A109620 a(n) = (1/3)*n^3 - n^2 - (1/3)*n - 1.

-1, -2, -3, -2, 3, 14, 33, 62, 103, 158, 229, 318, 427, 558, 713, 894, 1103, 1342, 1613, 1918, 2259, 2638, 3057, 3518, 4023, 4574, 5173, 5822, 6523, 7278, 8089, 8958, 9887, 10878, 11933, 13054, 14243, 15502, 16833, 18238, 19719, 21278, 22917, 24638, 26443, 28334, 30313, 32382, 34543, 36798

Offset: 0

Author

Creighton Dement, Aug 01 2005

Keywords

• easy
• sign

Comments

It is interesting that this sequence is generated using the same rules as those given for A108618 (version "jes"; the initial seed is the floretion given in program code, below). In reference to those rules, we have: **Loop 0** + .5'i + .5i' + .5'ik' + .5'ji' + e **Loop 1** + 1.5'i - .5'j + 1.5i' - .5k' + .5'ii' + 1.5'ik' + 1.5'ji' + .5'kj' + 3e **Loop 2** + 3.5'i - 2'j + 3.5i' - 2k' + 2'ii' + 3.5'ik' + 3.5'ji' + 2'kj' + 5e **Loop 3** + 6.5'i - 5.5'j + 6.5i' - 5.5k' + 5.5'ii' + 6.5'ik' + 6.5'ji' + 5.5'kj' + 7e **Loop 4** + 10.5'i - 12'j + 10.5i' - 12k' + 12'ii' + 10.5'ik' + 10.5'ji' + 12'kj' + 9e **Loop 5** + 15.5'i - 22.5'j + 15.5i' - 22.5k' + 22.5'ii' + 15.5'ik' + 15.5'ji' + 22.5'kj' + 11e **Loop 6** + 21.5'i - 38'j + 21.5i' - 38k' + 38'ii' + 21.5'ik' + 21.5'ji' + 38'kj' + 13e **Loop 7** + 28.5'i - 59.5'j + 28.5i' - 59.5k' + 59.5'ii' + 28.5'ik' + 28.5'ji' + 59.5'kj' + 15e **Loop 8** + 36.5'i - 88'j + 36.5i' - 88k' + 88'ii' + 36.5'ik' + 36.5'ji' + 88'kj' + 17e **Loop 9** + 45.5'i - 124.5'j + 45.5i' - 124.5k' + 124.5'ii' + 45.5'ik' + 45.5'ji' + 124.5'kj' + 19e. a(n) is calculated by adding the real number coefficients of 'i, 'j and 'k (which is always 0 here) from the n-th loop and multiplying the result by -2.

Links

• Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Crossrefs

Cf. A006527, A002061.

Programs

Maple

seriestolist(series((2*x-1)*(x^2+1)/(x-1)^4, x=0,50)); -or- Floretion Algebra Multiplication Program, FAMP Code: -2jessumseq[ + .5'i + .5i' + .5'ik' + .5'ji' + e], Sumtype: sum[Y[15]] = sum[ * ]. Note: 2ibasesumseq = A002061, apart from initial term, -2jbasesumseq = A006527.

Mathematica

Table[n^3/3-n^2-n/3-1,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{-1,-2,-3,-2},60] (* Harvey P. Dale, Jul 21 2013 *)

Formula

a(n) = A006527(n) - A002061(n+1), g.f. (2*x-1)*(x^2+1)/(x-1)^4

a(0)=-1, a(1)=-2, a(2)=-3, a(3)=-2, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 21 2013