A077043 "Three-quarter squares": a(n) = n^2 - A002620(n).
0, 1, 3, 7, 12, 19, 27, 37, 48, 61, 75, 91, 108, 127, 147, 169, 192, 217, 243, 271, 300, 331, 363, 397, 432, 469, 507, 547, 588, 631, 675, 721, 768, 817, 867, 919, 972, 1027, 1083, 1141, 1200, 1261, 1323, 1387, 1452, 1519, 1587, 1657, 1728, 1801, 1875, 1951
Offset: 0
Examples
G.f. = x + 3*x^2 + 7*x^3 + 12*x^4 + 19*x^5 + 27*x^6 + 37*x^7 + 48*x^8 + ... a(4)=12 since the compositions of floor(3*(4+1)/2) = 7 into exactly 3 positive integers each no more than 4 are 1+2+4, 1+3+3, 1+4+2, 2+1+4, 2+2+3, 2+3+3, 2+4+1, 3+1+3, 3+2+2, 3+3+1, 4+1+2, 4+2+1. From _Philippe Deléham_, Dec 17 2011: (Start) a(1) = 1 = 1^3; a(1) + a(3) = 1 + 7 = 2^3; a(1) + a(3) + a(5) = 1 + 7 + 19 = 3^3; a(1) + a(3) + a(5) + a(7) = 1 + 7 + 19 + 37 = 4^3; a(1) + a(3) + a(5) + a(7) + a(9) = 1 + 7 + 19 + 37 + 61 = 5^3; ... (End)
References
- John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Ralf Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 4.4.4.
- Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7 (2004), Article 04.1.6.
- Gabriel Nivasch and Eyal Lev, Nonattacking Queens on a Triangle, Mathematics Magazine, Vol. 78, No. 5 (Dec., 2005), pp. 399-403. See page 402.
- Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
- Index entries for sequences related to compositions.
Crossrefs
Column 3 of A195040. - Omar E. Pol, Sep 28 2011
Cf. A019298 (partial sums).
Programs
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Haskell
a077043 n = a077043_list !! n a077043_list = scanl (+) 0 a001651_list -- Reinhard Zumkeller, Jan 06 2014
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Magma
[Ceiling(n^2*3/4): n in [0..60]]; // Vincenzo Librandi, Jun 29 2011
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Maple
A077043:=n->ceil(3*n^2/4); seq(A077043(n), n=0..60); # Wesley Ivan Hurt, Jan 21 2014
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Mathematica
Table[Ceiling[(3n^2)/4], {n,0,60}] (* or *) LinearRecurrence[{2,0,-2,1}, {0,1,3,7}, 60] (* Harvey P. Dale, Dec 16 2012 *)
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PARI
{a(n) = n^2 - (n^2 \ 4)}; /* Michael Somos, Jun 29 2011 */
Formula
a(n) = ceiling(n^2*3/4) = A077042(n, 3); a(-n) = a(n).
Also can be computed from 1 * C(n,0) + 2 * C(n,1) + 2 * C(n,2) - Sum((-2)^(k-3) C(n, k)). - Joshua Zucker, Nov 10 2002
From Jon Perry, May 29 2003: (Start)
a(2k) = a(2k-2) + 6k - 3,
a(2k+1) = a(2k-1) + 6k,
a(4n) = 12n^2,
a(4n+1) = a(4n) + 6n + 1,
a(4n+2) = a(4n+1) + 6n + 2,
a(4n+3) = a(4n+2) + 6n + 4,
a(4n+4) = a(4n+3) + 6n + 5.
Differences between alternate terms give 3, 6, 9, 12, ... (End)
From R. J. Mathar, Nov 10 2008: (Start)
G.f.: x*(1+x+x^2)/((1+x)*(1-x)^3).
a(n) + a(n+1) = A005448(n+1).
The inverse binomial transform yields 0 followed by A141531. (End)
Euler transform of length 3 sequence [3, 1, -1]. - Michael Somos, Jun 29 2011
a(n) = 3*n^2/4 - ((-1)^n-1)/8. - Omar E. Pol, Sep 28 2011
Sum_{k=0..n} a(2k+1) = partial sums of A003215 = (n+1)^3 (see example). - Philippe Deléham, Dec 17 2011
a(0)=0, a(1)=1, a(2)=3, a(3)=7, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Dec 16 2012
a(0)=0, a(1)=1, a(n) = 3*(n-1) + a(n-2). - Reg Robson, Mar 08 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n)/2). - Wesley Ivan Hurt, Mar 12 2015
a(n) = (3*n)^2/12 for n even and a(n) = ((3*n)^2 + 3)/12 for n odd. - Miquel Cerda, Jun 30 2016
a(n) = Sum_{k=1..n} floor((n+k)/2). - Wesley Ivan Hurt, Mar 31 2017
0 = 1 +a(n)*(+a(n+1) -a(n+2)) +a(n+1)*(-3 -a(n+1) +a(n+2)) for all n in Z. - Michael Somos, Apr 02 2017
E.g.f.: (1/8)*exp(-x)*(-1 + exp(2*x)*(1 + 6*x + 6*x^2)). - Stefano Spezia, Nov 29 2019
Sum_{n>=1} 1/a(n) = Pi^2/18 + tanh(Pi/(2*sqrt(3)))*Pi/sqrt(3). - Amiram Eldar, Jan 16 2023
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