%I #113 Oct 02 2024 17:24:04

%S 0,6,30,84,180,330,546,840,1224,1710,2310,3036,3900,4914,6090,7440,

%T 8976,10710,12654,14820,17220,19866,22770,25944,29400,33150,37206,

%U 41580,46284,51330,56730,62496,68640,75174,82110,89460,97236,105450

%N a(n) = n*(n+1)*(2*n+1).

%C Original name: Areas of Pythagorean triangles (X, Y, Z = Y + 1) with X^2 + Y^2 = Z^2.

%C a(n) is the set of possible y values for 4*x^3 + x^2 = y^2 with the x values being A002378(n). - _Gary Detlefs_, Feb 22 2010

%C This sequence is related to A028896 by a(n) = n*A028896(n) - Sum_{i = 0..n-1} A028896(i) and this is the case d = 3 in the identity n*(d*(d+1)*n*(n+1)/4) - Sum_{i = 0..n-1} d*(d+1)*i*(i+1)/4 = d*(d+1)*n*(n+1)*(2*n+1)/12. - _Bruno Berselli_, Mar 31 2012

%C Also sums of rows of natural numbers (cf. A001477) seen as triangle with an odd numbers of terms per row, see example. - _Reinhard Zumkeller_, Jan 24 2013

%C Without mentioning the connection to Pythagorean triangles, Bolker (1967) gives it as an exercise to prove that these numbers are always divisible by 6. This is easy to prove from the formula that he gives, n(n - 1)(2n - 1): obviously either n or (n - 1) must be even; then, if n is congruent to 2 mod 3 it means that (2n - 1) is a multiple of 3, otherwise either n or (n - 1) is a multiple of 3; thus both prime divisors of 6 are accounted for in a(n). - _Alonso del Arte_, Oct 13 2013

%C a(n) = n*(n+1)*(n+(n+1)) is the product of two consecutive integers multiplied by the sum of those two consecutive integers. - _Charles Kusniec_, Sep 04 2022

%D Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.5.

%H Vincenzo Librandi, <a href="/A055112/b055112.txt">Table of n, a(n) for n = 0..1000</a>

%H R. Roy, <a href="http://www.jstor.org/stable/2690896">The discovery of the series formula for π by Leibniz, Gregory and Nilakantha</a>, Mathematics Magazine, 63 (5) 1990, 291-306.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = n*(n+1)*(2*n+1).

%F G.f.: 6*x*(1+x)/(1-x)^4. - _Bruno Berselli_, Mar 31 2012

%F From _Benoit Cloitre_, Apr 30 2002: (Start)

%F a(n) = 6*A000330(n) = A007531(2*n)/4 = 3*A000292(2*n-1)/2 = A005408(n)*A046092(n)/2 = A005408(n)*(A001844(n)-1)/2.

%F Sum_{n > 0} 1/a(n) = 3 - 4*log(2). (End)

%F a(n) = Sum_{i = 1..n} A033581(i). - _Jonathan Vos Post_, Mar 15 2006

%F a(n) = A000217(2*n)*A000217(2*n+1)/(2*n+1). - _Charlie Marion_, Feb 17 2012

%F a(n) = Sum_{i = 1..2*n + 1} (n^2 + (i-1)). - _Charlie Marion_, Sep 14 2012

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = Pi - 3, due to Nilakantha, circa 1500. See Roy p. 304. - _Peter Bala_, Feb 19 2015

%F a(n) = A002378(n) * (2n+1). - _Bruce J. Nicholson_, Aug 31 2017

%e . n A001477(n) as triangle with row lengths = 2*n+1 Row sums = a(n)

%e . 0 0 0

%e . 1 1 2 3 6

%e . 2 4 5 6 7 8 30

%e . 3 9 10 11 12 13 14 15 84

%e . 4 16 17 18 19 20 21 22 23 24 180

%e . 5 25 26 27 28 29 30 31 32 33 34 35 330

%e . 6 36 37 38 39 40 41 42 43 44 45 46 47 48 546

%e . 7 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 840 .

%e - _Reinhard Zumkeller_, Jan 24 2013

%t Table[n(n + 1)(2n + 1), {n, 0, 39}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 21 2010 *)

%t LinearRecurrence[{4,-6,4,-1},{0,6,30,84},50] (* _Harvey P. Dale_, Oct 02 2024 *)

%o (PARI) a(n)=n*(n+1)*(2*n+1);

%o (Python)

%o def A055112(n): return n*(n*((n<<1) + 3) + 1) # _Chai Wah Wu_, Nov 14 2022

%Y Cf. A005408 (X values), A046092 (Y values), A001844 (Z values), A002939 (perimeter), A033581.

%Y Similar sequences are listed in A316224.

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, Jun 15 2000