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A095794 a(n) = A005449(n) - 1, where A005449 = second pentagonal numbers.

%I A095794 #88 Jun 06 2023 08:17:42
%S A095794 1,6,14,25,39,56,76,99,125,154,186,221,259,300,344,391,441,494,550,
%T A095794 609,671,736,804,875,949,1026,1106,1189,1275,1364,1456,1551,1649,1750,
%U A095794 1854,1961,2071,2184,2300,2419,2541,2666,2794,2925,3059,3196,3336,3479,3625
%N A095794 a(n) = A005449(n) - 1, where A005449 = second pentagonal numbers.
%C A095794 Row sums of triangle A131414.
%C A095794 Equals binomial transform of (1,5,3,0,0,0,...). Equals A051340 * (1,2,3,...).
%C A095794 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
%C A095794 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
%H A095794 Leo Tavares, <a href="/A095794/a095794.jpg">Triangle/Square Pairs</a>
%H A095794 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A095794 a(n) = (3/2)*n^2 + (1/2)*n - 1 = (n+1)*(3*n-2)/2.
%F A095794 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
%F A095794 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
%F A095794 a(n) = n*A016777(n-1) - Sum_{i=1..n-2} A016777(i) - (n-1) = (n+1)*(3*n-2)/2. - _Bruno Berselli_, May 04 2010
%F A095794 a(n) = 3*n + a(n-1)-1, for n>1, a(1)=1. - _Vincenzo Librandi_, Nov 16 2010
%F A095794 a(n) = A115067(-n). - _Bruno Berselli_, Sep 02 2011
%F A095794 From _Wesley Ivan Hurt_, Dec 22 2015: (Start)
%F A095794 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
%F A095794 a(n) = Sum_{i=n..2n} (i-1). (End)
%F A095794 E.g.f.: 1 + exp(x)*(3*x^2 + 4*x - 2)/2. - _Stefano Spezia_, Jun 04 2021
%F A095794 From _Amiram Eldar_, Feb 22 2022: (Start)
%F A095794 Sum_{n>=1} 1/a(n) = Pi/(5*sqrt(3)) + 3*log(3)/5 + 2/5.
%F A095794 Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(5*sqrt(3)) + 4*log(2)/5 - 2/5. (End)
%F A095794 a(n) = A000217(n) + A000290(n) - 1. - _Leo Tavares_, Jun 02 2023
%e A095794 a(4) = 25 = A005449(4) - 1.
%e A095794 a(5) = 39 = (3/2)*5^2 + (1/2)*5 - 1.
%e A095794 a(7) = 76 = 3*56 - 3*39 + 25.
%e A095794 a(5) = 39 = right term of M^4 * [1 1 1] = [1 5 39].
%e A095794 For n = 8, a(8) = 8*22 - (1+4+7+10+13+16+19) - 7 = 99. - _Bruno Berselli_, May 04 2010
%p A095794 A005449 := proc(n) RETURN(n*(3*n+1)/2) ; end: A095794 := proc(n) RETURN(A005449(n)-1) ; end: for n from 1 to 100 do printf("%a,",A095794(n)) ; od: # _R. J. Mathar_, Jun 23 2006
%p A095794 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
%t A095794 FoldList[## + 2 &, 1, 3 Range@ 45] (* _Robert G. Wilson v_, Feb 03 2011 *)
%t A095794 LinearRecurrence[{3,-3,1},{1,6,14},50] (* _Harvey P. Dale_, Dec 09 2013 *)
%o A095794 (PARI) a(n)=(3/2)*n^2+(1/2)*n-1 \\ _Charles R Greathouse IV_, Sep 24 2015
%o A095794 (Magma) [(3/2)*n^2 + (1/2)*n - 1 : n in [1..50]]; // _Wesley Ivan Hurt_, Dec 22 2015
%Y A095794 Cf. A000217, A005449, A016777, A049777, A051340, A115067, A126890, A131414.
%Y A095794 Cf. A000290.
%K A095794 nonn,easy
%O A095794 1,2
%A A095794 _Gary W. Adamson_, Jun 06 2004, Jul 08 2007
%E A095794 Corrected and extended by _R. J. Mathar_, Jun 23 2006