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%I #45 Sep 08 2022 08:45:40
%S 5,25,75,175,350,630,1050,1650,2475,3575,5005,6825,9100,11900,15300,
%T 19380,24225,29925,36575,44275,53130,63250,74750,87750,102375,118755,
%U 137025,157325,179800,204600,231880,261800,294525,330225,369075,411255
%N a(n) = E(k)*C(n+k,k) = Euler(k)*binomial(n+k,k) for k=4.
%C a(n) = E(4)*binomial(n+4,4) where E(n) are the Euler number in the enumeration A122045.
%C a(n) is the special case k=4 in the sequence of diagonals in the triangle A153641.
%C a(n) is the 5th row in A093375.
%C a(n) is the 5th column in A103406.
%C a(n) is the 5th antidiagonal in A103283.
%C (a(n+1) - a(n))/5 are the pyramidal numbers A000292 (n>1).
%C (a(n+2) - 2a(n+1) + a(n))/5 are the triangular numbers A000217 (n>2).
%C (a(n+3) - 3a(n+2) + 3a(n+1) - a(n))/5 are the natural numbers A000027 (n > 3).
%C 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
%H G. C. Greubel, <a href="/A154286/b154286.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = (n+1)*(n+2)*(n+3)*(n+4)*5/24.
%F a(n) = a(n-1)*(n+4)/n (n>0), a(0)=5.
%F O.g.f.: 5/(1-x)^5.
%F E.g.f.: (5/24)*x*(24 + 36*x + 12*x^2 + x^3)*exp(x). - _G. C. Greubel_, Sep 09 2016
%F a(n) = 5*A000332(n+4). - _Michel Marcus_, Sep 10 2016
%p seq(euler(4)*binomial(n+4,4),n=0..32);
%t CoefficientList[Series[-5/(x - 1)^5, {x, 0, 35}], x] (* _Robert G. Wilson v_, Jan 29 2015 *)
%t Table[(n + 1)*(n + 2)*(n + 3)*(n + 4)*5/24, {n, 0, 25}] (* _G. C. Greubel_, Sep 09 2016 *)
%t LinearRecurrence[{5,-10,10,-5,1},{5,25,75,175,350},40] (* _Harvey P. Dale_, Nov 18 2021 *)
%o (Magma) [(n+1)*(n+2)*(n+3)*(n+4)*5 div 24: n in [0..40]]; // _Vincenzo Librandi_, Sep 10 2016
%o (PARI) x='x+O('x^99); Vec(5/(1-x)^5) \\ _Altug Alkan_, Sep 10 2016
%Y Cf. A000217, A153641, A000579.
%K easy,nonn
%O 0,1
%A _Peter Luschny_, Jan 06 2009
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