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A334251
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a(n) is the number of binary (0,1) sequences of length n that have at most two zeros in a window of seven consecutive symbols.
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3
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1, 2, 4, 7, 11, 16, 22, 29, 43, 66, 102, 157, 239, 358, 526, 777, 1159, 1740, 2619, 3942, 5923, 8870, 13259, 19822, 29667, 44451, 66641, 99912, 149745, 224338, 335993, 503199, 753720, 1129164, 1691796, 2534807, 3797721, 5689507, 8523275, 12768309, 19127928, 28655867, 42930562
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OFFSET
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0,2
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COMMENTS
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Application: Not all electronic devices connected to the Internet of Things (IoT) have batteries or are connected to the power cable. These self-contained devices must rely on the harvesting of energy of the signals sent by a transmitter. We investigate binary systems emitting 0's and 1's signals where it is assumed that the 1's carry the energy. A minimal number of 1's in transmitted sequences is required so as to carry sufficient energy within a prescribed time span. A binary sequence is said to obey the sliding-window (ell,t)-constraint if the number of 1's within any window of ell consecutive bits of that sequence is at least t, t<ell.
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LINKS
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FORMULA
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G.f.: (x^20 +x^19 +x^18 +2*x^17 +2*x^16 +x^15 -3*x^13 -4*x^12 -5*x^11 -7*x^10 -5*x^9 -3*x^8 -3*x^7 +2*x^6 +3*x^5 +3*x^4 +3*x^3 +2*x^2 +x +1) / (-x^21 -x^18 +x^15 +3*x^14 +x^12 +2*x^11 -3*x^7 -x^4 -x +1).
a(n) ~ c*r^n where c = 1.81880731105 and r = 1.498122533939865577.
a(n) = a(n - 1) + a(n - 4) + 3*a(n - 6) - 2*a(n - 10) - a(n - 12) - 3*a(n - 13) - a(n - 15) + a(n - 18) + a(n - 21), n >= 21. (End)
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EXAMPLE
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a(3) = 7 as there are 8 possible binary (0,1) sequences of length 3 but exactly one of them has more than 2 zero's in a window of seven consecutive symbols (the sequence (000)) leaving 8-1 = 7 such sequences. - David A. Corneth, Apr 20 2020
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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