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A154286
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a(n) = E(k)*C(n+k,k) = Euler(k)*binomial(n+k,k) for k=4.
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14
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5, 25, 75, 175, 350, 630, 1050, 1650, 2475, 3575, 5005, 6825, 9100, 11900, 15300, 19380, 24225, 29925, 36575, 44275, 53130, 63250, 74750, 87750, 102375, 118755, 137025, 157325, 179800, 204600, 231880, 261800, 294525, 330225, 369075, 411255
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OFFSET
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0,1
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COMMENTS
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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 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
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LINKS
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FORMULA
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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
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MAPLE
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seq(euler(4)*binomial(n+4, 4), n=0..32);
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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