|
| |
|
|
A081186
|
|
4th binomial transform of (1,0,1,0,1,......), A059841.
|
|
9
| |
|
|
1, 4, 17, 76, 353, 1684, 8177, 40156, 198593, 986404, 4912337, 24502636, 122336033, 611148724, 3054149297, 15265963516, 76315468673, 381534296644, 1907542343057, 9537324294796, 47685459212513, 238423809278164
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| Binomial transform of A007582.
a(n) is a companion to A005059(n): a(n) + A005059(n) = 5^n; e.g. a(4) = A005059(4) = 353 + 272 = 625 = 5^4. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 30 2006
a(n) is the number of sequences of length n from an alphabet of size 5 in which a chosen letter appears an even number of times. - James Mahoney, Feb 03 2012
|
|
|
FORMULA
| a(n) = 8a(n-1) - 15a(n-2), a(0)=1, a(1)=4.
G.f.: (1-4x)/((1-3x)(1-5x)).
a(n) = 3^n/2 + 5^n/2.
a(n) = sum{k=0..floor(n/2); C(n, 2k)4^(n-2k) }.
E.g.f. : exp(4x)cosh(x) - Paul Barry (pbarry(AT)wit.ie), Oct 06 2004
|
|
|
EXAMPLE
| Say the alphabet is {a,b,c,d,e} and we want to know how many sequences of length one and two contain c an even number of times. a(1) = 4, which we can see by the four sequences {(a),(b),(d),(e)} and a(2) = 17, which we can see by the seventeen sequences {(a,a), (a,b), (a,d), (a,e), (b,a), (b,b), (b,d), (b,e), (c,c), (d,a), (d,b), (d,d), (d,e), (e,a), (e,b), (e,d), (e,e)}. - James Mahoney, Feb 03 2012
|
|
|
CROSSREFS
| Cf. A005059.
Sequence in context: A117439 A081910 A026773 * A005572 A202879 A081922
Adjacent sequences: A081183 A081184 A081185 * A081187 A081188 A081189
|
|
|
KEYWORD
| easy,nonn,changed
|
|
|
AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Mar 11 2003
|
| |
|
|
