cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A098595 Roman numerals version of the Look and Say sequence (original at A005150).

Original entry on oeis.org

1, 11, 111, 1111, 151, 111511, 111115111, 51151111, 1511115151, 111515115111511, 111115111511115111115111, 51151111151511551151111, 15111155115111511111511115151, 111515111511115111115511515115111511, 1111151115111115151155111511115111511115111115111, 511511111551151115111115111115151151111151511551151111
Offset: 1

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Author

Mark Hudson (mrmarkhudson(AT)hotmail.com), Sep 17 2004

Keywords

Comments

This sequence is suggested at the link given in point 7 at the bottom of the page, but it is given as: I, II, III, IIII, IIV, IIIIV, IVIIV, ... The 5th term seems to use the "number" followed by "frequency" method (one appears 4 times -> "I"//"IV") but the 6th and 7th terms use the usual "frequency"//"number" method.
The analog of Conway's constant for this sequence is C=1.09807850157..., the largest real root of x^17+x^16-x^14-x^13+x^11+x^10-x^8-x^7-x^6+x^4-x^2-x-1. - Ercole S. Banani, Jul 27 2023

Examples

			a(1)=I=1; a(2)="one I"="II"=11; a(3)=111; a(4)=1111; a(5)="four Is"="IVI"=151; and so on.
		

Crossrefs

Cf. A005150 for original Look and Say sequence and a number of references.

Programs

Formula

We write the Roman numerals as I, II, III, IV, V and convert the letters into 1's and 5's for the inclusion in the encyclopedia.

Extensions

More terms from David Wasserman, Feb 25 2008

A100108 Primes in A005150.

Original entry on oeis.org

11, 312211, 13112221
Offset: 1

Views

Author

Jorge Coveiro, Dec 26 2004

Keywords

Crossrefs

Cf. A005150.

Extensions

When further terms have been found, we may add the sequence "Numbers n such that A005150(n) is prime". It begins 2, 6, 7.

A356008 A variant of Look and Say sequence (A005150) based on exponents in prime factorization of n (see Comments section for precise definition).

Original entry on oeis.org

1, 6, 105, 12, 315, 18, 945, 24, 525, 6006, 2835, 420, 8505, 42042, 735, 48, 25515, 1050, 76545, 12012, 440895, 294294, 229635, 840, 1575, 2060058, 2625, 84084, 688905, 54, 2066715, 96, 5731635, 14420406, 2205, 36, 6200145, 100942842, 74511255, 24024, 18600435
Offset: 1

Views

Author

Rémy Sigrist, Jul 23 2022

Keywords

Comments

To compute a(n):
- a(1) = 1,
- for n > 1:
- consider the prime factorization of n:
n = Product_{i = 1..k} prime(i)^e_i
(where e_k > 0 and prime(i) denotes the i-th prime number),
- apply the Look and Say procedure to the list (e_k, ..., e_1),
- the result, say (f_m, ..., f_1), gives the prime exponents for a(n):
a(n) = Product_{i = 1..m} prime(i)^f_i.
There are only two fixed points: a(1) = 1 and a(36) = 36.
All terms are distinct and belong to A244990 (but some terms of A244990, like 210 = 7*5*3*2, do not appear here).

Examples

			For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- applying the Look and Say procedure, we obtain: 1 1 2 0 1 2 1 0,
- so a(99) = 19^1 * 17^1 * 13^2 * 11^0 * 7^1 * 5^2 * 3^1 * 2^0 = 28658175.
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(n) = n mod 2.
A007814(a(n)) = A007814(n).
a(prime(n)) = 7*5*3^(n-1) for any n > 1.
a(A002110(n)) = 2*3^n = A008776(n) for any n > 0.

A045998 Binary Gleichniszahlen-Reihe (BGR) sequence: describe previous term (cf. A005150), reduce number of digits seen mod 2 (then for the purposes of this data-base, discard leading zeros).

Original entry on oeis.org

1, 11, 1, 1011, 111001, 110011, 10001, 10111011, 1110111001, 1110110011, 1110010001, 1100111011, 100111001, 101100110011, 11100100010001, 11001110111011, 1001110111001, 1011001110110011, 111001001110010001
Offset: 0

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Author

Keywords

Comments

Terms with a leading zero: a(2), a(6), a(12), a(16), a(20), a(28), a(32), a(36), a(40), a(44), a(48), a(60), ...

