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A108352
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a(n) = primal code characteristic of n, which is the least positive integer, if any, such that (n o)^k = 1, otherwise equal to 0. Here "o" denotes the primal composition operator, as illustrated in A106177 and A108371 and (n o)^k = n o ... o n, with k occurrences of n.
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20
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1, 0, 2, 2, 2, 0, 2, 2, 0, 0, 2, 0, 2, 0, 2, 2, 2, 0, 2, 3, 2, 0, 2, 3, 2, 0, 2, 3, 2, 0, 2, 2, 2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, 3, 0, 0, 2, 3, 2, 0, 2, 3, 2, 0, 2, 3, 2, 0, 2, 0, 2, 0, 2, 2, 2, 0, 2, 3, 2, 0, 2, 0, 2, 0, 3, 3, 2, 0, 2, 3, 2, 0, 2, 0, 2, 0, 2, 3, 2, 0, 2, 3, 2, 0, 2, 3, 2, 0, 0, 2
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OFFSET
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1,3
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LINKS
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Table of n, a(n) for n=1..100.
Jon Awbrey, Primal Code Characteristic, n = 1 to 1000
Jon Awbrey, Primal Code Characteristic, n = 1001 to 2000
Jon Awbrey, Primal Code Characteristic, n = 2001 to 3000
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EXAMPLE
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a(1) = 1 because (1 o)^1 = ({ } o)^1 = 1.
a(2) = 0 because (2 o)^k = (1:1 o)^k = 2, for all positive k.
a(3) = 2 because (3 o)^2 = (2:1 o)^2 = 1.
a(4) = 2 because (4 o)^2 = (1:2 o)^2 = 1.
a(5) = 2 because (5 o)^2 = (3:1 o)^2 = 1.
a(6) = 0 because (6 o)^k = (1:1 2:1 o)^k = 6, for all positive k.
a(7) = 2 because (7 o)^2 = (4:1 o)^1 = 1.
a(8) = 2 because (8 o)^2 = (1:3 o)^1 = 1.
a(9) = 0 because (9 o)^k = (2:2 o)^k = 9, for all positive k.
a(10) = 0 because (10 o)^k = (1:1 3:1 o)^k = 10, for all positive k.
Detail of calculation for compositional powers of 12:
(12 o)^2 = (1:2 2:1) o (1:2 2:1) = (1:1 2:2) = 18
(12 o)^3 = (1:1 2:2) o (1:2 2:1) = (1:2 2:1) = 12
Detail of calculation for compositional powers of 20:
(20 o)^2 = (1:2 3:1) o (1:2 3:1) = (3:2) = 25
(20 o)^3 = (3:2) o (1:2 3:1) = 1
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CROSSREFS
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Cf. A055231, A061396, A062504, A062537, A062860, A106177, A106178.
Cf. A108353, A108370, A108371, A108372, A108373, A108374, A111801.
Sequence in context: A138319 A217864 A002100 * A215883 A036476 A104994
Adjacent sequences: A108349 A108350 A108351 * A108353 A108354 A108355
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KEYWORD
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nonn
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AUTHOR
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Jon Awbrey, May 31 2005, revised Jun 01 2005
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EXTENSIONS
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Links and cross-references added, Aug 19 2005
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STATUS
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approved
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