A055112 a(n) = n*(n+1)*(2*n+1).

0, 6, 30, 84, 180, 330, 546, 840, 1224, 1710, 2310, 3036, 3900, 4914, 6090, 7440, 8976, 10710, 12654, 14820, 17220, 19866, 22770, 25944, 29400, 33150, 37206, 41580, 46284, 51330, 56730, 62496, 68640, 75174, 82110, 89460, 97236, 105450

Offset: 0

Author

Henry Bottomley, Jun 15 2000

Keywords

• nonn
• easy

Comments

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

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

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

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

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

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

Examples

			.  n   A001477(n) as triangle with row lengths = 2*n+1   Row sums = a(n)
.  0                         0                                  0
.  1                      1  2  3                               6
.  2                   4  5  6  7  8                           30
.  3                9 10 11 12 13 14 15                        84
.  4            16 17 18 19 20 21 22 23 24                    180
.  5         25 26 27 28 29 30 31 32 33 34 35                 330
.  6      36 37 38 39 40 41 42 43 44 45 46 47 48              546
.  7   49 50 51 52 53 54 55 56 57 58 59 60 61 62 63           840 .
- _Reinhard Zumkeller_, Jan 24 2013
		

References

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• R. Roy, The discovery of the series formula for π by Leibniz, Gregory and Nilakantha, Mathematics Magazine, 63 (5) 1990, 291-306.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

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

Similar sequences are listed in A316224.

Cf. A000217, A000290, A005898, A083374.

Programs

Mathematica

Table[n(n + 1)(2n + 1), {n, 0, 39}] (* Vladimir Joseph Stephan Orlovsky, Nov 21 2010 *)
LinearRecurrence[{4,-6,4,-1},{0,6,30,84},50] (* Harvey P. Dale, Oct 02 2024 *)

PARI

a(n)=n*(n+1)*(2*n+1);

Python

def A055112(n): return n*(n*((n<<1) + 3) + 1) # Chai Wah Wu, Nov 14 2022

Formula

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

G.f.: 6*x*(1+x)/(1-x)^4. - Bruno Berselli, Mar 31 2012

From Benoit Cloitre, Apr 30 2002: (Start)

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.

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

a(n) = Sum_{i = 1..n} A033581(i). - Jonathan Vos Post, Mar 15 2006

a(n) = A000217(2*n)*A000217(2*n+1)/(2*n+1). - Charlie Marion, Feb 17 2012

a(n) = Sum_{i = 1..2*n + 1} (n^2 + (i-1)). - Charlie Marion, Sep 14 2012

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

a(n) = A002378(n) * (2n+1). - Bruce J. Nicholson, Aug 31 2017

a(n) = Sum_{k=n^2..(n+1)^2-1} k. - Darío Clavijo, Jan 31 2025

E.g.f.: exp(x)*x*(6 + 9*x + 2*x^2). - Stefano Spezia, Feb 02 2025

a(n) = A005898(n) - A005408(n) = A083374(n+1) - A083374(n). - J.S. Seneschal, Jul 08 2025