A071232 a(n) = (n^6 + n^3)/2.

0, 1, 36, 378, 2080, 7875, 23436, 58996, 131328, 266085, 500500, 886446, 1493856, 2414503, 3766140, 5697000, 8390656, 12071241, 17009028, 23526370, 32004000, 42887691, 56695276, 74024028, 95558400, 122078125, 154466676, 193720086, 240956128, 297423855, 364513500

Offset: 0

Author

N. J. A. Sloane, Jun 11 2002

Keywords

• nonn
• easy

Comments

Number of unoriented rows of length 6 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)=36, there are 2^6=64 oriented arrangements of two colors. Of these, 2^3=8 are achiral. That leaves (64-8)/2=28 chiral pairs. Adding achiral and chiral, we get 36. - Robert A. Russell, Nov 14 2018

For n > 0, a(2n+1) is the number of non-isomorphic 8C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 16 2019

References

• C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60 (2001), 85-96.
• T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
• T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..2000
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Row 6 of A277504.

Cf. A001014 (oriented), A085744 (chiral), A000578 (achiral).

Programs

GAP

List([0..50], n -> (n^6 + n^3)/2); # G. C. Greubel, Nov 15 2018

Magma

[(n^6 + n^3)/2: n in [0..50]]; // Vincenzo Librandi, Jun 14 2011

Mathematica

Table[(n^6+n^3)/2,{n,0,40}] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,36,378,2080,7875,23436},40] (* Harvey P. Dale, Nov 06 2011 *)

PARI

vector(50, n, n--; (n^6 + n^3)/2) \\ G. C. Greubel, Nov 15 2018

Sage

[(n^6 + n^3)/2 for n in range(50)] # G. C. Greubel, Nov 15 2018

Formula

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7); a(0)=0, a(1)=1, a(2)=36, a(3)=378, a(4)=2080, a(5)=7875, a(6)=23436. - Harvey P. Dale, Nov 06 2011

G.f.: x*(28*x^4 + 155*x^3 + 147*x^2 + 29*x + 1)/(1-x)^7. - Colin Barker, Oct 12 2012

From Robert A. Russell, Nov 14 2018: (Start)

a(n) = (A001014(n) + A000578(n)) / 2 = (n^6 + n^3) / 2.

a(n) = A001014(n) - A085744(n) = A085744(n) + A000578(n).

G.f.: (Sum_{j=1..6} S2(6,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..3} S2(3,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.

G.f.: x*Sum_{k=0..5} A145882(6,k) * x^k / (1-x)^7.

E.g.f.: (Sum_{k=1..6} S2(6,k)*x^k + Sum_{k=1..3} S2(3,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.

For n>6, a(n) = Sum_{j=1..7} -binomial(j-8,j) * a(n-j). (End)

E.g.f.: x*(2 +34*x +91*x^2 +65*x^3 +15*x^4 +x^5)*exp(x)/2. - G. C. Greubel, Nov 15 2018

a(n) = A000217(n^3), sum of the integers up to the n'th cube. - R. J. Mathar, Mar 11 2025