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A034494
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a(n) = (7^n+1)/2.
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12
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1, 4, 25, 172, 1201, 8404, 58825, 411772, 2882401, 20176804, 141237625, 988663372, 6920643601, 48444505204, 339111536425, 2373780754972, 16616465284801, 116315256993604, 814206798955225, 5699447592686572
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OFFSET
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0,2
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COMMENTS
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Number of compositions of even natural numbers into n parts <=6. [Adi Dani, May 28 2011]
a(n)+(a(n)+1)+...+(a(n+1)-7^n-1)=(a(n+1)-7^n)+...+(a(n+1)-1). Let
S(2n) and S(2n+1) be the sets of addends on the left- and right-hand
sides, respectively, of the preceding equations. Then, since the
intersection of any 2 different S(i) is null and the union of all of
them is the positive integers, {S(i)} forms a partition of the
positive integers. See also A034659.
In general, for k>0, let b(n)=((4k+3)^n+1)/2. Then b(n)+(b(n)+1)+ ...
+(b(n+1)-(4k+3)^n-1)=k*((b(n+1)-(4k+3)^n)+ ... +(b(n+1)-1)). Then,
for each k, the set of addends on the two sides of these equations
also forms a partition of the positive integers. Also, with b(0)=1,
b(n)=(4k+3)*b(n-1)-(2k+1).
For k>0, let c(0)=1 and, for n>0, c(n)=(2*(2k+1))^n/2. Then the
sequence b(0),b(1),... is the binomial transform of the sequence
c(0),c(1),....
For k>0, let d(2n)=(2k+1)^(2n) and d(2n+1)=0. Then the sequence
b(0),b(1),... is the (2k+2)nd binomial transform of the sequence
d(0),d(1),.... (End)
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LINKS
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FORMULA
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E.g.f.: exp(4*x)*cosh(3*x). - Paul Barry, Apr 20 2003
a(n) = 7a(n-1) - 3, a(0) = 1.
a(n) = ((4+sqrt(9))^n+(4-sqrt(9))^n)/2. [Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008]
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EXAMPLE
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a(2)=25: there are 25 compositions of even numbers into 2 parts <=6:
(0,0)
(0,2),(2,0),(1,1)
(0,4),(4,0),(1,3),(3,1),(2,2)
(0,6),(6,0),(1,5),(5,1),(2,4),(4,2),(3,3)
(2,6),(6,2),(3,5),(5,3),(4,4)
(4,6),(6,4),(5,5)
(6,6)
(end)
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MAPLE
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PROG
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(PARI) Vec((1-4*x)/((1-x)*(1-7*x)) + O(x^100)) \\ Altug Alkan, Nov 01 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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