A221369 -id:A221369 - cp's OEIS frontend

A209233 A two-digit Look-and-Say sequence starting with 11: each term summarizes the increasing two-digit substrings of the previous term.

11, 111, 211, 111121, 311112121, 311212221131, 211212113221222231, 211312113421422123131132, 311212413114421122123331132134242, 411412313114421122123224331132233134141342144, 411312413414321322323124431232233234441242143244

Offset: 0

Reinhard Zumkeller, Jan 13 2013

a(16) is the first term containing a zero; this is due to the fact that a(15) is the first term having exactly 10 occurrences of a two-digit number, namely 10 x 31.

			a(0) = 11: 1 x 11 --> a(1) = 111;
a(1) = 111: 2 x 11 --> a(2) = 211;
a(2) = 211: 1 x 11 and 1 x 21 --> a(3) = 111121;
a(3) = 111121: 3 x 11, 1 x 12 and 1 x 21 --> a(4) = 311112121.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842.

Cf. A209234 (start=10), A221368 (start=12), A221369 (start=13), A221372 (start=19), A221373 (start=99).

Haskell
```
-- See Link.
```

A209234 A two-digit Look-and-Say sequence starting with 10: each term summarizes the increasing two-digit substrings of the previous term.

Original entry on oeis.org

10, 110, 110111, 101110311, 101103210311131, 101203310311113121231132, 101203210411312213120121123431132133, 201103104210311512413220421122123331232133134141143, 101203204310411412313214115220421222223124431232333134341242143151

Offset: 0

Reinhard Zumkeller, Jan 13 2013

			a(0) = 10: 1x10 --> a(1)=110;
a(1) = 110: 1x10 and 1x11 --> a(2)=110111;
a(2) = 110111: 1x01, 1x10 and 3x11 -> a(3)=101110311;
a(3) = 101110311: 1x01, 1x03, 2x10, 3x11 and 1x31 -> a(4)=101103210311131.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842, A001155.

Cf. A209233 (start=11), A221368 (start=12), A221369 (start=13), A221372 (start=19), A221373 (start=99).

Haskell
```
-- See Link.
```

A221368 A two-digit Look-and-Say sequence starting with 12: each term summarizes the increasing two-digit substrings of the previous term.

Original entry on oeis.org

12, 112, 111112, 411112, 311112141, 311112114121131141, 611212113214221231241, 211412113114421122123124131132141142161, 611412313414116621122123124331132341242144161, 411512213314216321122323224331132133234541142143144261162166

Offset: 0

Reinhard Zumkeller, Jan 13 2013

a(36) is the first term containing a zero; this is due to the fact that a(35) is the first term having exactly 10 occurrences of a two-digit number, namely 10 x 42.

			a(0) = 12: 1 x 12 --> a(1) = 112;
a(1) = 112: 1 x 11 ana 1 x 12 --> a(2) = 111112;
a(2) = 111112: 4 x 11 and 1 x 12 --> a(4) = 411112;
a(3) = 411112: 3 x 11, 1 x 12 and 1 x 41 --> a(4) = 311112141.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842.

Cf. A209234 (start=10), A209233 (start=11), A221369 (start=13), A221372 (start=19), A221373 (start=99).

Haskell
```
-- See Link.
```

A221372 A two-digit Look-and-Say sequence starting with 19: each term summarizes the increasing two-digit substrings of the previous term.

Original entry on oeis.org

19, 119, 111119, 411119, 311119141, 311114119131141191, 611113214219231241291, 311212113114119221123124129131132141142161191192, 911512313314116319521122123124129431132341142161291292

Offset: 0

Reinhard Zumkeller, Jan 13 2013

a(16) is the first term containing a zero; this is due to the fact that a(15) is the first term having exactly 10 occurrences of a two-digit number, namely 10 x 51.

			a(0) = 19: 1 x 19 --> a(1) = 119;
a(1) = 119: 1 x 11 and 1 x 19 --> a(2) = 111119;
a(2) = 111119: 4 x 11 and 1 x 19 --> a(3) = 411119;
a(3) = 411119: 3 x 11, 1 x 19 and 1 x 41 --> a(4) = 311119141.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842.

Cf. A209234 (start=10), A209233 (start=11), A221368 (start=12), A221369 (start=13), A221373 (start=99).

Haskell
```
-- See Link.
```

A221373 A two-digit Look-and-Say sequence starting with 99: each term summarizes the increasing two-digit substrings of the previous term.

