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(1)=A001652(1)=3 and z(1)=A001653(1)=5; cf. A000290, A118058-A118061.
Sequence found by reading the segment (1, 21) together with the line from 21, in the direction 21, 57, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 04 2011
LINKS
Harvey P. Dale, 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+18*x-3*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012
a(n)+(a(n)+1)+...+(a(n)+8n+5)=(a(n)+8n+6)+...+a(n+1)-1; a(n+1)-1=a(n)+16n+3.
a(n)+(a(n)+1)+...+(a(n)+8n+5)=(4n-1)(4n+1)(4n+3); e.g., 21+22+...+56=693=7*9*11.
a(n) = 16*n+a(n-1)-12 (with a(1)=1). - Vincenzo Librandi, Nov 13 2010
EXAMPLE
a(3)=8*3^2-4*3-3=57, a(4)=8*4^2-4*4-3=109 and 57+58+...+86=87+...+108.
MATHEMATICA
Table[8n^2-4n-3, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 21, 57}, 50] (* Harvey P. Dale, Sep 18 2012 *)
PROG
(PARI) a(n)=8*n^2-4*n-3 \\ Charles R Greathouse IV, Oct 07 2015
(Magma) [8*n^2-4*n-3 : n in [1..60]]; // Wesley Ivan Hurt, Jan 28 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Charlie Marion, Apr 26 2006
STATUS
approved