A118059 288*n^2 - 168*n - 119.

1, 697, 1969, 3817, 6241, 9241, 12817, 16969, 21697, 27001, 32881, 39337, 46369, 53977, 62161, 70921, 80257, 90169, 100657, 111721, 123361, 135577, 148369, 161737, 175681, 190201, 205297, 220969, 237217, 254041, 271441, 289417, 307969

Offset

1,2

Comments

In general, all sequences of equations which contain every positive integer in order exactly once (a pairwise equal summed, ordered partition of the positive integers) may be defined as follows: For all k, let x(k)=A001652(k) and z(k)=A001653(k). Then if we define a(n) to be (x(k)+z(k))n^2-(z(k)-1)n-x(k), the following equation is true: a(n)+(a(n)+1)+...+(a(n)+(x(k)+z(k))n+(2x(k)+z(k)-1)/2)=(a(n)+(x(k)+z(k))n+(2x(k)+z(k)+1)/2)+...+(a(n)+2(x(k)+z(k))n+x(k)); a(n)+2(x(k)+z(k))n+x(k))=a(n+1)-1; e.g., in this sequence, x(3)=A001652(3)=119 and z(3)=A001653(3)=169; cf. A000290, A118057-A118058, A118060-A118061.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: x*(1+694*x-119*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012

a(n)+(a(n)+1)+...+(a(n)+288n+203)=(a(n)+288n+204)+...+a(n+1)-1; a(n+1)-1=a(n)+576n+119.

a(n)+(a(n)+1)+...+(a(n)+288n+203)=6(24n-7)(24n+5)(24n+17); e.g., 1969+1970+...+3036=2672670=6*65*77*89.

Example

a(3)=288*3^2-168*3-119=337, a(4)=288*4^2-168*4-119=3817 and 1969+1970+...+3036=3037+...+3816

Mathematica

Table[288*n^2 - 168*n - 119, {n, 100}] (* Vincenzo Librandi, Jul 08 2012 *)

Prog

(Magma) [288*n^2 - 168*n - 119: n in [1..50]]; // Vincenzo Librandi, Jul 08 2012
(PARI) a(n)=288*n^2-168*n-119 \\ Charles R Greathouse IV, Jun 17 2017

Crossrefs

Sequence in context: A137559 A306097 A185377 * A028500 A133251 A243837

Adjacent sequences: A118056 A118057 A118058 * A118060 A118061 A118062

Keyword

nonn,easy,less

Author

Charlie Marion, Apr 26 2006

Extensions

Corrected by T. D. Noe, Nov 13 2006

Status

approved