login
A005476
a(n) = n*(5*n - 1)/2.
25
0, 2, 9, 21, 38, 60, 87, 119, 156, 198, 245, 297, 354, 416, 483, 555, 632, 714, 801, 893, 990, 1092, 1199, 1311, 1428, 1550, 1677, 1809, 1946, 2088, 2235, 2387, 2544, 2706, 2873, 3045, 3222, 3404, 3591
OFFSET
0,2
COMMENTS
a(n) is half the number of ways to divide an n X n square into 3 rectangles whose side-lengths are integers. See Matthew Scroggs link. - George Witty, Feb 06 2024
FORMULA
a(n) = C(5*n,2)/5 for n>=0. - Zerinvary Lajos, Jan 02 2007
a(n) = A033991(n) - A000326(n). - Zerinvary Lajos, Jun 11 2007
a(n) = a(n-1) + 5*n - 3 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010
a(n) = A000217(n) + A000384(n) = A000290(n) + A000326(n). - Omar E. Pol, Jan 11 2013
a(n) = A130520(5*n+1). - Philippe Deléham, Mar 26 2013
a(n) = A033994(n) - A033994(n-1). - J. M. Bergot, Jun 12 2013
From Bruno Berselli, Oct 17 2016: (Start)
G.f.: x*(2 + 3*x)/(1 - x)^3.
a(n) = A000217(3*n-1) - A000217(2*n-1). (End)
E.g.f.: x*(4 + 5*x)*exp(x)/2. - G. C. Greubel, Jul 30 2019
Sum_{n>=1} 1/a(n) = 2 * A294833. - Amiram Eldar, Nov 16 2020
From Leo Tavares, Nov 20 2021: (Start)
a(n) = A016754(n) - A133694(n+1). See Triangulated Diamonds illustration.
a(n) = A000290(n) + A000217(n) + 2*A000217(n-1)
a(n) = 2*A000217(n) + 3*A000217(n-1). (End)
MAPLE
[seq(binomial(5*n, 2)/5, n=0..40)]; # Zerinvary Lajos, Jan 02 2007
MATHEMATICA
Table[n(5n-1)/2, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)
PROG
(PARI) a(n)=n*(5*n-1)/2 \\ Charles R Greathouse IV, Sep 24 2015
(Magma) [Binomial(5*n, 2)/5: n in [0..40]]; // G. C. Greubel, Jul 30 2019
(Sage) [binomial(5*n, 2)/5 for n in (0..40)] # G. C. Greubel, Jul 30 2019
(GAP) List([0..40], n-> Binomial(5*n, 2)/5); # G. C. Greubel, Jul 30 2019
CROSSREFS
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. similar sequences listed in A022288.
Sequence in context: A041963 A298912 A368187 * A316430 A131476 A023549
KEYWORD
nonn,easy
STATUS
approved