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A246047
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Irregular triangle read by rows, of partial serial probabilities T(n,k)_{2,3} (see "comments" for definitions and explanation).
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0
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1, 0, 2, 4, 4, 16, 8, 16, 48, 16, 96, 128, 64, 32, 384, 320, 512, 64, 1280, 256, 768, 2560, 128, 3840, 2560, 1792, 10240, 1024, 256, 10752, 15360, 4096, 35840, 12288, 512, 28672, 71680, 4096, 9216, 114688, 86016, 1024, 73728, 286720, 57344
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OFFSET
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0,3
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COMMENTS
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Define "serial probability" as the probability that n will occur as a partial sum in an infinite sequence of numbers drawn randomly from set J = {j1,j2,..jz}, where 1 <= j1 < j2< ... < jz and z is the number of members in set J. Generally, serial probabilities are found by the recurrence equation: a(n) = (z^(j1-1)*a(n-j1) + z^(j2-1)*a(n-j2) + z^(j3-1)*a(n-j3) + ... + z^(jz-1)*a(n-jz))/z^n, where a(0)=1 and a(n)=0 when n < 0.
Denote the recurrence sequence for set J as S(n)_{J}, and denote serial probability (P) for set J as P(n)_{J}, such that P = S(n)_{J}/z^n. For example, S(n)_{2,3} = 2*a(n-2) + 4*a(n-3); therefore P(n)_{2,3} = (2*a(n-2) + 4*a(n-3))/2^n. This also is equivalent to A176739(n)/2^n; so for example, since A176739(9) = 192, the probability that 9 will occur as a partial sum in a randomly-generated infinite sequence of 2s and 3s is 192/512 = 3/8. That is, P(9)_{2,3} = 3/8.
Define "partial serial probability" (P'') as the probability that n would occur given the different ways to sort the compositions (ordered partitions) of n into j1's..jz's; and let T(n,k)_{J} be the triangle of partial serial probabilities for set J, such that P'' = T(n,k)_{J}/z^n. Denote these probabilities as P''(n,k)_{J}.
This triangle therefore is T(n,k)_{2,3}, and P''(n,k)_{2,3} = T(n,k)_{2,3}/2^n.
In general, row sums of T(n,k)_{J} are S(n)_{J}; thus, the row sums of T(n,k)_{2,3} are A176739(n) and sums of P''(n,k)_{2,3} are A176739(n)/2^n.
For T(n,k)_{2,3}: there are [(n-3*(n mod 2)-6k)/2] sorts of 2s, and [2k+(n mod 2)] sorts of 3s. So taking again example A176739(9) = 192, the probability that 9 will occur as a partial sum with three sorts of 2s and one sort of 3s is 128/512 = 1/4 (n=9, k=0), and with zero sorts of 2s and three sorts of 3s is 64/512 = 1/8 (n=9, k=1), totaling 192/512 = 3/8. That is, P''(9,0)_{2,3} = 1/4 and P''(9,1)_{2,3} = 1/8.
Given n, maximum k for T(n,k)_{2,3} is A103221(n)-1. That is, row lengths are floor(n/6)+1 unless n == 1 (mod 6); if n == 1 (mod 6), row length is floor(n/6).
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LINKS
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FORMULA
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T(n,k) = binomial((n - (n mod 2) - 2*k)/2, 2*k + (n mod 2)) * 2^((n + (n mod 2) + 2*k)/2).
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EXAMPLE
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Triangle starts:
1;
0;
2;
4;
4;
16;
8, 16;
48;
16, 96;
128, 64;
32, 384;
320, 512;
64, 1280, 256;
768, 2560;
128, 3840, 2560;
1792, 10240, 1024;
E.g., T(13,0) = 768 because 768/2^13 (3/32) is the probability that 13 will occur as a partial sum in a randomly-generated infinite sequence of 2s and 3s, where the compositions of 13 are into five sorts of 2s and one sort of 3s. In other words, P''(13,0)_{2,3} = 3/32. The sorts are 5 and 1, respectively, because (13 - 3*(13 mod 2) - 6*0)/2 = 5 and 2*0 + (13 mod 2) = 1.
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PROG
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(PARI) tabf(nn) = {for (n=0, nn, for (k=0, max(0, (n+2)\2 - (n+2)\3 - 1), tnk = binomial((n - (n % 2) -2*k)/2, 2*k + (n % 2)) * 2^((n + (n % 2) + 2*k)/2); print1(tnk, ", "); ); print(); ); } \\ Michel Marcus, Sep 26 2014
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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