 42, ...). - Gary W. Adamson, May 15 2009
Sum of coefficients of expansion of (1 + x + x^2 + x^3)^n.
a(n) is number of compositions of natural numbers into n parts less than 4. For example, a(2) = 16 since there are 16 compositions of natural numbers into 2 parts less than 4.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 4-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Row sums of Pascal's triangle using the rule that going left increases the value by a factor of k = 3. For example, the first three rows are {1}, {3, 1}, and {9, 6, 1}. Using this rule gives row sums as (k+1)^n. - Jon Perry, Oct 11 2012
Sum of all peak heights in Dyck paths of semilength n+1. - David Scambler, Apr 22 2013
a(n) is equal to 1 plus the sum for 0 < k < 2^n of the numerators and denominators of the reduced fractions k/2^n. - J. M. Bergot, Jul 13 2015
Number of nodes at level n regular 4-ary tree.
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
Also the number of connected dominating sets in the (n+1)-barbell graph. - Eric W. Weisstein, Jun 29 2017
Side length of the cells at level n in a pyramid scheme where a square grid is decomposed into overlapping 2 X 2 blocks (cf. Kropatsch, 1985). - Felix Fröhlich, Jul 04 2019
a(n-1) is the number of 3-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
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REFERENCES
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H. W. Gould, Combinatorial Identities, 1972, eq. (1.93), p. 12.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, eq. (5.39), p. 187.
D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
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LINKS
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FORMULA
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a(n) = 4^n.
a(0) = 1; a(n) = 4*a(n-1).
G.f.: 1/(1-4*x).
E.g.f.: exp(4*x).
a(n) = Sum_{k = 0..n} binomial(2k, k) * binomial(2(n - k), n - k). - Benoit Cloitre, Jan 26 2003 [See Graham et al., eq. (5.39), p. 187. - Wolfdieter Lang, Aug 16 2019]
1 = Sum_{n >= 1} 3/a(n) = 3/4 + 3/16 + 3/64 + 3/256 + 3/1024, ...; with partial sums: 3/4, 15/16, 63/64, 255/256, 1023/1024, ... - Gary W. Adamson, Jun 16 2003
a(n) = Sum_{j = 0..n} 2^(n - j)*binomial(n + j, j). - Peter C. Heinig (algorithms(AT)gmx.de), Apr 06 2007
a(n) = 6*Stirling2(n+1, 4) + 6*Stirling2(n+1, 3) + 3*Stirling2(n+1, 2) + 1 = 2*Stirling2(2^n, 2^n - 1) + Stirling2(n+1, 2) + 1. - Ross La Haye, Jun 26 2008
a(n) = Sum_{k = 0..n} binomial(2*n+1, k). - Mircea Merca, Jun 25 2011
Sum_{n >= 1} Mobius(n)/a(n) = 0.1710822479183... - R. J. Mathar, Aug 12 2012
a(n) = Sum_{k = 0..n} binomial(2*k + x, k)*binomial(2*(n - k) - x, n - k) for every real number x. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
a(n) = (2*n+1) * binomial(2*n,n) * Sum_{j=0..n} (-1)^j/(2*j+1)*binomial(n,j). - Vaclav Kotesovec, Sep 15 2013
a(n) = (1/2) * Product_{k = 0..n} (1 + (2*n + 1)/(2*k + 1)). - Peter Bala, Mar 06 2018
a(n) = denominator(zeta_star({2}_(n + 1))/zeta(2*n + 2)) where zeta_star is the multiple zeta star values and ({2}_n) represents (2, ..., 2) where the multiplicity of 2 is n. - Roudy El Haddad, Feb 22 2022
a(n) = 1 + 3*Sum_{k=0..n} binomial(2*n, n+k)*(k|9), where (k|9) is the Jacobi symbol. - Greg Dresden, Oct 11 2022
a(n) = Sum_{k = 0..n} binomial(2*n+1, 2*k) = Sum_{k = 0..n} binomial(2*n+1, 2*k+1). - Sela Fried, Mar 23 2023
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MAPLE
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for n from 0 to 10 do sum(2^(n-j)*binomial(n+j, j), j=0..n); od; # Peter C. Heinig (algorithms(AT)gmx.de), Apr 06 2007
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MATHEMATICA
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CoefficientList[Series[1/(1 - 4 x), {x, 0, 50}], x] (* Vincenzo Librandi, May 29 2014 *)
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PROG
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(Haskell)
a000302 = (4 ^)
(Scala) (List.fill(20)(4: BigInt)).scanLeft(1: BigInt)(_ * _) // Alonso del Arte, Jun 22 2019
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CROSSREFS
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KEYWORD
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easy,nonn,nice,core
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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