Examples

			1, 11, 01, 1011, 111001, 110011, 010001, ... (after 110011, next term is 212021 -> 010001 -> 10001).
		

References

  • N. Worrick, S. Lewis and B. Shrader, A possible formula for the length of BGR sequences, Graph Theory Notes of New York, XXXVI (1999), p. 25.

Crossrefs

Extensions

More terms from Patrick De Geest, Jun 15 1999

A087283 An infinite audioactive word, one of three in the cycle that results from the limit of the 'Look and Say' sequence using method A with an initial term of 1 (A005150).

Original entry on oeis.org

3, 1, 1, 3, 1, 1, 2, 2, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 3, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1, 1, 2, 1, 3, 2, 1, 1, 3, 1, 2, 1, 1, 1, 3, 2, 2, 2, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 2, 3, 1, 1
Offset: 1

Views

Author

Paul D. Hanna, Aug 28 2003

Keywords

Comments

Performing 'Look and Say' once generates A087284, twice results in A087282, while three operations yield the original infinite word.

Crossrefs

A087284 An infinite audioactive word, one of three in the cycle that results from the limit of the 'Look and Say' sequence using method A with an initial term of 1 (A005150).

Original entry on oeis.org

1, 3, 2, 1, 1, 3, 2, 1, 3, 2, 2, 1, 1, 3, 3, 1, 1, 2, 1, 3, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 2, 1, 1, 2, 1, 3, 2, 1, 1, 3, 3, 1, 1, 2, 1, 3, 2, 1, 1, 2, 3, 1, 2, 3, 2, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 1, 1, 2, 1, 3, 2
Offset: 1

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Author

Paul D. Hanna, Aug 28 2003

Keywords

Comments

Performing 'Look and Say' once generates A087282, twice results in A087283, while three operations yield the original infinite word.

Crossrefs

A098097 Terms of sequence A005150 interpreted as numbers written in base 4 (written here in base ten).

Original entry on oeis.org

1, 5, 9, 101, 1385, 3493, 30121, 358885, 225859433, 521661150629, 1537769393977769, 248286184577158309349, 2349690596380762356487460713, 7261901690904608168867803518651168165, 76840829143792845266223009702402770017026931113
Offset: 1

Views

Author

Axel Harvey, Sep 14 2004

Keywords

Comments

The terms in A005150 contain no digits other than 1, 2, and 3, so they can be read as numbers in base 4.

Examples

			A005150(5) = 111221, and 111221_4 = 1385.
		

Crossrefs

Formula

A005150(n) = A007090(a(n)). - Michel Marcus, Jun 22 2024

Extensions

a(13)-a(15) added by Andrew Howroyd, Feb 12 2020

A253677 Product of decimal digits of n-th term of the Look and Say sequence A005150.

Original entry on oeis.org

1, 1, 2, 2, 4, 12, 24, 36, 216, 1296, 62208, 746496, 107495424, 46438023168, 240734712102912, 8423789045905096704, 5589622068988418728132608, 133524176512370516060034538930176, 581306137722148449693374999786222908342272
Offset: 1

Views

Author

Fabian Nedic, Jan 08 2015

Keywords

Examples

			a(5) = A007954(A005150(5)) = A007954(111221) = 4.
		

Crossrefs

Formula

a(n) = A007954(A005150(n)).

A056635 Difference between length (A005341) and sum of digits (A004977) of n-th term in Look and Say Sequence (A005150).

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 5, 6, 9, 12, 18, 22, 30, 40, 54, 72, 93, 120, 157, 203, 271, 364, 473, 612, 806, 1062, 1388, 1804, 2349, 3057, 4001, 5224, 6812, 8874, 11582, 15065, 19661, 25647, 33393, 43509, 56738, 73989, 96469, 125774, 163943, 213683, 278605
Offset: 1