Original entry on oeis.org

99, 199, 119199, 111219191199, 311112319121291199, 411312219121123129231291199, 311512113219221122223229331141291192199, 511312113114115319421522123229231232133141151191292193199, 611412313214315419521222323229631232133241142251152153191292193194199

Offset: 0

Reinhard Zumkeller, Jan 13 2013

a(22) is the first term containing a zero; this is due to the fact that a(21) is the first term having exactly 10 occurrences of a two-digit number, namely 10 x 32.

			a(0) = 11: 1x99 --> a(1)=199;
a(1) = 199: 1x19 and 1x99 --> a(2)=119199;
a(2) = 119199: 1x11, 2x19, 1x91 and 1x99 --> a(3)=111219191199;
a(3) = 111219191199: 3x11, 1x12, 3x19, 1x21, 2x91 and 1x99 --> a(4)=311112319121291199.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842.

Cf. A209234 (start=10), A209233 (start=11), A221368 (start=12), A221369 (start=13), A221372 (start=19).

Haskell
```
-- See Link.
```

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

Original entry on oeis.org

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4,0,1).

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

More terms from Michel Marcus, Feb 21 2025

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(5 - sqrt(21)).

Original entry on oeis.org

1, 3, 1, 21, 1, 108, 1, 525, 1, 2523, 1, 12096, 1, 57963, 1, 277725, 1, 1330668, 1, 6375621, 1, 30547443, 1, 146361600, 1, 701260563, 1, 3359941221, 1, 16098445548, 1, 77132286525, 1, 369562987083, 1, 1770682648896, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 5. See also A221364 (N = 3), A221366 (N = 7) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(5 - sqrt(21)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...].

			F(1/2*(5 - sqrt(21))) = 1.25274 83510 08359 27965 ... = 1 + 1/(3 + 1/(1 + 1/(21 + 1/(1 + 1/(108 + 1/(1 + 1/(525 + ...))))))).
F((1/2*(5 - sqrt(21)))^2) = 1.04545 84663 16495 30047 ... = 1 + 1/(21 + 1/(1 + 1/(525 + 1/(1 + 1/(12096 + 1/(1 + 1/(277725 + ...))))))).
F((1/2*(5 - sqrt(21)))^3) = 1.00917 43188 83793 73068 ... = 1 + 1/(108 + 1/(1 + 1/(12096 + 1/(1 + 1/(1330668 + 1/(1 + 1/(146361600 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-6,0,1).

Cf. A004254, A030221, A054493, A174500 (N = 4), A221364 (N = 3), A221366 (N = 7), A221369 (N = 9).

LinearRecurrence[{0,6,0,-6,0,1},{1,3,1,21,1,108},40] (* Harvey P. Dale, Jun 06 2023 *)

a(2*n-1) = (1/2*(5 + sqrt(21)))^n + (1/2*(5 - sqrt(21)))^n - 2 = 3*A054493(n); a(2*n) = 1.
a(4*n+1) = 3*(A030221(n))^2; a(4*n-1) = 21*(A004254(n))^2.
a(n) = 6*a(n-2)-6*a(n-4)+a(n-6). G.f.: -(x^4+3*x^3-5*x^2+3*x+1) / ((x-1)*(x+1)*(x^4-5*x^2+1)). - Colin Barker, Jan 20 2013

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

Original entry on oeis.org

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 7. See also A221364 (N = 3), A221365 (N = 5) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(7 - sqrt(45)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).

Cf. A004187, A033890, A049682, A081070.

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

cp's OEIS Frontend

A209233 A two-digit Look-and-Say sequence starting with 11: each term summarizes the increasing two-digit substrings of the previous term.

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A209234 A two-digit Look-and-Say sequence starting with 10: each term summarizes the increasing two-digit substrings of the previous term.

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A221368 A two-digit Look-and-Say sequence starting with 12: each term summarizes the increasing two-digit substrings of the previous term.

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Haskell

A221372 A two-digit Look-and-Say sequence starting with 19: each term summarizes the increasing two-digit substrings of the previous term.

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Haskell

A221373 A two-digit Look-and-Say sequence starting with 99: each term summarizes the increasing two-digit substrings of the previous term.

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Haskell

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(3 - sqrt(5)).

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A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(5 - sqrt(21)).

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A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).

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A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(5 - sqrt(21)).

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).