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 A034868 Left half of Pascal's triangle. 16
 1, 1, 1, 2, 1, 3, 1, 4, 6, 1, 5, 10, 1, 6, 15, 20, 1, 7, 21, 35, 1, 8, 28, 56, 70, 1, 9, 36, 84, 126, 1, 10, 45, 120, 210, 252, 1, 11, 55, 165, 330, 462, 1, 12, 66, 220, 495, 792, 924, 1, 13, 78, 286, 715, 1287, 1716, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 15 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS T(n,k) = A034869(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012 LINKS Reinhard Zumkeller, Rows n=0..150 of triangle, flattened EXAMPLE 1; 1; 1, 2; 1, 3; 1, 4,  6; 1, 5, 10; 1, 6, 15, 20; ... MATHEMATICA Flatten[ Table[ Binomial[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* Robert G. Wilson v, May 28 2005 *) PROG (Haskell) a034868 n k = a034868_tabf !! n !! k a034868_row n = a034868_tabf !! n a034868_tabf = map reverse a034869_tabf -- Reinhard Zumkeller, improved Dec 20 2015, Jul 27 2012 (PARI) for(n=0, 14, for(k=0, floor(n/2), print1(binomial(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Mar 31 2017 (Python) import math from sympy import binomial for n in range(15):     print([binomial(n, k) for k in range(int(math.floor(n/2)) + 1)]) # Indranil Ghosh, Mar 31 2017 CROSSREFS Cf. A007318, A107430, A062344, A122366, A027306 (row sums). Cf. A008619. Cf. A225860. Cf. A126257. Cf. A034869 (right half), A014413, A014462, A265848. Sequence in context: A295885 A329035 A082904 * A050382 A197956 A054072 Adjacent sequences:  A034865 A034866 A034867 * A034869 A034870 A034871 KEYWORD nonn,tabf,easy AUTHOR STATUS approved

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Last modified June 2 08:16 EDT 2020. Contains 334767 sequences. (Running on oeis4.)