Revisions by Michel Dekking
(See also Michel Dekking's wiki page)
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Showing entries 1-10
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#18 by Michel Dekking at Wed Jun 26 08:12:07 EDT 2024
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Discussion
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Wed Jun 26
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| Michel Marcus: I can't see your file; can you upload again (and do NOT change the filename chosen by system)
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| 08:35
| Michel Marcus: what is the name of your file on your system ?
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#17 by Michel Dekking at Wed Jun 26 08:07:09 EDT 2024
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Conjecture: it appears that the asymptotic frequencies of terms 1, 2 and 3 are 1/2, 1/(2*phi^2) and 1/(2*phi) respectively, where phi = (1+sqrt(5))/2 is the golden ratio. - Vladimir Reshetnikov, Mar 17 2022For a proof of this conjecture see my link to A095276-runs.pdf.2022
For a proof of this conjecture see my link to A095276-runs.pdf.
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Michel Dekking, <a href="/A095276/A095276-runs/..pdf">Proof of conjecture on frequencies</a>
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Discussion
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Wed Jun 26
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| Michel Dekking: Here is the result. I am not sure whether the link is correctly added. The Guide Text is not very clear about that.
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#16 by Michel Dekking at Wed Jun 26 08:05:00 EDT 2024
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Conjecture: it appears that the asymptotic frequencies of terms 1, 2 and 3 are 1/2, 1/(2*phi^2) and 1/(2*phi) respectively, where phi = (1+sqrt(5))/2 is the golden ratio. - Vladimir Reshetnikov, Mar 17 2022For a proof of this conjecture see my link to 2022A095276-runs.pdf.
From Michel Dekking, Jun 25 2024: (Start)
Proof of this conjecture, based on my 2019 comment in A095076. We exploit the analysis of the return words of 1 in the sequence x = 1244343143112..., the fixed point of the morphism sigma be given by sigma(1) = 12, sigma(2) = 4, sigma(3) = 1, sigma(4) = 43,
which yields the sequence A095076 = 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0....
after applying the letter-to-letter map lambda given by
lambda(1) = 0, lambda(2) = 1, lambda(3) = 0, lambda(4) = 1.
The return words partioning x are given by a coding gamma:
A:=1, B:=12, C:=124, D:=143, E:=1243, F:=12443, G:=1244343.
Moreover, sigma induces a morphism rho on the alphabet {A,B,C,D,E,F,G} given by
rho(A)=B, rho(B)=C, rho(C)=F, rho(D)=EA, rho(E)=FA, rho(F)=GA, rho(G)=GDA.
Let y = GDAEABFA... be the unique fixed point of rho.
Then gamma(y)=x, so lambda(gamma(y)) = A095076. We have
zA:=lambda(gamma(rho(A)))=lambda(gamma(B))=lambda(12) = 01,
zB:=lambda(gamma(rho(B)))=lambda(gamma(C))=lambda(124) = 011,
zC:=lambda(gamma(rho(C)))=lambda(gamma(F))=lambda(12443) = 01110,
zD:=lambda(gamma(rho(D)))=lambda(gamma(EA))=lambda(12431) = 01100,
zE:=lambda(gamma(rho(E)))=lambda(gamma(FA))=lambda(124431) = 011100,
zF:=lambda(gamma(rho(F)))=lambda(gamma(GA))=lambda(12443431) = 01110100,
zG:=lambda(gamma(rho(G)))=lambda(gamma(GDA))=lambda(1244343431431) = 01110100100.
The crucial observation is that zA, zB,...,zG all have prefix 01 so that we can read off the runs rA,rB,...,rG in the sequence A095076 from these words.
For example zF = 01110100, ignoring the prefix 0 we see runs of length 3,1,1. The suffix 00 of zF is the first part of a run of length 3. This yields rF = 3113.
We obtain rA = 11, rB = 21, rC = 32, rD = 23, rE = 33, rF = 3113, rG = 311213.
Here we skip the first letter of A095076.
To prove the conjecture, let LR be the length of the prefix of (a(n)) that is induced by a prefix y[0,L] of y of length L. Then
LR = 2*N_A[0,L] + 2*N_B[0,L] + ... + 4*N_F[0,L] + 6*N_G[0,L],
where N_A[0,L] is the number of letters A in y[0,L], etc.
