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A023811
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Largest metadrome (number with digits in strict ascending order) in base n.
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20
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0, 1, 5, 27, 194, 1865, 22875, 342391, 6053444, 123456789, 2853116705, 73686780563, 2103299351334, 65751519677857, 2234152501943159, 81985529216486895, 3231407272993502984, 136146740744970718253, 6106233505124424657789, 290464265927977839335179
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OFFSET
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1,3
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COMMENTS
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LINKS
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FORMULA
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a(n) = Sum_{j=1...n-1} j*n^(n-1-j).
lim_{n->infinity} a(n)/a(n-1) - a(n-1)/a(n-2) = exp(1). - Conjectured by Gerald McGarvey, Sep 26 2004. Follows from the formula below and lim_{n->infinity} (1+1/n)^n = e. - Franklin T. Adams-Watters, Jan 25 2010
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EXAMPLE
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a(5) = 1234[5] (in base 5) = 1*5^3 + 2*5^2 + 3*5 + 4 = 125 + 50 + 15 + 4 = 194.
a(10) = 123456789 (in base 10).
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MAPLE
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0, seq((n^n-n^2+n-1)/(n-1)^2, n=2..100); # Robert Israel, Dec 13 2015
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MATHEMATICA
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Table[Total[(#1 n^#2) & @@@ Transpose@ {Range[n - 1], Reverse@ (Range[n - 1] - 1)}], {n, 20}] (* Michael De Vlieger, Jul 24 2015 *)
Table[Sum[(b - k)*b^(k - 1), {k, b - 1}], {b, 30}] (* Clark Kimberling, Aug 22 2015 *)
Table[FromDigits[Range[0, n - 1], n], {n, 20}] (* L. Edson Jeffery, Dec 13 2015 *)
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PROG
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(PARI) {for(i=1, 18, cuo=0; for(j=1, i-1, cuo=cuo+j*i^(i-j-1)); print1(cuo, ", "))} \\\ Douglas Latimer, May 16 2012
(Magma) [0] cat [(n^n-n^2+n-1)/(n-1)^2: n in [2..20]]; // Vincenzo Librandi, May 22 2012
(Haskell)
a023811 n = foldl (\val dig -> val * n + dig) 0 [0 .. n - 1]
(Python)
def a(n): return (n**n - n**2 + n - 1)//((n - 1)**2) if n > 1 else 0
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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