Views

Author

Robert G. Wilson v, Aug 08 2000

Keywords

Programs

  • Mathematica
    RunLengthEncode[ x_List ] := (Through[ {First, Length}[ #1 ] ] &) /@ Split[ x ]; LookAndSay[ n_, d_:1 ] := NestList[ Flatten[ Reverse /@ RunLengthEncode[ # ] ] &, {d}, n - 1 ]; F[ n_ ] := LookAndSay[ n, 1 ][ [ n ] ]; Table[ Apply[ Plus, F[ n ] ]-Length[ F[ n ] ], {n, 1, 53} ] (* Eric Weisstein *)
    p={8,-10,11,-28,33,-2,-11,-34,11,23,3,-10,0,-3,35,-43,-5,46,11,0,-1,-3,-13,-14,-28,24,20,56,-76,0,-5,-29,59,-9,-26,18,-28,55,-1,-34,-33,-21,51,48,-16,-28,-23,24,31,-37,-9,-9,31,21,-16,-19,-11,5,14,3,-1,-4,-1,6,0,-4,-3,0,5,1,-1,-1,-1,0,0,0,1,0,0}; q={6,-9,9,-18,16,-11,14,-8,1,-5,7,2,8,-14,-5,-5,19,3,-6,-7,-6,16,-7,8,-22,17,-12,7,5,7,-8,4,-7,-9,13,-4,-6,14,-14,19,-7,-13,2,-4,18,0,-1,-4,-12,8,-5,0,8,1,7,-8,-5,-2,3,3,0,0,0,0,-2,-1,0,3,1,-1,-1,-1,1}; gf=Fold[x #1+#2&,0,p]/Fold[x #1+#2&,0,q]; CoefficientList[Series[gf,{x,0,99}],x] (* Peter J. C. Moses, Jun 24 2013 *)

A334054 Lexicographically earliest numbers containing only digits 1,2,3 that appear again in the iterative cycle of their own 'Look and Say' description (Cf. A005150).

Original entry on oeis.org

1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 311, 312, 321, 322, 331, 332, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1213, 1221, 1222, 1223, 1231, 1232, 1311, 1312, 1321, 1322, 1331, 1332, 2111
Offset: 1

Views

Author

Scott R. Shannon, Sep 07 2020

Keywords

Comments

Take a number that contains only digits 1,2,3 and then describe the given number with its 'Look and Say' string, see A005150. Repeat this process until the starting number appears in the resulting string. The sequence lists the numbers that eventually reappear.
A334055 gives the number of iterations for the starting number to reappear and details of the final string for all recurrent numbers.
The number 233 is the first number that does not appear in its iterative Look and Say cycle. See examples below.

Examples

			1 is a term as the Look and Say description of 1 is '11', which contains '1' as a substring.
111 is a term as the Look and Say description of 111 is '31'. Repeating this process leads to '1311' and then '111321', which contains '111' as a substring.
1112 is a term as the Look and Say description of 1112 is '3112'. Repeating this process leads to '132112', '1113122112', '311311222112', '13211321322112', '1113122113121113222112', and then '31131122211311123113322112', which contains '1112' as a substring.
233 is not a term. If it were, it would have to have a parent string that could produce 233 from its Look and Say description, and one can show that that is not possible. The '33' part cannot be interpreted as 'three 3's' as '333' cannot be inside the parent string. To prove that, consider that the string '333' can be interpreted in two ways. 1. "x number of 3's followed by three 3's" which is not possible as that would be written as "x+three 3's". 2. "three 3's followed by three x's", where x cannot be a 3, else it would be written as "six 3's". So it has to be either '333111' or '333222'. But the first case has to be interpreted as either "x 3's followed by three 3's followed by one 1 ..." which is not possible as that would be written as "x+three 3's", else it has to be interpreted as "three 3's followed by three 1's followed by one 1", which is not possible as that would be written as "... four 1's". Likewise for '333222', hence '333' has no valid parent string. Therefore '233' must be interpreted as "two 3's followed by 3 x's", where the x's cannot be 3's. Thus the two possible strings are '33111' or '33222'. But the '33' part of these strings cannot be "three 3's" as shown, so the first string has to be interpreted as "x 3's followed by three 1's followed by one 1", but that is not possible as the 1's would be combined. Similarly for the second string. Hence '233' has no parent string, thus it cannot reappear in the iterative process of this sequence.
A similar argument can prove that numbers like 313, 323, 1111, 2232, 2313, and with more effort, 22213, 112321, 213321, do not have any parent string and thus do not appear in this sequence.
		

Crossrefs

Previous Showing 11-20 of 151 results. Next