|
| |
|
|
A055112
|
|
a(n) = n*(n+1)*(2*n+1).
|
|
11
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
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
|
|
|
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).
|
|
|
FORMULA
|
a(n) = n*(n+1)*(2*n+1).
G.f.: 6*x*(1+x)/(1-x)^4. - Bruno Berselli, Mar 31 2012
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). - Benoit Cloitre, Apr 30 2002
a(n) = Sum_{i = 1..n} 6*i^2 = 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
a(n) = n(n + 1)(2n + 1) = 2n^3 + 3n^2 + n. With a different offset, a(n) = n(n - 1)(2n - 1) = 2n^3 - 3n^2 + n. - Alonso del Arte, Oct 13 2013
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
|
|
|
EXAMPLE
|
. 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
In this triangle, the center column is A002378 (oblong numbers) or 2 * A000217 (triangular numbers). - Bruce J. Nicholson, Aug 31 2017
|
|
|
MATHEMATICA
|
Table[n(n + 1)(2n + 1), {n, 0, 39}] (* Vladimir Joseph Stephan Orlovsky, Nov 21 2010 *)
|
|
|
PROG
|
(PARI) a(n)=n*(n+1)*(2*n+1);
|
|
|
CROSSREFS
|
Cf. A005408 (X values), A046092 (Y values), A001844 (Z values), A002939 (perimeter), A033581.
Similar sequences are listed in A316224.
Sequence in context: A259918 A002444 A152788 * A094143 A217260 A009775
Adjacent sequences: A055109 A055110 A055111 * A055113 A055114 A055115
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Henry Bottomley, Jun 15 2000
|
|
|
STATUS
|
approved
|
| |
|
|
