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288*n^2 - 168*n - 119.
3

%I #24 Sep 08 2022 08:45:24

%S 1,697,1969,3817,6241,9241,12817,16969,21697,27001,32881,39337,46369,

%T 53977,62161,70921,80257,90169,100657,111721,123361,135577,148369,

%U 161737,175681,190201,205297,220969,237217,254041,271441,289417,307969

%N 288*n^2 - 168*n - 119.

%C 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.

%H Vincenzo Librandi, <a href="/A118059/b118059.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F 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

%F 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.

%F 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.

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

%t Table[288*n^2 - 168*n - 119, {n, 100}] (* _Vincenzo Librandi_, Jul 08 2012 *)

%o (Magma) [288*n^2 - 168*n - 119: n in [1..50]]; // _Vincenzo Librandi_, Jul 08 2012

%o (PARI) a(n)=288*n^2-168*n-119 \\ _Charles R Greathouse IV_, Jun 17 2017

%K nonn,easy,less

%O 1,2

%A _Charlie Marion_, Apr 26 2006

%E Corrected by _T. D. Noe_, Nov 13 2006