OFFSET
0,3
COMMENTS
With a different offset, number of n-permutations of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly one u. - Zerinvary Lajos, Dec 28 2007
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..500
Frank Ellermann, Illustration of binomial transforms
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 756.
Index entries for linear recurrences with constant coefficients, signature (10,-25).
FORMULA
a(n) = Sum_{k=0..n} 5^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - Paul Barry, Oct 15 2004
a(n) = 10*a(n-1) - 25*a(n-2); n>1; a(0)=0, a(1)=1.
Fourth binomial transform of n (starting 0, 1, 10...) Convolution of powers of 5.
G.f.: x/(1-5*x)^2; E.g.f.: x*exp(5*x). - Paul Barry, Jul 22 2003
a(n) = - 25^n * a(-n) for all n in Z. - Michael Somos, Jun 26 2017
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 5*log(5/4).
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(6/5). (End)
MATHEMATICA
Join[{a=0, b=1}, Table[c=10*b-25*a; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
Table[n*5^(n-1), {n, 0, 20}] (* or *) LinearRecurrence[{10, -25}, {0, 1}, 30] (* Harvey P. Dale, Jul 22 2014 *)
PROG
(PARI) {a(n) = n*5^(n-1)}; /* Michael Somos, Sep 12 2005 */
(Sage) [lucas_number1(n, 10, 25) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009
(Magma) [n*(5^(n-1)): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Barry E. Williams, Jan 13 2000
EXTENSIONS
More terms from James A. Sellers, Feb 02 2000
STATUS
approved