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A247608
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a(n) = Sum_{k=0..3} binomial(6,k)*binomial(n,k).
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8
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1, 7, 28, 84, 195, 381, 662, 1058, 1589, 2275, 3136, 4192, 5463, 6969, 8730, 10766, 13097, 15743, 18724, 22060, 25771, 29877, 34398, 39354, 44765, 50651, 57032, 63928, 71359, 79345, 87906, 97062, 106833, 117239, 128300, 140036, 152467, 165613, 179494
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OFFSET
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0,2
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LINKS
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FORMULA
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G.f.: (1+3*x+6*x^2+10*x^3)/(1-x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
a(n) = (6+31*n-15*n^2+20*n^3)/6.
a(n) = 1+6*Binomial(n,1)+15*Binomial(n,2)+20*Binomial(n,3).
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MATHEMATICA
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Table[(6 + 31 n - 15 n^2 + 20 n^3)/6, {n, 0, 50}] (* or *) CoefficientList[Series[(1 + 3 x + 6 x^2 + 10 x^3)/(1-x)^4, {x, 0, 50}], x]
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PROG
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(Magma) [(6+31*n-15*n^2+20*n^3)/6: n in [0..40]]; /* or */ [1+6*Binomial(n, 1)+15*Binomial(n, 2)+20*Binomial(n, 3): n in [0..40]]; /* or */ I:=[1, 7, 28, 84]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]
(PARI) Vec((1+3*x+6*x^2+10*x^3)/(1-x)^4 + O (x^50)) \\ Michel Marcus, Sep 22 2014
(Sage) m=3; [sum((binomial(2*m, k)*binomial(n, k)) for k in (0..m)) for n in (0..40)] # Bruno Berselli, Sep 22 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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