The total number N_1[0,LR] of 1's in the prefix a[0,LR] of (a(n)) is equal to
N_1[0,LR] = 2*N_A[0,L] + N_B[0,L] + ... + 2*N_F[0,L] + 3*N_G[0,L].
The asymptotic frequency mu(1) of the letter 1 is equal to the limit as LR tends to infinity of the quotient N_1[0,LR]/LR.
This is the same as the quotient
(2*N_A[0,L] + N_B[0,L] + ... + 2*N_F[0,L] + 3*N_G[0,L])/
(2*N_A[0,L] + 2* N_B[0,L] + ... + 4*N_F[0,L] + 6*N_G[0,L]).
Dividing both denominator and numerator by L, we see that we have to compute the frequencies of the letters A, B,...,G in y = GDAEABFA....
It is standard that these frequencies nu(A), ..., nu(G) are given by the normalized left eigenvector of the incidence matrix M of the morphism rho.
It is easily checked that V := (phi, 1, 1/phi, 1, 1/phi, 1-1/phi, 1/phi, 1) satisfies V M = phi M.
The final conclusion is that
mu(1) = (2*V1 + V2 + 2*V6 + 3*V7)/
(2*V1 + 2*V2 + 2*V3 + 2*V4 + 2*V5 + 4*V6 + 6*V7)
= (2*phi + 1 + 2/phi + 3)/(2*phi + 2 + 2/phi + 2/phi + 2-2/phi + 4/phi + 6)
= (4 + 2 phi + 2/phi )/(8 + 4 phi + 4/phi) = 1/2.
Here one uses that 1/phi = phi - 1.
Similarly, one finds that mu(2) = 1/(2*phi^2), and mu(3) = 1/(2*phi).
(End)
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Michel Dekking, <a href="/A095276/a095276.-runs/.pdf">Proof of conjecture on frequencies</a>
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#15 by Michel Dekking at Wed Jun 26 08:00:04 EDT 2024
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Michel Dekking, <a href="/A095276/a095276.pdf">Proof of conjecture on frequencies</a>
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| STATUS
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proposed
editing
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#12 by Michel Dekking at Tue Jun 25 01:56:56 EDT 2024
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#11 by Michel Dekking at Tue Jun 25 01:52:43 EDT 2024
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From Michel Dekking, Jun 25 2024: (Start) Proof of this conjecture, based on my 2019 comment in A095076. We exploit the analysis of the return words of 1 in the sequence x = 1244343143112..., the fixed point of the morphism sigma be given by sigma(1) = 12, sigma(2) = 4, sigma(3) = 1, sigma(4) = 43,
which yields the sequence A095076 = 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0....
after applying the letter-to-letter map lambda given by
lambda(1) = 0, lambda(2) = 1, lambda(3) = 0, lambda(4) = 1.
The return words partioning x are given by a coding gamma:
A:=1, B:=12, C:=124, D:=143, E:=1243, F:=12443, G:=1244343.
Moreover, sigma induces a morphism rho on the alphabet {A,B,C,D,E,F,G} given by
rho(A)=B, rho(B)=C, rho(C)=F, rho(D)=EA, rho(E)=FA, rho(F)=GA, rho(G)=GDA.
Let y = GDAEABFA... be the unique fixed point of rho.
Then gamma(y)=x, so lambda(gamma(y)) = A095076. We have
zA:=lambda(gamma(rho(A)))=lambda(gamma(B))=lambda(12) = 01,
zB:=lambda(gamma(rho(B)))=lambda(gamma(C))=lambda(124) = 011,
zC:=lambda(gamma(rho(C)))=lambda(gamma(F))=lambda(12443) = 01110,
zD:=lambda(gamma(rho(D)))=lambda(gamma(EA))=lambda(12431) = 01100,
zE:=lambda(gamma(rho(E)))=lambda(gamma(FA))=lambda(124431) = 011100,
zF:=lambda(gamma(rho(F)))=lambda(gamma(GA))=lambda(12443431) = 01110100,
zG:=lambda(gamma(rho(G)))=lambda(gamma(GDA))=lambda(1244343431431) = 01110100100.
The crucial observation is that zA, zB,...,zG all have prefix 01 so that we can read off the runs rA,rB,...,rG in the sequence A095076 from these words.
