14,2

a(n) = A110555(n+1,14). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 27 2005

Product of 14 consecutive numbers divided by 14!. - Artur Jasinski (grafix(AT)csl.pl), Dec 02 2007

In this sequence there are no primes - Artur Jasinski (grafix(AT)csl.pl), Dec 02 2007

With a different offset, number of n-permutations (n>=14) of 2 objects: u,v, with repetition allowed, containing exactly (14) u's. Example: n=14, a(0)=1 because we have uuuuuuuuuuuuuu n=15, a(1)=15 because we have uuuuuuuuuuuuuuv, uuuuuuuuuuuuuvu, uuuuuuuuuuuuvuu, uuuuuuuuuuuvuuu, uuuuuuuuuuvuuuu, uuuuuuuuuvuuuuu, uuuuuuuuvuuuuuu, uuuuuuuvuuuuuuu, uuuuuuvuuuuuuuu, uuuuuvuuuuuuuuu, uuuuvuuuuuuuuuu, uuuvuuuuuuuuuuu, uuvuuuuuuuuuuuu, uvuuuuuuuuuuuuu, vuuuuuuuuuuuuuu. [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 06 2008]

With a different offset, number of n-permutations (n>=14) of 2 objects: u,v, with repetition allowed, containing exactly 14 u's. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 03 2008

Milan Janjic, Two Enumerative Functions

a(n+13)=n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)(n+11)(n+12)(n+13)/14! - Artur Jasinski (grafix(AT)csl.pl), Dec 02 2007, R. J. Mathar, Jul 07 2009.

Gf.: x^14/(1-x)^15. a(n)=C(n,14),n>=14 [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 06 2008, R. J. Mathar, Jul 07 2009]

seq(binomial(n, 14), n=14..37); [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 06 2008]

Table[n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)(n+11)(n+12)(n+13)/14!\ , {n, 1, 100}] - Artur Jasinski (grafix(AT)csl.pl), Dec 02 2007

Sequence in context: A139615 A027484 A126898 this_sequence A022580 A081079 A138424

Adjacent sequences: A010964 A010965 A010966 this_sequence A010968 A010969 A010970

nonn

N. J. A. Sloane (njas(AT)research.att.com).

Some formulas rewritten for the correct offset by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2009