A032527 - OEIS

%I #67 Jan 16 2023 08:17:41

%S 0,1,5,11,20,31,45,61,80,101,125,151,180,211,245,281,320,361,405,451,

%T 500,551,605,661,720,781,845,911,980,1051,1125,1201,1280,1361,1445,

%U 1531,1620,1711,1805,1901,2000,2101,2205,2311,2420,2531,2645,2761,2880,3001

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

%C Also A033429 and A062786 interleaved. - _Omar E. Pol_, Sep 28 2011

%C Partial sums of A047209. - _Reinhard Zumkeller_, Jan 07 2012

%C From _Wolfdieter Lang_, Aug 06 2013: (Start)

%C 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).

%C 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)

%H Vincenzo Librandi, <a href="/A032527/b032527.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F a(n) = 5*n^2/4+((-1)^n-1)/8. - _Omar E. Pol_, Sep 28 2011

%F G.f.: x*(1+3*x+x^2)/(1-2*x+2*x^3-x^4). - _Colin Barker_, Jan 06 2012

%F 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

%F a(n) = Sum_{j=1..n} Sum{i=1..n} ceiling((i+j-n+1)/2). - _Wesley Ivan Hurt_, Mar 12 2015

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

%e From _Omar E. Pol_, Sep 28 2011 (Start):

%e Illustration of initial terms (In a precise representation the pentagons should appear strictly concentric):

%e .

%e .                                             o

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%e .                            o            o   o   o

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%e .               o        o   o   o    o   o   o   o   o

%e .             o   o    o   o   o   o   o   o     o   o

%e .      o    o   o   o   o   o o   o     o   o o o   o

%e .    o   o   o     o     o       o       o         o

%e . o   o o     o o o       o o o o         o o o o o

%e .

%e . 1    5        11          20                31

%e .

%e (End)

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

%t Table[Round[5n^2/4], {n, 0, 39}] (* _Alonso del Arte_, Sep 28 2011 *)

%o (PARI) a(n)=5*n^2>>2 \\ _Charles R Greathouse IV_, Sep 28 2011

%o (Magma) [5*n^2/4+((-1)^n-1)/8: n in [0..50]]; // _Vincenzo Librandi_, Sep 29 2011

%o (Haskell)

%o a032527 n = a032527_list !! n

%o a032527_list = scanl (+) 0 a047209_list

%o -- _Reinhard Zumkeller_, Jan 07 2012

%o (Python)

%o def A032527(n): return 5*n**2>>2  # _Chai Wah Wu_, Jul 30 2022

%Y Cf. A000290, A032528, A077043, A195041. Column 5 of A195040. [_Omar E. Pol_, Sep 28 2011]

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_

%E New name from _Omar E. Pol_, Sep 28 2011