A071235 a(n) = (n^12 + n^6)/2.

0, 1, 2080, 266085, 8390656, 122078125, 1088414496, 6920702425, 34359869440, 141215033961, 500000500000, 1569215074141, 4458051717120, 11649044974645, 28346959952416, 64873174640625, 140737496743936, 291311130683665, 578415707719200, 1106657483056021

Offset: 0

Author

N. J. A. Sloane, Jun 12 2002

Keywords

• nonn
• easy

Comments

Number of unoriented rows of length 12 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)=2080, there are 2^12=4096 oriented arrangements of two colors. Of these, 2^6=64 are achiral. That leaves (4096-64)/2=2016 chiral pairs. Adding achiral and chiral, we get 2080. - 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 (13, -78, 286, -715, 1287, -1716, 1716, -1287, 715, -286, 78, -13, 1).

Crossrefs

Row 12 of A277504.

Cf. A008456 (oriented), A001014 (achiral).

Programs

GAP

List([0..40], n -> (n^12 + n^6)/2); # G. C. Greubel, Nov 15 2018

Magma

[n^6*(n^2+1)*(n^4-n^2+1)/2: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011

Mathematica

Table[(n^12 + n^6)/2, {n,0,30}] (* Robert A. Russell, Nov 13 2018 *)

PARI

vector(40, n, n--; ) \\ G. C. Greubel, Nov 15 2018

Python

for n in range(0,20): print(int((n**12 + n**6)/2), end=', ') # Stefano Spezia, Nov 15 2018

Sage

[n^6*(1 + n^6)/2 for n in range(40)] # G. C. Greubel, Nov 15 2018

Formula

a(n) = n^6*(n^2 + 1)*(n^4 - n^2 + 1)/2.

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

a(n) = (A008456(n) + A001014(n)) / 2 = (n^12 + n^6) / 2.

G.f.: (Sum_{j=1..12} S2(12,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..6} S2(6,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.

G.f.: x*Sum_{k=0..11} A145882(12,k) * x^k / (1-x)^13.

E.g.f.: (Sum_{k=1..12} S2(12,k)*x^k + Sum_{k=1..6} S2(6,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.

For n>12, a(n) = Sum_{j=1..13} -binomial(j-14,j) * a(n-j). (End)

From G. C. Greubel, Nov 15 2018: (Start)

G.f.: x*(1 +2067*x +239123*x^2 +5093505*x^3 [phone]*x^4 + 81255642*x^5 [phone]*x^6 [phone]*x^7 +5093505*x^8 +239123*x^9 +2067*x^10 +x^11)/( 1-x)^13.

E.g.f.: x*(2 +2078*x +86616*x^2 +611566*x^3 +1379415*x^4 +*1323653*x^5 + 627396*x^6 +159027*x^7 +22275*x^8 +1705*x^9 +66*x^10 +x^11)*exp(x)/2. (End)

Extensions

New name from G. C. Greubel, Nov 15 2018