A226575 Ordered excesses of internal lattice point counts of scaled up primitive Pythagorean triangles (PPT's) (see comments).
4, 24, 48, 72, 160, 168, 180, 300, 448, 504, 520, 768, 784, 900, 1080, 1152, 1176, 1320, 1584, 1620, 1920, 2200, 2232, 2268, 2548, 2904, 3108, 3744, 3784, 3808, 3840, 4416, 4680, 4732, 5508, 5880, 5880, 5928, 6624, 6720, 6732, 7600, 8568, 8760, 9280, 9900
Offset: 1
Keywords
Examples
a(6) = 168 as the PPT (20,21,29) when scaled by 29 to (580,609,841) has a lattice point count of 176002 (config. 1) and 176170 (config. 2). Hence E = 168 and it is the 6th occurrence.
Links
- Stanley Rabinowitz, Oblique Pythagorean Lattice Triangles, Pi Mu Epsilon Journal, 9(1989), 26-29.
- Eric W. Weisstein, MathWorld: Pick's Theorem
- Wikipedia, Pick's theorem
Programs
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Mathematica
getpairs[k_] := Reverse[Select[IntegerPartitions[k, {2}], GCD[#[[1]], #[[2]]]==1 &]]; getlist[j_] := (newlist=getpairs[j]; Table[(newlist[[m]][[1]]^2+newlist[[m]][[2]]^2-1)(newlist[[m]][[1]]-newlist[[m]][[2]])(newlist[[m]][[2]]), {m, 1, Length[newlist]}]); maxterms=10; table=Sort@Flatten@Table[getlist[2p+1], {p, 1, maxterms}][[1;;maxterms]]
Formula
For config. 1 the internal lattice count I = (c^2*a*b-c*(a+b+1)+2)/2. For config. 2 the internal lattice count I = (c^2*a*b-(a+b+c^2)+2)/2. So the excess of config. 2 over 1 is E = (c-1)*(a+b-c)/2.
Comments