A187801 Pascal's triangle construction method applied to {1,1,2} as an initial term.
1, 1, 2, 1, 2, 3, 2, 1, 3, 5, 5, 2, 1, 4, 8, 10, 7, 2, 1, 5, 12, 18, 17, 9, 2, 1, 6, 17, 30, 35, 26, 11, 2, 1, 7, 23, 47, 65, 61, 37, 13, 2, 1, 8, 30, 70, 112, 126, 98, 50, 15, 2, 1, 9, 38, 100, 182, 238, 224, 148, 65, 17, 2, 1, 10, 47, 138, 282, 420, 462, 372
Offset: 2
Examples
Triangle begins: 1,1,2; 1,2,3,2; 1,3,5,5,2; 1,4,8,10,7,2; 1,5,12,18,17,9,2; 1,6,17,30,35,26,11,2; 1,7,23,47,65,61,37,13,2; 1,8,30,70,112,126,98,50,15,2; 1,9,38,100,182,238,224,148,65,17,2; 1,10,47,138,282,420,462,372,213,82,19,2; 1,11,57,185,420,702,882,834,585,295,101,21,2; 1,12,68,242,605,1122,1584,1716,1419,880,396,122,23,2; 1,13,80,310,847,1727,2706,3300,3135,2299,1276,518,145,25,2; From _Philippe Deléham_, Mar 10 2013: (Start) Row 2: 1+x+2*x^2 Row 3: (1+x+2*x^2)*(1+x) = 1+2*x+3*x^2+2*x^3 Row 4: (1+x+2*x^2)*(1+x)^2 = 1+3*x+5*x^2+5*x^3+2*x^4 Row 5: (1+x+2*x^2)*(1+x)^3 = 1+4*x+8*x^2+10*x^3+7*x^4+2*x^5 (End)
Links
- T. D. Noe, Rows n = 2..50 of triangle, flattened
- Eric W. Weisstein, MathWorld: Triangle Graph
- Eric W. Weisstein, MathWorld: Star Graph
Programs
-
Mathematica
c = {1, 1, 2}; Join[{c}, t = Table[c = Append[c, 0]; c = c + RotateRight[c], {9}]]; Flatten[t] (* T. D. Noe, Mar 11 2013 *)
Formula
For the selection of the initial term: neighborhood decomposition of graph.
For sequence: Pascal's triangle construction method applied to selected initial term.
Row sums: A000079(n+2) = (4, 8, 16, 32, 64, ...). - Philippe Deléham, Mar 10 2013
Comments