A071231 a(n) = (n^8 + n^4)/2.

0, 1, 136, 3321, 32896, 195625, 840456, 2883601, 8390656, 21526641, 50005000, 107186761, 215001216, 407879641, 737913736, 1281470625, 2147516416, 3487920481, 5510032776, 8491846681, 12800080000, 18911526921, 27438053896, 39155632561, 55037822976

Offset: 0

Author

N. J. A. Sloane, Jun 11 2002

Keywords

• nonn
• easy

Comments

Number of unoriented rows of length 8 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=136, there are 2^8=256 oriented arrangements of two colors. Of these, 2^4=16 are achiral. That leaves (256-16)/2=120 chiral pairs. Adding achiral and chiral, we get 136. - Robert A. Russell, Nov 13 2018

References

• T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..2000
• Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Crossrefs

Row 8 of A277504.

Cf. A001016 (oriented), A000583 (achiral).

Cf. A000583, A001016.

Cf. A008277, A145882.

Subsequence of A000217 (triangular numbers).

Programs

GAP

List([0..50], n -> (n^4 + n^8)/2); # G. C. Greubel, Nov 15 2018

Magma

[(n^8 + n^4)/2: n in [0..50]]; // Vincenzo Librandi, Jun 14 2011

Mathematica

Table[(n^8+n^4)/2,{n,0,30}] (* or *) LinearRecurrence[{9,-36,84,-126,126, -84,36,-9,1},{0,1,136,3321,32896,195625,840456,2883601,8390656},30] (* Harvey P. Dale, Oct 11 2011 *)

PARI

vector(50, n, n--; (n^8 + n^4)/2) \\ G. C. Greubel, Nov 15 2018

Sage

[(n^4 + n^8)/2 for n in range(50)] # G. C. Greubel, Nov 15 2018

Formula

From Harvey P. Dale, Oct 11 2011: (Start)

a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9); a(0)=0, a(1)=1, a(2)=136, a(3)=3321, a(4)=32896, a(5)=195625, a(6)=840456, a(7)=2883601, a(8)=8390656.

G.f.: -((x*(x+1)*(x*(x*(x*(x*(x*(x+126)+2007)+5812)+2007)+126)+1))/ (x-1)^9). (End)

From Robert A. Russell, Nov 13 2018: (Start)

a(n) = (A001016(n) + A000583(n)) / 2 = (n^8 + n^4) / 2.

G.f.: (Sum_{j=1..8} S2(8,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.

G.f.: x*Sum_{k=0..7} A145882(8,k) * x^k / (1-x)^9.

E.g.f.: (Sum_{k=1..8} S2(8,k)*x^k + Sum_{k=1..4} S2(4,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.

For n>8, a(n) = Sum_{j=1..9} -binomial(j-10,j) * a(n-j). (End)

E.g.f.: x*(2 + 134*x + 972*x^2 + 1702*x^3 + 1050*x^4 + 266*x^5 + 28*x^6 + x^7)*exp(x)/2. - G. C. Greubel, Nov 15 2018