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A305459
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a(0) = 1, a(1) = 3, a(n) = 3*n*a(n-1) + a(n-2).
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4
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1, 3, 19, 174, 2107, 31779, 574129, 12088488, 290697841, 7860930195, 236118603691, 7799774851998, 281028013275619, 10967892292601139, 460932504302523457, 20752930585906156704, 996601600627798045249, 50847434562603606464403
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OFFSET
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0,2
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COMMENTS
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Let S(i,j,n) denote a sequence of the form a(0) = 1, a(1) = i, a(n) = i*n*a(n-1) + j*a(n-2). Then S(i,j,n) = Sum_{k=0..floor(n/2)} ((n-k)!/k!)*binomial(n-k,k)*i^(n-2*k)*j^k.
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LINKS
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FORMULA
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a(n) = Sum_{k=0..floor(n/2)} ((n-k)!/k!)*binomial(n-k,k)*3^(n-2*k).
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MAPLE
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a:=proc(n) option remember: if n=0 then 1 elif n=1 then 3 elif n>=2 then 3*n*procname(n-1)-procname(n-2) fi; end:
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MATHEMATICA
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RecurrenceTable[{a[0]==1, a[1]==3, a[n]==3n a[n-1]+a[n-2]}, a, {n, 20}] (* Harvey P. Dale, Aug 27 2019 *)
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PROG
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(PARI) {a(n) = sum(k=0, n/2, ((n-k)!/k!)*binomial(n-k, k)*3^(n-2*k))}
(GAP) List([0..20], n->Sum([0..Int(n/2)], k->((Factorial(n-k))/(Factorial(k))*Binomial(n-k, k)*3^(n-2*k)))); # Muniru A Asiru, Jun 01 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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