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The sum of the first n integers, with every third integer taken negative.
4

%I #92 Mar 15 2024 10:12:19

%S 1,3,0,4,9,3,10,18,9,19,30,18,31,45,30,46,63,45,64,84,63,85,108,84,

%T 109,135,108,136,165,135,166,198,165,199,234,198,235,273,234,274,315,

%U 273,316,360,315,361,408,360,409,459,408,460,513,459,514,570,513,571,630

%N The sum of the first n integers, with every third integer taken negative.

%C The partial sum for the first 10^k terms are 76, 57256, 55722556, 55572225556, 55557222255556,..., i.e., the palindrome 5{k}2{k-1}5{k} plus 1+2*10^(2*k-1). - _R. J. Cano_, Mar 10 2013, edited by _M. F. Hasler_, Mar 25 2013

%H Vincenzo Librandi, <a href="/A181482/b181482.txt">Table of n, a(n) for n = 1..1000</a>

%H Wolfram Alpha, <a href="http://www.wolframalpha.com/input/?i=Sum%5Bn*%28-1%29%5E%28%28n%2B1%29Mod3%29%2Cn%3D1..k%5D">WA Query</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,2,-2,0,-1,1).

%F From _R. J. Mathar_, Oct 23 2010: (Start)

%F a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7).

%F G.f.: -x*(1+2*x+2*x^3+x^4-3*x^2) / ( (1+x+x^2)^2*(x-1)^3 ).

%F a(n) = 2*A061347(n+1)/9 +4/9 + n*(n+1)/6 + 2*b(n)/3 where b(3k+1) = 0, b(3k) = -3k - 1 and b(3k+2) = 3k + 3. (End)

%F a(n) = sum((i+1)*A131561(i), i=0..n-1) = A000217(n)-6*A000217(floor(n/3)). [_Bruno Berselli_, Dec 10 2010]

%F a(0) = 0, a(n) = a(n-1) + (-1)^((n + 1) mod 3)*n - _Jon Perry_, Feb 17 2013

%F a(n) = n*(n+1)/2-3*floor(n/3)*(floor(n/3)+1). - _R. J. Cano_, Mar 01 2013 [Same as Berselli's formula. - Ed.]

%F a(3k) = 3k(k-1)/2. - _Jon Perry_, Mar 01 2013

%F a(0) = 0, a(n) = a(n-1) + (1 - ((n+1) mod 3 mod 2) * 2) * n. - _Jon Perry_, Mar 03 2013

%e a(7) = 1 + 2 - 3 + 4 + 5 - 6 + 7 = 10.

%t a[n_] := Sum[If[Mod[j, 3] == 0, -j, j], {j, 1, n}]; Table[a[i], {i, 1, 50, 1}] (* _Jon Perry_ *)

%t tri[n_] := n (n + 1)/2; f[n_] := tri@ n - 6 tri@ Floor[n/3]; Array[f, 63] (* _Robert G. Wilson v_, Oct 24 2010 *)

%t CoefficientList[Series[-(1 + 2*x + 2*x^3 + x^4 - 3*x^2)/((1 + x + x^2)^2*(x - 1)^3), {x, 0,30}], x] (* _Vincenzo Librandi_, Feb 17 2013 *)

%t Table[Sum[k * (-1)^Boole[Mod[k, 3] == 0], {k, n}], {n, 60}] (* _Alonso del Arte_, Feb 24 2013 *)

%t With[{nn=20},Accumulate[Times@@@Partition[Riffle[Range[3nn],{1,1,-1}],2]]] (* _Harvey P. Dale_, Feb 09 2015 *)

%o (JavaScript) c = 0; for (i = 1; i < 100; i++) {c += Math.pow(-1, (i + 1) % 3)*i; document.write(c, ", ");} // _Jon Perry_, Feb 17 2013

%o (JavaScript) c=0; for (i = 1; i < 100; i++) { c += (1 - (i + 1) % 3 % 2 * 2) * i; document.write(c + ", "); } // _Jon Perry_, Mar 03 2013

%o (Magma) I:=[1,3,0,4,9,3,10]; [n le 7 select I[n] else Self(n-1)+2*Self(n-3)-2*Self(n-4)-Self(n-6)+Self(n-7): n in [1..60]]; // _Vincenzo Librandi_, Feb 17 2013

%o (PARI) a(n)=sum(k=1,n,k*((-1)^(k%3==0)) ) \\ _R. J. Cano_, Feb 26 2013

%o (PARI) a(n)={my(y=n\3);n*(n+1)\2-3*y*(y+1)} \\ _R. J. Cano_, Feb 28 2013

%o (Haskell)

%o a181482 n = a181482_list !! (n-1)

%o a181482_list = scanl1 (+) $ zipWith (*) [1..] $ cycle [1, 1, -1]

%o -- _Reinhard Zumkeller_, Nov 23 2014

%Y Cf. A213203, A000217.

%K nonn,easy

%O 1,2

%A _Jon Perry_, Oct 23 2010

%E More terms added by _R. J. Mathar_, Oct 23 2010