For example zF = 01110100, ignoring the prefix 0 we see runs of length 3,1,1. The suffix 00 of zF is the first part of a run of length 3. This yields rF = 3113.
We obtain rA = 11, rB = 21, rC = 32, rD = 23, rE = 33, rF = 3113, rG = 311213.
Here we skip the first letter of A095076.
To prove the conjecture, let LR be the length of the prefix of (a(n)) that is induced by a prefix y[0,L] of y of length L. Then
LR = 2*N_A[0,L] + 2*N_B[0,L] + ... + 4*N_F[0,L] + 6*N_G[0,L],
where N_A[0,L] is the number of letters A in y[0,L], etc.
The total number N_1[0,LR] of 1's in the prefix a[0,LR] of (a(n)) is equal to
N_1[0,LR] = 2*N_A[0,L] + N_B[0,L] + ... + 2*N_F[0,L] + 3*N_G[0,L].
The asymptotic frequency mu(1) of the letter 1 is equal to the limit as LR tends to infinity of the quotient N_1[0,LR]/LR.
This is the same as the quotient
(2*N_A[0,L] + N_B[0,L] + ... + 2*N_F[0,L] + 3*N_G[0,L])/
(2*N_A[0,L] + 2* N_B[0,L] + ... + 4*N_F[0,L] + 6*N_G[0,L]).
Dividing both denominator and numerator by L, we see that we have to compute the frequencies of the letters A, B,...,G in y = GDAEABFA....
It is standard that these frequencies nu(A), ..., nu(G) are given by the normalized left eigenvector of the incidence matrix M of the morphism rho.
It is easily checked that V := (phi, 1, 1/phi, 1, 1/phi, 1-1/phi, 1/phi, 1) satisfies V M = phi M.
The final conclusion is that
mu(1) = (2*V1 + V2 + 2*V6 + 3*V7)/
(2*V1 + 2*V2 + 2*V3 + 2*V4 + 2*V5 + 4*V6 + 6*V7)
= (2*phi + 1 + 2/phi + 3)/(2*phi + 2 + 2/phi + 2/phi + 2-2/phi + 4/phi + 6)
= (4 + 2 phi + 2/phi )/(8 + 4 phi + 4/phi) = 1/2.
Here one uses that 1/phi = phi - 1.
Similarly, one finds that mu(2) = 1/(2*phi^2), and mu(3) = 1/(2*phi).
(End)
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(a(n)) is a morphic sequence. Let y = GDAEABFA... be the unique fixed point of the morphism rho given by rho(A) = B, rho(B) =C, rho(C) = F, rho(D) = EA, rho(E) = FA, rho(F) = GA, rho(G) = GDA on the alphabet {A,B,C,D,E,F,G}. Then (a(n+1)) is the image of y under the morphism A->11, B->21, C->32, D->23, E->33, F->3113, G->311213. - Michel Dekking, Jun 25 2024
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approved
editing
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#17 by Michel Dekking at Wed Jun 19 09:42:15 EDT 2024
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#16 by Michel Dekking at Wed Jun 19 09:41:56 EDT 2024
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#15 by Michel Dekking at Wed Jun 19 09:29:57 EDT 2024
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Let (v(n)) = 1,4,5,8,9,12,... be the sequence of values taken by (a(n)). Then it follows directly from the Lucas formula for (a(n)) that v(n) = A042948(n) (where A042948 has been given offset 1, as it should; see also the comment by Jianing Song in A042948). - Michel Dekking, Jun 19 2024
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For example if n=1: L(2)=3, L(3)=4, L(4)=7, so a(3) = a(4) = 5, and a(5) = a(6) = 8.
Let (v(n)) = 1,4,5,8,9,12,... be the sequence of values taken by (a(n)). Then it follows directly from the Lucas formula for (a(n)) that v(n) = A042948(n) (where A042948 has been given offset 1, as it should; see also the comment by Jianing Song in A042948).
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| EXAMPLE
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For n=1: L(2)=3, L(3)=4, L(4)=7, so a(3) = a(4) = 5, and a(5) = a(6) = 8.- Michel Dekking, Jun 19 2024
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proposed
editing
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Discussion
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Wed Jun 19
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| Michel Dekking: O.K. I thought about that, but I decided not do it like that because it is in conflict with the basic structure of an OEIS item.....
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#10 by Michel Dekking at Wed Jun 19 01:02:28 EDT 2024
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