A309361
Numbers n such that the number of interior intersection points A091908(n) of the n-intersected triangle increases exactly by 1 when the subdivision of the triangle is refined from n-1 to n cutting line segments.
Original entry on oeis.org
1, 3, 5, 7, 9, 11, 13, 17, 21, 25, 27, 31, 33, 37, 43, 49, 51, 53, 55, 57, 61, 67, 73, 81, 93, 97, 101, 107, 113, 115, 121, 123, 127, 133, 137, 141, 145, 147, 157, 163, 173, 177, 183, 185, 193, 201, 205, 211, 213, 217, 235, 241, 243, 249, 253, 257
Offset: 1
a(1) = 1 corresponds to change from the triangle without cutting line segments and correspondingly A091908(1)=0 interior intersection points to the triangle where the sides are divided into 2 equal pieces and the 3 line segments connecting the midpoints of the sides with the opposite vertices cutting each other in one common point, the center of gravity. (A091908(2)=1). Thus A091908(2) - A091908(1) = 1 -> a(1) = 1.
a(2) = 3 because the trisected triangle has one less interior intersection point (A091908(3) = 12) than the 4-sected triangle (A091908(4) = 13) -> a(2) = 3.
A125641
Square of the (3,1)-entry of the 3 X 3 matrix M^n, where M = [1,0,0; 1,1,0; 1,i,1].
Original entry on oeis.org
1, 5, 18, 52, 125, 261, 490, 848, 1377, 2125, 3146, 4500, 6253, 8477, 11250, 14656, 18785, 23733, 29602, 36500, 44541, 53845, 64538, 76752, 90625, 106301, 123930, 143668, 165677, 190125, 217186, 247040, 279873, 315877, 355250, 398196, 444925
Offset: 1
a(5)=25 because M^5 = [1,0,0; 5,1,0; 5+10i, 5i, 1] and |5+10i|^2 = 125.
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- César Eliud Lozada, Counting regions [Warning: Although the drawings here appear to be correct for n <= 5, the generalization to higher n fails - see Comment above and A332953. - _N. J. A. Sloane_, Mar 04 2020]
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
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List([1..40],n-> n^2*(n^2-2*n+5)/4); # Muniru A Asiru, Feb 22 2019
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[n^2*(n^2-2*n+5)/4: n in [1..40]]; // G. C. Greubel, Feb 22 2019
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b[1]:=1: b[2]:=2+I: b[3]:=3+3*I: for n from 4 to 45 do b[n]:=3*b[n-1]-3*b[n-2]+b[n-3] od: seq(abs(b[j])^2,j=1..45);
with(linalg): M[1]:=matrix(3,3,[1,0,0,1,1,0,1,I,1]): for n from 2 to 45 do M[n]:=multiply(M[1],M[n-1]) od: seq(abs(M[j][3,1])^2,j=1..45);
seq(sum((binomial(n,m))^2,m=1..2),n=1..37); # Zerinvary Lajos, Jun 19 2008
# alternative Maple program:
a:= n-> abs((<<1|0|0>, <1|1|0>, <1|I|1>>^n)[3,1])^2:
seq(a(n), n=1..40); # Alois P. Heinz, Mar 09 2020
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Table[n^2(n^2-2n+5)/4,{n,40}] (* Vincenzo Librandi, Feb 14 2012 *)
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vector(40, n, n^2*(n^2-2*n+5)/4) \\ G. C. Greubel, Feb 22 2019
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[n^2*(n^2-2*n+5)/4 for n in (1..40)] # G. C. Greubel, Feb 22 2019
A356119
Irregular table read by rows: T(n,k) is the number of k-sided polygons formed, for k>=3, in an equilateral triangle when straight line segments connect the three corner vertices to the points dividing the sides into n equal parts.
Original entry on oeis.org
1, 6, 12, 3, 3, 1, 24, 6, 36, 9, 9, 7, 48, 24, 6, 72, 21, 15, 19, 84, 48, 12, 6, 108, 51, 33, 25, 132, 78, 18, 18, 168, 69, 51, 43, 180, 120, 48, 18, 216, 135, 57, 61, 252, 156, 66, 36, 294, 159, 105, 67, 312, 234, 84, 48, 372, 225, 117, 103, 408, 264, 138, 60, 456, 291, 159, 121, 486, 372, 138, 84
Offset: 1
The table begins:
1;
6;
12, 3, 3, 1;
24, 6;
36, 9, 9, 7;
48, 24, 6;
72, 21, 15, 19;
84, 48, 12, 6;
108, 51, 33, 25;
132, 78, 18, 18;
168, 69, 51, 43;
180, 120, 48, 18;
216, 135, 57, 61;
252, 156, 66, 36;
294, 159, 105, 67;
312, 234, 84, 48;
372, 225, 117, 103;
408, 264, 138, 60;
456, 291, 159, 121;
486, 372, 138, 84;
.
.
See the attached text file for more examples.
Comments