A302337 Triangle read by rows: T(n,k) is the number of 2k-cycles in the n X n grid graph (2 <= k <= floor(n^2/2), n >= 2).
1, 4, 4, 5, 9, 12, 26, 52, 76, 32, 6, 16, 24, 61, 164, 446, 1100, 2102, 2436, 1874, 900, 226, 25, 40, 110, 332, 1070, 3504, 11144, 32172, 77874, 146680, 217470, 255156, 233786, 158652, 69544, 13732, 1072, 36, 60, 173, 556, 1942, 7092, 26424, 97624, 346428, 1136164, 3313812, 8342388, 18064642, 33777148, 54661008, 76165128, 89790912, 86547168, 64626638, 34785284, 12527632, 2677024, 255088
Offset: 2
Examples
Triangle begins: 1; 4, 4, 5; 9, 12, 26, 52, 76, 32, 6; 16, 24, 61, 164, 446, 1100, 2102, 2436, 1874, 900, 226; ... So for example, the 3 X 3 grid graph has 4 4-cycles, 4 6-cycles, and 5 8-cycles.
Links
- Seiichi Manyama, Rows n = 2..9, flattened
- Eric Weisstein's World of Mathematics, Graph Cycle
- Eric Weisstein's World of Mathematics, Grid Graph
Crossrefs
Programs
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Mathematica
Flatten[Table[Tally[Length /@ FindCycle[GridGraph[{n, n}], Infinity, All]][[All, 2]], {n, 6}]] (* Eric W. Weisstein, Mar 26 2021 *)
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Python
# Using graphillion from graphillion import GraphSet import graphillion.tutorial as tl def A302337(n): universe = tl.grid(n - 1, n - 1) GraphSet.set_universe(universe) cycles = GraphSet.cycles() return [cycles.len(2 * k).len() for k in range(2, n * n // 2 + 1)] print([i for n in range(2, 8) for i in A302337(n)]) # Seiichi Manyama, Mar 29 2020
Formula
Row sums equal A140517(n).
Length of row n equals A047838(n) = floor(n^2/2) - 1.
T(n,2) = 1 - 2*n + n^2 = (n-1)^2.
T(n,3) = 4 - 6*n + 2*n^2 = A046092(n-2).
T(n,4) = 26 - 28*n + 7*n^2 for n > 2.
T(n,5) = 164 - 140*n + 28*n^2 for n > 3.
T(n,6) = 1046 - 740*n + 124*n^2 for n > 4.
T(n,floor(n^2/2)) = A301648(n).
T(n,n^2/2) = A003763(n) for n even.