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A016123
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a(n) = (11^(n+1) - 1)/10.
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58
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1, 12, 133, 1464, 16105, 177156, 1948717, 21435888, 235794769, 2593742460, 28531167061, 313842837672, 3452271214393, 37974983358324, 417724816941565, 4594972986357216, 50544702849929377, 555991731349223148
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OFFSET
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0,2
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COMMENTS
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11^a(n) is highest power of 11 dividing (11^(n+1))!.
Partial sums of powers of 11 (A001020).
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=11, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=12, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
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LINKS
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Eric Weisstein's World of Mathematics, Repunit
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FORMULA
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a(n) = Sum_{k=0..n} 11^k = (11^(n+1) - 1)/10.
G.f.: (1/(1-11*x) - 1/(1-x))/(10*x) = 1/((1-11*x)*(1-x)).
a(0)=0, a(1)=1, a(n) = 12*a(n-1) - 11*a(n-2). - Harvey P. Dale, Apr 05 2012
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MATHEMATICA
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(11^Range[0, 20]-1)/10 (* or *) LinearRecurrence[{12, -11}, {0, 1}, 20] (* Harvey P. Dale, Apr 05 2012 *)
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PROG
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(Sage) [lucas_number1(n, 12, 11) for n in range(1, 19)] # Zerinvary Lajos, Apr 27 2009
(Sage) [gaussian_binomial(n, 1, 11) for n in range(1, 19)] # Zerinvary Lajos, May 28 2009
(Maxima) A016123(n):=(11^(n+1)-1)/10$
(Magma) [(11^(n+1)-1)/10: n in [0..30]]; // G. C. Greubel, Feb 21 2024
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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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EXTENSIONS
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STATUS
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approved
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