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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). 8
 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, 1, 5, 9, 5, -5, -9, -5, -1, 1, 6, 14, 14, 0, -14, -14, -6, -1, 1, 7, 20, 28, 14, -14, -28, -20, -7, -1, 1, 8, 27, 48, 42, 0, -42, -48, -27, -8, -1, 1, 9, 35, 75, 90, 42, -42, -90 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,12 COMMENTS 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 REFERENCES A. A. Kirillov, Variations on the triangular theme, Amer. Math. Soc. Transl., (2), Vol. 169, 1995, pp. 43-73, see p. 71. LINKS Ivan Neretin, Table of n, a(n) for n = 0..5150 Tewodros Amdeberhan, Moa Apagodu, Doron Zeilberger, Wilf's "Snake Oil" Method Proves an Identity in The Motzkin Triangle, arXiv:1507.07660 [math.CO], 2015. Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4. Pedro J. Miana, Hideyuki Ohtsuka, Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016 (see 1.2). FORMULA T(n, k) = T(n-1, k-1)+T(n-1, k); T(0, 0)=0, T(1, 0)=1, T(1, 1)=-1. T(n, k) = C(n, k)-C(n, k-1) where C = binomial coefficient A007318. G.f.: (1-y) / (1-x-x*y). - Ralf Stephan, Jan 23 2005 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 a(n) = -A008482(n). - Michael Somos, May 24 2015 Sum of positive terms of the row n is the central binomial coefficient A001405(n-1).- Ivan Neretin, Apr 26 2016 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 EXAMPLE Triangle begins: 0; 1, -1; 1, 0, -1; 1, 1, -1, -1; 1, 2, 0, -2, -1; 1, 3, 2, -2, -3, -1; ... MAPLE T(n, k):=piecewise(n=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 CROSSREFS Skew analog of Pascal's triangle A007318. Equals -A008482. Elements near the center give Catalan numbers A000108 repeated, cf. formula. Apart from initial initial term, same as A080232. Sequence in context: A061398 A080232 A008482 * A112467 A112466 A166348 Adjacent sequences: A037009 A037010 A037011 * A037013 A037014 A037015 KEYWORD sign,easy,tabl AUTHOR N. J. A. Sloane, Michael Somos STATUS approved

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Last modified August 8 12:42 EDT 2024. Contains 375021 sequences. (Running on oeis4.)