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%I #75 Sep 03 2018 23:00:35
%S 1,0,2,4,4,16,8,16,48,16,96,128,64,32,384,320,512,64,1280,256,768,
%T 2560,128,3840,2560,1792,10240,1024,256,10752,15360,4096,35840,12288,
%U 512,28672,71680,4096,9216,114688,86016,1024,73728,286720,57344
%N Irregular triangle read by rows, of partial serial probabilities T(n,k)_{2,3} (see "comments" for definitions and explanation).
%C 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.
%C 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.
%C 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}.
%C This triangle therefore is T(n,k)_{2,3}, and P''(n,k)_{2,3} = T(n,k)_{2,3}/2^n.
%C 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.
%C 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.
%C 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).
%F T(n,k) = binomial((n - (n mod 2) - 2*k)/2, 2*k + (n mod 2)) * 2^((n + (n mod 2) + 2*k)/2).
%e Triangle starts:
%e 1;
%e 0;
%e 2;
%e 4;
%e 4;
%e 16;
%e 8, 16;
%e 48;
%e 16, 96;
%e 128, 64;
%e 32, 384;
%e 320, 512;
%e 64, 1280, 256;
%e 768, 2560;
%e 128, 3840, 2560;
%e 1792, 10240, 1024;
%e 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.
%o (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
%Y Cf. A176739, A007318 (binomial(n,k)), A103221, A128099 (related sequence).
%K nonn,tabf
%O 0,3
%A _Bob Selcoe_, Aug 26 2014
%E More terms from _Michel Marcus_, Sep 26 2014
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