editing

proposed

Michel Marcus: I can't see your file; can you upload again (and do NOT change the filename chosen by system)

Michel Marcus: what is the name of your file on your system ?

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.

Michel Dekking, <a href="/A095276/A095276-runs/..pdf">Proof of conjecture on frequencies</a>

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.

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)

Michel Dekking, <a href="/A095276/a095276.-runs/.pdf">Proof of conjecture on frequencies</a>

Michel Dekking, <a href="/A095276/a095276.pdf">Proof of conjecture on frequencies</a>

proposed

editing

editing

proposed

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)

(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

approved

editing

editing

proposed

Cf. A055778, A130600, A055778, A133775, A042948.

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

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).

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

proposed

editing

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.....

editing

proposed