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A338573 Array read by ascending antidiagonals: T(m,n) (m, n >= 1) is the minimum number of unit resistors needed to produce resistance m/n. 10

%I #56 Nov 15 2020 12:49:13

%S 1,2,2,3,1,3,4,3,3,4,5,2,1,2,5,6,4,4,4,4,6,7,3,4,1,4,3,7,8,5,2,5,5,2,

%T 5,8,9,4,5,3,1,3,5,4,9,10,6,5,5,5,5,5,5,6,10,11,5,3,2,5,1,5,2,3,5,11,

%U 12,7,6,6,5,5,5,5,6,6,7,12,13,6,6,4,6,4,1,4,6,4,6,6,13

%N Array read by ascending antidiagonals: T(m,n) (m, n >= 1) is the minimum number of unit resistors needed to produce resistance m/n.

%C Karnofsky (2004, p. 5): "[...] if some circuit has resistance m/n then some other circuit likely has n/m. In fact, for 9 or fewer resistors, this symmetry is perfect. However, for 10 resistors the following values are achieved, but not their inverses: 95/106, 101/109, 98/103, 97/98, 103/101, 97/86, 110/91, 103/83, 130/101, 103/80, 115/89, 106/77, 109/77, 98/67, 101/67". That means, that T(m,n) = T(n,m), if T(m,n) <= 9.

%C This starts with the values of A113881, but the Karnofsky comment says that T(n,m) is not symmetric, whereas the count of tiles in A113881 is. - _R. J. Mathar_, Nov 06 2020

%C The first difference where T(m,n) = T(n,m), but differs from the corresponding entry of A113881 occurs for (n,m) = (154,167) and (n,m) = (167,154), both representable by networks with non-planar graphs of 11 resistors, whereas A113881 counts 12 tiles. See Pfoertner link for illustration of more differences. - _Hugo Pfoertner_, Nov 13 2020

%D Technology Review's Puzzle Corner, How many different resistances can be obtained by combining 10 one ohm resistors? Oct 3, 2003.

%H Joel Karnofsky, <a href="http://cs.nyu.edu/~gottlieb/tr/overflow/2003-dec-2.pdf">Solution of problem from Technology Review's Puzzle Corner Oct 3, 2003</a>, Feb 23 2004.

%H Hugo Pfoertner, <a href="/A338573/a338573.pdf">Where A338573 differs from A113881</a>, x,y <= 380.

%H <a href="/index/Res#resistances">Index to sequences related to resistances</a>.

%e T(1,2) = 2: at least 2 unit resistors in parallel are needed for resistance 1/2.

%e T(2,1) = 2: at least 2 unit resistors in series are needed for resistance 2 = 2/1.

%e T(11,13) = 6: the following "bridge" has resistance Bri(Par(1,1),1,1,1,1) = 11/13 (see A337516 for definitions):

%e .

%e (+)

%e / \

%e ---* \

%e / / \

%e (1)(1) (1)

%e \ | |

%e \| |

%e *--(1)--*

%e \ /

%e (1) (1)

%e \ /

%e (-)

%e .

%e T(13,11) = 6: Bri(Ser(1,1),1,1,1,1) = 13/11.

%e T(95,106) = 10, but T(106,95) > 10: Karnofsky (2004, p. 5), see comment.

%Y Cf. A048211, A113881, A180414, A174283, A337516, A337517.

%Y Non-reciprocal ratios: A338601/A338602 (10 resistors), A338581/A338591 (11 resistors), A338582/A338592 (12 resistors).

%K tabl,nonn,hard

%O 1,2

%A _Rainer Rosenthal_, Nov 05 2020

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