OFFSET
0,2
COMMENTS
Generalization: T(n,i) = A000217(i-1+n) - A000217(i-1) = i*n + A000217(n-1) (corrected by Zak Seidov, Jun 21 2012); in this case is i=18.
For i = 11..16, Milan Janjic observed that if we define f(n,b,i) = Sum_{k=0..n-b} binomial(n,k)*Stirling1(n-k,b)*Product_{j=0..k-1} (-i - j), then T(n-1,i) = -f(n,n-1,i) for n >= 1.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = (17+n)*(18+n)/2 - 17*18/2 = 18*n + (n-1)*n/2 = n*(n+35)/2.
G.f.: x*(18-17*x)/(1-x)^3. - Bruno Berselli, Jun 21 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 10 2012
a(n) = 18*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
From Amiram Eldar, Jan 11 2021: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/35 - 102126365345729/2527062175638000. (End)
E.g.f.: exp(x)*x*(36 + x)/2. - Elmo R. Oliveira, Dec 12 2024
MATHEMATICA
Table[-18 (18 - 1)/2 + (18 + n) (17 + n)/2, {n, 0, 100}]
LinearRecurrence[{3, -3, 1}, {0, 18, 37}, 60] (* Harvey P. Dale, Jun 09 2024 *)
PROG
(Magma) [n*(n+35)/2: n in [0..48]]; // Bruno Berselli, Jun 21 2012
(PARI) a(n)=n*(n+35)/2 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jesse Han, May 16 2012
STATUS
approved