OFFSET
0,2
COMMENTS
Suppose the vertices of a triangle are (S(n), S(n+j)), (S(n+2*j), S(n+3*j)) and (S(n+4*j), S(n+5*j)) where S(n) is the n-th square number, A000290(n). Then the area of this triangle will be a(j). A generalization follows: let P(k,n) = the n-th k-gonal number and suppose the vertices of a triangle are (P(k,n), P(k,n+j)), (P(k,n+2*j), P(k,n+3*j)) and (P(k,n+4*j), P(k,n+5*j)). Then the area of this triangle will be (2*k-4)^2*j^4. See also A141046 for k = 3. - Charlie Marion, Mar 26 2021
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
G.f.: 16*x*(x+1)*(x^2+10*x+1)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
E.g.f.: 16*x*(1 + 7*x + 6*x^2 + x^3)*exp(x). - G. C. Greubel, Sep 15 2018
From Amiram Eldar, Oct 10 2020: (Start)
Sum_{n>=1} 1/a(n) = Pi^4/1440 (conjecturally A258945).
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/11520. (End)
MAPLE
MATHEMATICA
Table[(2*n)^4, {n, 0, 30}] (* G. C. Greubel, Sep 15 2018 *)
PROG
(Magma) [(2*n)^4: n in [0..40]]; // Vincenzo Librandi, Sep 05 2011
(PARI) vector(30, n, n--; (2*n)^4) \\ G. C. Greubel, Sep 15 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved