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A037012 Triangle read by rows; row 0 is 0; the n-th row for n>0 contains the coefficients in the expansion of (1-x)*(1+x)^(n-1).

%I A037012 #50 Jul 08 2025 10:39:32
%S A037012 0,1,-1,1,0,-1,1,1,-1,-1,1,2,0,-2,-1,1,3,2,-2,-3,-1,1,4,5,0,-5,-4,-1,
%T A037012 1,5,9,5,-5,-9,-5,-1,1,6,14,14,0,-14,-14,-6,-1,1,7,20,28,14,-14,-28,
%U A037012 -20,-7,-1,1,8,27,48,42,0,-42,-48,-27,-8,-1,1,9,35,75,90,42,-42,-90
%N A037012 Triangle read by rows; row 0 is 0; the n-th row for n>0 contains the coefficients in the expansion of (1-x)*(1+x)^(n-1).
%C A037012 The greatest term in the row n is reached when k is the nearest integer to (n - sqrt(n+1))/2. When n is one less than a square, and consequently this formula gives a half-integer, the maximum is reached twice. - _Ivan Neretin_, Apr 26 2016
%D A037012 A. A. Kirillov, Variations on the triangular theme, Amer. Math. Soc. Transl., (2), Vol. 169, 1995, pp. 43-73, see p. 71.
%H A037012 Ivan Neretin, <a href="/A037012/b037012.txt">Table of n, a(n) for n = 0..5150</a>
%H A037012 Tewodros Amdeberhan, Moa Apagodu, and Doron Zeilberger, <a href="http://arxiv.org/abs/1507.07660">Wilf's "Snake Oil" Method Proves an Identity in The Motzkin Triangle</a>, arXiv:1507.07660 [math.CO], 2015.
%H A037012 Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, <a href="https://www.emis.de/journals/JIS/VOL21/Falcao/falcao2.html">Combinatorial Identities Associated with a Multidimensional Polynomial Sequence</a>, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
%H A037012 Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, <a href="http://arxiv.org/abs/1602.04347">Sums of powers of Catalan triangle numbers</a>, arXiv:1602.04347 [math.NT], 2016 (see 1.2).
%F A037012 T(n, k) = T(n-1, k-1)+T(n-1, k); T(0, 0)=0, T(1, 0)=1, T(1, 1)=-1.
%F A037012 T(n, k) = C(n, k)-C(n, k-1) where C = binomial coefficient A007318.
%F A037012 G.f.: (1-y) / (1-x-x*y). - _Ralf Stephan_, Jan 23 2005
%F A037012 T(n,k) = binomial(n-1,k) - binomial(n-1,k-1), for n >= k. T(n,k)=0, for n < k. T(n,k) = Sum_{i=-k..k} (-1)^i*binomial(n-1,k+i)*binomial(n+1,k-i), for n > 0. Row sums are 0. - _Mircea Merca_, Apr 28 2012
%F A037012 a(n) = -A008482(n). - _Michael Somos_, May 24 2015
%F A037012 Sum of positive terms of the row n is the central binomial coefficient A001405(n-1).- _Ivan Neretin_, Apr 26 2016
%F A037012 T(n, n-k) = - T(n, k); T(n, 0) = 1; T(n, 1) = n-2; T(n, 2) = (n-3)(n-4)/2; T(2k,n) = 0; T(2k, k-1) = T(2k+1, k) = A000108(k). - _M. F. Hasler_, Feb 11 2019
%e A037012 Triangle begins:
%e A037012   0;
%e A037012   1, -1;
%e A037012   1,  0, -1;
%e A037012   1,  1, -1, -1;
%e A037012   1,  2,  0, -2, -1;
%e A037012   1,  3,  2, -2, -3, -1;
%e A037012   ...
%p A037012 T(n,k):=piecewise(n<k,0,k<=n,binomial(n-1,k)-binomial(n-1,k-1)) # _Mircea Merca_, Apr 28 2012
%t A037012 T[ n_, k_] := If[ n < 1, 0, Coefficient[ (1 - x) (1 + x)^(n - 1), x, k]]; (* _Michael Somos_, May 24 2015 *)
%t A037012 Flatten@NestList[Join[{1}, Most@# + Rest@#, {-1}] &, {0}, 11] (* _Ivan Neretin_, Apr 26 2016 *)
%o A037012 (PARI) {T(n, k) = if( n<1, 0, polcoeff( (1-x) * (1+x)^(n-1), k))}
%o A037012 (PARI) A037012(n, k)={if(k>=n-k, if(k>n-k, -A037012(n, n-k)), k>2, A037012(n-1, k-1)+A037012(n-1, k), k>1, (n-2)*(n-3)\2-1, k, n-2, 1)} \\ _M. F. Hasler_, Feb 11 2019
%Y A037012 Skew analog of Pascal's triangle A007318. Equals -A008482.
%Y A037012 Elements near the center give Catalan numbers A000108 repeated, cf. formula.
%Y A037012 Apart from initial term, same as A080232.
%Y A037012 Cf. A001405.
%K A037012 sign,easy,tabl
%O A037012 0,12
%A A037012 _N. J. A. Sloane_, _Michael Somos_