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a(n) = 49n^2 - 28n - 20.
3

%I #23 Sep 08 2022 08:45:24

%S 1,120,337,652,1065,1576,2185,2892,3697,4600,5601,6700,7897,9192,

%T 10585,12076,13665,15352,17137,19020,21001,23080,25257,27532,29905,

%U 32376,34945,37612,40377,43240,46201,49260,52417,55672,59025,62476,66025

%N a(n) = 49n^2 - 28n - 20.

%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(2)=A001652(2) and z(2)=A001653(2)=29; cf. A000290,A118057,A118059-A118061.

%H Vincenzo Librandi, <a href="/A118058/b118058.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)+(a(n)+1)+...+(a(n)+98n+34)=7(7n-2)(7n+5)(14n+3)/2; e.g., 337+338+...+518=77805=7*19*26*45/2.

%F a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: x*(1+117*x-20*x^2)/(1-x)^3. - _Colin Barker_, Jun 30 2012

%e a(3)=49*3^2-28*3-20=337, a(4)=49*4^2-28*4-20=652 and 337+338+...+518=519+...+651.

%t Table[49*n^2 - 28*n - 20, {n, 10}] (* _Vincenzo Librandi_, Jul 08 2012 *)

%o (Magma) [49*n^2-28*n-20: n in [1..50]]; // _Vincenzo Librandi_, Jul 08 2012

%o (PARI) a(n)=49*n^2-28*n-20 \\ _Charles R Greathouse IV_, Jun 17 2017

%K nonn,easy

%O 1,2

%A _Charlie Marion_, Apr 26 2006

%E Corrected by _Franklin T. Adams-Watters_ and _T. D. Noe_, Oct 25 2006