OFFSET
0,1
COMMENTS
a(n) = E(4)*binomial(n+4,4) where E(n) are the Euler number in the enumeration A122045.
a(n) is the special case k=4 in the sequence of diagonals in the triangle A153641.
a(n) is the 5th row in A093375.
a(n) is the 5th column in A103406.
a(n) is the 5th antidiagonal in A103283.
(a(n+1) - a(n))/5 are the pyramidal numbers A000292 (n>1).
(a(n+2) - 2a(n+1) + a(n))/5 are the triangular numbers A000217 (n>2).
(a(n+3) - 3a(n+2) + 3a(n+1) - a(n))/5 are the natural numbers A000027 (n > 3).
Number of orbits of Aut(Z^7) as function of the infinity norm (n+4) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 107520. - Philippe A.J.G. Chevalier, Dec 28 2015
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = (n+1)*(n+2)*(n+3)*(n+4)*5/24.
a(n) = a(n-1)*(n+4)/n (n>0), a(0)=5.
O.g.f.: 5/(1-x)^5.
E.g.f.: (5/24)*x*(24 + 36*x + 12*x^2 + x^3)*exp(x). - G. C. Greubel, Sep 09 2016
a(n) = 5*A000332(n+4). - Michel Marcus, Sep 10 2016
MAPLE
seq(euler(4)*binomial(n+4, 4), n=0..32);
MATHEMATICA
CoefficientList[Series[-5/(x - 1)^5, {x, 0, 35}], x] (* Robert G. Wilson v, Jan 29 2015 *)
Table[(n + 1)*(n + 2)*(n + 3)*(n + 4)*5/24, {n, 0, 25}] (* G. C. Greubel, Sep 09 2016 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {5, 25, 75, 175, 350}, 40] (* Harvey P. Dale, Nov 18 2021 *)
PROG
(Magma) [(n+1)*(n+2)*(n+3)*(n+4)*5 div 24: n in [0..40]]; // Vincenzo Librandi, Sep 10 2016
(PARI) x='x+O('x^99); Vec(5/(1-x)^5) \\ Altug Alkan, Sep 10 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter Luschny, Jan 06 2009
STATUS
approved