A032527 Concentric pentagonal numbers: floor( 5*n^2 / 4 ).

0, 1, 5, 11, 20, 31, 45, 61, 80, 101, 125, 151, 180, 211, 245, 281, 320, 361, 405, 451, 500, 551, 605, 661, 720, 781, 845, 911, 980, 1051, 1125, 1201, 1280, 1361, 1445, 1531, 1620, 1711, 1805, 1901, 2000, 2101, 2205, 2311, 2420, 2531, 2645, 2761, 2880, 3001

Offset

0,3

Comments

Also A033429 and A062786 interleaved. - Omar E. Pol, Sep 28 2011

Partial sums of A047209. - Reinhard Zumkeller, Jan 07 2012

From Wolfdieter Lang, Aug 06 2013: (Start)

a(n) = -N(-floor(n/2),n) with the N(a,b) = ((2*a+b)^2 - b^2*5)/4, the norm for integers a + b*omega(5), a, b rational integers, in the quadratic number field Q(sqrt(5)), where omega(5) = (1 + sqrt(5))/2 (golden section).

a(n) = max({|N(a,n)|,a = -n..+n}) = |N(-floor(n/2),n)| = n^2 + n*floor(n/2) - floor(n/2)^2 = floor(5*n^2/4) (the last eq. checks for even and odd n). (End)

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Formula

a(n) = 5*n^2/4+((-1)^n-1)/8. - Omar E. Pol, Sep 28 2011

G.f.: x*(1+3*x+x^2)/(1-2*x+2*x^3-x^4). - Colin Barker, Jan 06 2012

a(n) = a(-n); a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>0, a(-1) = 1, a(0) = 0, a(1) = 1, a(2) = 5, n >= 3. (See the Bruno Berselli recurrence and a general comment for primes 1 (mod 4) under A227541). - Wolfdieter Lang, Aug 08 2013

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

Sum_{n>=1} 1/a(n) = Pi^2/30 + tan(Pi/(2*sqrt(5)))*Pi/sqrt(5). - Amiram Eldar, Jan 16 2023

Example

From Omar E. Pol, Sep 28 2011 (Start):
Illustration of initial terms (In a precise representation the pentagons should appear strictly concentric):
.
.                                             o
.                                           o   o
.                            o            o   o   o
.                          o   o        o   o   o   o
.               o        o   o   o    o   o   o   o   o
.             o   o    o   o   o   o   o   o     o   o
.      o    o   o   o   o   o o   o     o   o o o   o
.    o   o   o     o     o       o       o         o
. o   o o     o o o       o o o o         o o o o o
.
. 1    5        11          20                31
.
(End)

Maple

A032527:=n->5*n^2/4+((-1)^n-1)/8: seq(A032527(n), n=0..100); # Wesley Ivan Hurt, Mar 12 2015

Mathematica

Table[Round[5n^2/4], {n, 0, 39}] (* Alonso del Arte, Sep 28 2011 *)

Prog

(PARI) a(n)=5*n^2>>2 \\ Charles R Greathouse IV, Sep 28 2011
(Magma) [5*n^2/4+((-1)^n-1)/8: n in [0..50]]; // Vincenzo Librandi, Sep 29 2011
(Haskell)
a032527 n = a032527_list !! n
a032527_list = scanl (+) 0 a047209_list
-- Reinhard Zumkeller, Jan 07 2012
(Python)
def A032527(n): return 5*n**2>>2  # Chai Wah Wu, Jul 30 2022

Crossrefs

Cf. A000290, A032528, A077043, A195041. Column 5 of A195040. [Omar E. Pol, Sep 28 2011]

Sequence in context: A331133 A110208 A034308 * A212978 A026038 A080957

Adjacent sequences: A032524 A032525 A032526 * A032528 A032529 A032530

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

New name from Omar E. Pol, Sep 28 2011

Status

approved