

A227541


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


2



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
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OFFSET

0,2


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 (n1)/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 13gons (the dot numbers of each side for these 13gons are 2, 4, 6, ..., n) for even n>=2. There is no figure for n = 0. For odd n one has (n1)/2 concentric dotted 13gons (the dot numbers of each side for these 13gons are 3, 5, 7, ..., n) and an extra dotted 3gon 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)/((1x)^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)^n1)/8, n >= 0 (which checks for even and odd n). Because the denominator of the o.g.f. is 12*x+2*x^3x^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


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).


FORMULA

a(n) = 13*n^2/4+((1)^n1)/8, n >= 0 (use even or odd n to prove it).
G.f.: x*(3+7*x+3*x^2)/((1x)^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(n1)  2*a(n3) + a(n4).  Bruno Berselli, Aug 08 2013
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+jn+5)/2).  Wesley Ivan Hurt, Mar 12 2015


EXAMPLE

Counting dots in the concentric dotted 13gons described above in a comment:
a(2*k), k >= 1: (21)*13 = 13, (1+(41))*13 = 52, (1+3+(61))*13 = 117, (1+3+5+7)*13 = 208, ... a(2*k+1), k >= 0: 3, 3+(31)*13 = 29, 3+(2+(51))*13 = 81, 3+2*(1+2+3)*13 = 159, ... (a dotted triangle is put into the middle of the k concentric 13gons).


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 *)


PROG

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


CROSSREFS

Cf. A032527 (case for prime 5).
Sequence in context: A031378 A144391 A024836 * A023553 A268184 A183436
Adjacent sequences: A227538 A227539 A227540 * A227542 A227543 A227544


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Aug 07 2013


STATUS

approved



