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A095794
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a(n) = A005449(n) - 1, where A005449 = second pentagonal numbers.
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26
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1, 6, 14, 25, 39, 56, 76, 99, 125, 154, 186, 221, 259, 300, 344, 391, 441, 494, 550, 609, 671, 736, 804, 875, 949, 1026, 1106, 1189, 1275, 1364, 1456, 1551, 1649, 1750, 1854, 1961, 2071, 2184, 2300, 2419, 2541, 2666, 2794, 2925, 3059, 3196, 3336, 3479, 3625
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OFFSET
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1,2
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COMMENTS
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Equals binomial transform of (1,5,3,0,0,0,...). Equals A051340 * (1,2,3,...).
a(n) is essentially the case -1 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} (k-2)*i-(k-3). Thus P_{-1}(n) = n*(5-3*n)/2 and a(n) = -P_{-1}(n+2). - Peter Luschny, Jul 08 2011
Beginning with n=2, a(n) is the falling diagonal starting with T(1,3) in A049777 (as a square array). - Bob Selcoe, Oct 27 2014
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LINKS
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FORMULA
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a(n) = (3/2)*n^2 + (1/2)*n - 1 = (n+1)*(3*n-2)/2.
a(n) = A126890(n+1,n-2) for n>1. - Reinhard Zumkeller, Dec 30 2006, corrected by Jason Bandlow (jbandlow(AT)math.upenn.edu), Feb 28 2009
G.f.: x*(-1-3*x+x^2)/(-1+x)^3 = 1 - 3/(-1+x)^3 - 4/(-1+x)^2. - R. J. Mathar, Nov 19 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
a(n) = Sum_{i=n..2n} (i-1). (End)
Sum_{n>=1} 1/a(n) = Pi/(5*sqrt(3)) + 3*log(3)/5 + 2/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(5*sqrt(3)) + 4*log(2)/5 - 2/5. (End)
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EXAMPLE
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a(5) = 39 = (3/2)*5^2 + (1/2)*5 - 1.
a(7) = 76 = 3*56 - 3*39 + 25.
a(5) = 39 = right term of M^4 * [1 1 1] = [1 5 39].
For n = 8, a(8) = 8*22 - (1+4+7+10+13+16+19) - 7 = 99. - Bruno Berselli, May 04 2010
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MAPLE
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a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]-a[n-2]-3 od: seq(-a[n], n=2..50); # Zerinvary Lajos, Feb 18 2008
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MATHEMATICA
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LinearRecurrence[{3, -3, 1}, {1, 6, 14}, 50] (* Harvey P. Dale, Dec 09 2013 *)
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PROG
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(Magma) [(3/2)*n^2 + (1/2)*n - 1 : n in [1..50]]; // Wesley Ivan Hurt, Dec 22 2015
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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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