A227541 a(n) = floor(13*n^2/4).

0, 3, 13, 29, 52, 81, 117, 159, 208, 263, 325, 393, 468, 549, 637, 731, 832, 939, 1053, 1173, 1300, 1433, 1573, 1719, 1872, 2031, 2197, 2369, 2548, 2733, 2925, 3123, 3328, 3539, 3757, 3981, 4212, 4449, 4693, 4943, 5200, 5463, 5733, 6009

Offset: 0

Author

Wolfdieter Lang, Aug 07 2013

Keywords

• nonn
• easy

Comments

This generalizes A032527, which uses 5, to 13 (the next prime 1 (mod 4)). The figures in A032527 use n/2 concentric dotted pentagons for even n, and (n-1)/2 concentric dotted pentagons plus an extra dot in the middle if n is odd. In the present case one can take n/2 concentric dotted 13-gons (the dot numbers of each side for these 13-gons are 2, 4, 6, ..., n) for even n>=2. There is no figure for n = 0. For odd n one has (n-1)/2 concentric dotted 13-gons (the dot numbers of each side for these 13-gons are 3, 5, 7, ..., n) and an extra dotted 3-gon in the middle. See the example section below for the counting.

a(n) = -N(-floor(n/2),n) with the N(a,b) = ((2*a+b)^2 - b^2*13)/4, the norm for integers a + b*omega(13), a, b rational integers, in the quadratic number field Q(sqrt(13)), where omega(13) = (1 + sqrt(13))/2.

a(n) = max({|N(a,n)|,a = -n..+n}) = |N(-floor(n/2),n)| = 3*n^2 + n*floor(n/2) - floor(n/2)^2 = floor(13*n^2/4) (the last eq. checks for even and odd n).

In the general case one has for primes 1 (mod 4), p(k) = A002144(k), k >= 1, a(p(k);n) = floor(p(k)*n^2/4) with o.g.f. G(p(k);x) = x*(A(k)*(1+x^2) + B(k)*x)/((1-x)^3*(1+x)), where A(k) = A005098(k) = (p(k)-1)/4 and B(k) = A119681(k) = (p(k)+1)/2. This follows from the alternative formula a(p(k),n) = p(k)*n^2/4 + ((-1)^n-1)/8, n >= 0 (which checks for even and odd n). Because the denominator of the o.g.f. is 1-2*x+2*x^3-x^4 the recurrence given by Bruno Berselli below holds for all a(k;n) sequences with inputs for n = -1, 0, 1, 2 given by (p(k)-1)/4, 0, (p(k)-1)/4, p(k), respectively.

The dot counting in the concentric p(k)-gons is similar to the one described for p = 5 in A032527 and for p=13 here. For odd n one puts an additional dotted A(k)-gon into the center. - Wolfdieter Lang, Aug 08 2013

Examples

			Counting dots in the concentric dotted 13-gons described above in a comment:
a(2*k), k >= 1: (2-1)*13 = 13, (1+(4-1))*13 = 52, (1+3+(6-1))*13 = 117, (1+3+5+7)*13 = 208, ... a(2*k+1), k >= 0: 3, 3+(3-1)*13 = 29, 3+(2+(5-1))*13 = 81, 3+2*(1+2+3)*13 = 159, ... (a dotted triangle is put into the middle of the k concentric 13-gons).
		

Links

• Ivan Panchenko, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A032527 (case for prime 5).

Cf. A002144, A005098, A119681.

Programs

Magma

[ Floor(13*n^2/4) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

Maple

A227541:=n->floor(13*n^2/4); seq(A227541(n), n=0..50); # Wesley Ivan Hurt, Jun 09 2014

Mathematica

Table[Floor[13*n^2/4], {n, 0, 50}] (* Wesley Ivan Hurt, Jun 09 2014 *)

Formula

a(n) = 13*n^2/4+((-1)^n-1)/8, n >= 0 (use even or odd n to prove it).

G.f.: x*(3+7*x+3*x^2)/((1-x)^3*(1+x)).

a(2*k) = k^2*13, k >= 0.

a(2*k+1) = 3 + k*(k+1)*13, k >= 0.

a(n) = a(-n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Bruno Berselli, Aug 08 2013

a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+5)/2). - Wesley Ivan Hurt, Mar 12 2015

Sum_{n>=1} 1/a(n) = Pi^2/78 + tan(Pi/(2*sqrt(13)))*Pi/sqrt(13). - Amiram Eldar, Jul 30 2024