A246047 Irregular triangle read by rows, of partial serial probabilities T(n,k)_{2,3} (see "comments" for definitions and explanation).

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

Offset: 0

Author

Bob Selcoe, Aug 26 2014

Keywords

• nonn
• tabf

Comments

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)A176739(n)/2%5En;%20so%20for%20example,%20since%20A176739(9)%20=%20192,%20the%20probability%20that%209%20will%20occur%20as%20a%20partial%20sum%20in%20a%20randomly-generated%20infinite%20sequence%20of%202s%20and%203s%20is%20192/512%20=%203/8.%20That%20is,%20P(9)">{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)A176739(n)%20and%20sums%20of%20P''(n,k)">{2,3} are A176739(n) and sums of P''(n,k){2,3} are A176739(n)/2^n.

For T(n,k)A176739(9)%20=%20192,%20the%20probability%20that%209%20will%20occur%20as%20a%20partial%20sum%20with%20three%20sorts%20of%202s%20and%20one%20sort%20of%203s%20is%20128/512%20=%201/4%20(n=9,%20k=0),%20and%20with%20zero%20sorts%20of%202s%20and%20three%20sorts%20of%203s%20is%2064/512%20=%201/8%20(n=9,%20k=1),%20totaling%20192/512%20=%203/8.%20That%20is,%20P''(9,0)">{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).

Examples

			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.
		

Crossrefs

Cf. A176739, A007318 (binomial(n,k)), A103221, A128099 (related sequence).

Programs

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

Formula

T(n,k) = binomial((n - (n mod 2) - 2*k)/2, 2*k + (n mod 2)) * 2^((n + (n mod 2) + 2*k)/2).

Extensions

More terms from Michel Marcus, Sep 26 2014