A007223 Number of distinct perforation patterns for deriving (v,b) = (n+2,n) punctured convolutional codes from (2,1).
1, 2, 8, 24, 85, 286, 1008, 3536, 12618, 45220, 163504, 594320, 2173197, 7983990, 29465440, 109174560, 405995326, 1514797020, 5669021488, 21275014800, 80047272578, 301892460012, 1141069157408, 4321730134624, 16399422757300
Offset: 2
Keywords
References
- Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Guy Bégin and David Haccoun, High rate punctured convolutions codes: Structure properties and construction techniques, IEEE Transactions on Communications 37(12) (1989), 1381-1385.
- David Haccoun and Guy Bégin, High rate punctured convolutional codes for Viterbi and sequential coding, IEEE Transactions on Communications, 37(11) (1989), 1113-1125; see Section II.
Programs
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Maple
with(numtheory):P:=proc(b,v0) local k: RETURN(add(phi(k)*(1+z^k)^(v0*(b/k)),k=divisors(b))/b): end; seq(coeff(P(b,2),z,b+2),b=2..40); # Pab Ter
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Mathematica
A[x_] = (1 - Sqrt[1 - 4x])/(2x) - 1; CoefficientList[(A[x]^2 + A[x^2])/(2 x^2) + O[x]^25, x] (* Jean-François Alcover, Apr 30 2023, after R. J. Mathar's proven conjecture *)
Formula
Conjecture: Expansion of [A(x)^2 + A(x^2)]/2, where A(x) = A000108(x) - 1. - R. J. Mathar, Jul 19 2016
From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code, this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+2,n) and (v0,1) = (2,1), so
a(n) = (1/n)*Sum_{k|gcd(n+2,n)} phi(k)*binomial(2*n/k, (n+2)/k).
This simplifies to
a(n) = (1/n)*(binomial(2*n, n+2) + [(n mod 2) == 0]*binomial(n, (n/2) + 1)).
It follows from my comments in A275206 that R. J. Mathar's conjecture is correct and that
a(n) = (-2*c(n) + c(n+1) + [(n mod 2) == 0]*c(n/2))/2 for n >= 1, where c = A000108. (End)
D-finite with recurrence -(11*n-30)*(n+2)*(n+1) *a(n) +10*(n+1) *(7*n^2-22*n+6) *a(n-1) -60*(n-2)*(n^2-5*n+1) *a(n-2) -40*(n-2) *(7*n^2-22*n+6) *a(n-3) +16*(2*n-7) *(n-3) *(13*n-22) *a(n-4)=0. - R. J. Mathar, Mar 21 2021
Extensions
More terms from Pab Ter (pabrlos2(AT)yahoo.com), Nov 13 2005
a(2) = 1 prepended by R. J. Mathar, Jul 19 2016
Comments