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(4)=A001652(4)=696 and z(4)=A001653(4)=985; cf. A000290, A118057-A118059, 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+4057*x-696*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012
a(n)+(a(n)+1)+...+(a(n)+1681n+1188) = (a(n)+1681n+1189)+ ... +a(n+1)-1; a(n+1)-1 = a(n)+3362n+696.
a(n)+(a(n)+1)+...+(a(n)+1681n+1188)=41(41n-12)(41n+29)(82n+17)/2; e.g., 11481+11482+...+17712=90965388=41*111*152*263/2.
EXAMPLE
a(3)=1681*3^2-984*3-696=11481, a(4)=1681*4^2-984*4-696=22264 and 11481+11482+...+17712=17713+...+22263
MATHEMATICA
CoefficientList[Series[(1+4057*x-696*x^2)/(1-x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 09 2012 *)
LinearRecurrence[{3, -3, 1}, {1, 4060, 11481}, 40] (* Harvey P. Dale, Oct 28 2016 *)
PROG
(Magma) [1681*n^2 - 984*n - 696: n in [1..40]]; // Vincenzo Librandi, Jul 09 2012
(PARI) a(n)=1681*n^2-984*n-696 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy,less
AUTHOR
Charlie Marion, Apr 26 2006
EXTENSIONS
Corrected by T. D. Noe, Nov 13 2006
STATUS
approved