A214292 Triangle read by rows: T(n,k) = T(n-1,k-1) + T(n-1,k), 0 < k < n with T(n,0) = n and T(n,n) = -n.
0, 1, -1, 2, 0, -2, 3, 2, -2, -3, 4, 5, 0, -5, -4, 5, 9, 5, -5, -9, -5, 6, 14, 14, 0, -14, -14, -6, 7, 20, 28, 14, -14, -28, -20, -7, 8, 27, 48, 42, 0, -42, -48, -27, -8, 9, 35, 75, 90, 42, -42, -90, -75, -35, -9, 10, 44, 110, 165, 132, 0, -132, -165, -110, -44, -10
Offset: 0
Examples
The triangle begins:
0: 0
1: 1 -1
2: 2 0 -2
3: 3 2 -2 -3
4: 4 5 0 -5 -4
5: 5 9 5 -5 -9 -5
6: 6 14 14 0 -14 -14 -6
7: 7 20 28 14 -14 -28 -20 -7
8: 8 27 48 42 0 -42 -48 -27 -8
9: 9 35 75 90 42 -42 -90 -75 -35 -9
10: 10 44 110 165 132 0 -132 -165 -110 -44 -10
11: 11 54 154 275 297 132 -132 -297 -275 -154 -54 -11 .
Links
Crossrefs
Programs
-
Haskell
a214292 n k = a214292_tabl !! n !! k a214292_row n = a214292_tabl !! n a214292_tabl = map diff $ tail a007318_tabl where diff row = zipWith (-) (tail row) row
-
Mathematica
row[n_] := Table[Binomial[n, k], {k, 0, n}] // Differences; T[n_, k_] := row[n + 1][[k + 1]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 31 2018 *)
Formula
T(n,k) = A007318(n+1,k+1) - A007318(n+1,k), 0<=k<=n, i.e. first differences of rows in Pascal's triangle;
T(n,k) = -T(n,k);
row sums and central terms equal 0, cf. A000004;
sum of positive elements of n-th row = A014495(n+1);
T(n,0) = n;
T(n,3) = A005587(n-6) for n > 5;
T(n,4) = A005557(n-9) for n > 8;
T(n,5) = A064059(n-11) for n > 10;
T(n,6) = A064061(n-13) for n > 12;
T(n,7) = A124087(n) for n > 14;
T(n,8) = A124088(n) for n > 16;
T(2*n+1,n) = T(2*n+2,n) = A000108(n+1), Catalan numbers;
T(2*n+3,n) = A000245(n+2);
T(2*n+4,n) = A002057(n+1);
T(2*n+5,n) = A000344(n+3);
T(2*n+6,n) = A003517(n+3);
T(2*n+7,n) = A000588(n+4);
T(2*n+8,n) = A003518(n+4);
T(2*n+9,n) = A001392(n+5);
T(2*n+10,n) = A003519(n+5);
T(2*n+11,n) = A000589(n+6);
T(2*n+12,n) = A090749(n+6);
T(2*n+13,n) = A000590(n+7).