A346638 a(n) is the number of 6-tuples (a_1,a_2,a_3,a_4,a_5,a_6) having all terms in {1,...,n} such that there exists a hexagon with these side-lengths.

0, 1, 64, 729, 4096, 15619, 46614, 117481, 261640, 530181, 997228, 1766017, 2975688, 4808791, 7499506, 11342577, 16702960, 24026185, 33849432, 46813321, 63674416, 85318443, 112774222, 147228313, 190040376, 242759245, 307139716, 385160049, 479040184, 591260671

Offset: 0

Author

Giovanni Corbelli, Jul 26 2021

Keywords

• nonn
• easy

Comments

The existence of such a six-sided polygon implies that every element of the sextuple is less than the sum of the other elements.

Links

• Giovanni Corbelli, Visual Basic routine for generating number of six-sided polygons
• Giovanni Corbelli Proof of the formula: Number of k-tuples with elements in {1,2,...,N} corresponding to k-sided polygons
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Cf. A006003, A346636, A346637.

Programs

Mathematica

CoefficientList[Series[x (1+57x+302x^2+302x^3+51x^4+x^5)/(1-x)^7,{x,0,40}],x] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,64,729,4096,15619,46614},40] (* Harvey P. Dale, Oct 30 2024 *)

Formula

a(n) = n^6 - 6*binomial(n+1,6) = n^6 - (n+1)*binomial(n,5).

General formula for k-tuples: a_k(n) = n^k - k*binomial(n+1,k) = n^k - (n+1)*binomial(n,k-1).

G.f.: x*(1 + 57*x + 302*x^2 + 302*x^3 + 51*x^4 + x^5)/(1 - x)^7. - Stefano Spezia, Sep 27 2021