OEIS – and a new (?) idea

(English translation after the French part)

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Hello M.

Une « nouvelle » idée ? Elle se baserait sur k(n), lequel k(n) serait la somme des facteurs premiers distincts d’un entier quelconque 'n'.

Je ne trouve pas trace de ce « concept » dans l’OEIS. [correction par M. quelques heures après lui avoir envoyé ce message : on évoque bien k(n) ici dans la suite baptisée sopf https://oeis.org/A008472 ; cette idée n’est donc pas nouvelle].

On aurait donc : 

k(15) = 8 (car 8 est la somme des facteurs premiers distincts de 15, lesquels sont 3 et 5)

k(16) = 2 (car 16 = 2*2*2*2, avec 2 comme unique facteur premier distinct) k(17) = 17 (car 17 est le seul diviseur premier de 17) k(18) = 5 (car 18 = 2*3*3 avec 2 et 3 comme facteurs premiers distincts, lesquels ont pour somme 5). etc. 

Notons qu’aucun 'n' > 0 n’a pour k(n) 1, 4 ou 6.

a) une première suite S à proposer serait celle des k(n) des entiers naturels 'n' ; voici 'n' sur la première ligne et les k(n) qui leur correspondent juste dessous :

   n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,...

k(n) = 0, 2, 3, 2, 5, 5, 7, 2, 3,  7, 11,  5, 13,  9,  8,  2, 17,  5, 19,  7,...

… on a ici les k(n) des 20 premiers entiers, avec 0 quand il n’y a pas de k(n). [Cette suite est bien la sopf, mentionnée plus haut, elle figure donc depuis longtemps dans l’OEIS – en revanche les points b), c) et d) ci-dessous semblent originaux].

b) on voit ci-dessus que certains termes sont répétés (2 par exemple, ou 3, ou 5, ou 7). Il faudrait donc proposer, je pense, une vingtaine de suites :

 — suite des entiers qui ont 2 pour k(n) ; elle existe déjà (puissances de 2), il faut peut-être juste lui ajouter une note au sujet de k(n) : 2, 4, 8, 16, 32, 64,… 

 — suite des entiers qui ont 3 pour k(n) ; elle existe aussi (puissances de 3), on pourrait lui ajouter une note aussi : 3, 9, 27, 81, 243, 729, … — suite des entiers qui ont 5 pour k(n) ; elle est originale (et affreuse à calculer à la main) : 5, 6, 12, 18, 24, 25, … — suite des entiers ayant 7 pour k(n) ; elle est originale aussi [comme toutes celles qui vont suivre : nombres ayant 8 pour k(n), nombres ayant 9 pour k(n), nombres ayant 10 pour k(n) etc.] Toutes sont affreuses à calculer à la main... [On pourrait proposer à l’OEIS les suites allant jusqu’aux nombres ayant 20 pour k(n)].

c) suite _infinie_ et lexicographically earliest de termes distincts > 0 such that a(n) est le k(n) de a(n+1) :  2, 8, 15, 26, 69, 134, 393, 8043, 553173, … Attention, il n’est pas évident de trouver le plus petit des a(n+1)… Par ailleurs le mot _infinie_ ci-dessus est important — sinon la suite s’arrêterait immédiatement : 2, 4. 

Cette suite infinie « explose » vite…

d) tout a commencé (pour moi) avec la volonté de fabriquer la lexicographically suite E de termes > 0 distincts suivante :

« Remplacer chaque terme a(n) de E par son k(n) ; la concaténation des a(n) est identique à la concaténation des k(n) »

C’est très amusant à faire à la main — car le lexico-critère de termes distincts demande que l’on soit (très) astucieux ! J’obtiens :

E = (mettre ici la suite des nombres premiers de 2 à 53 suivie de) 54, 1218, 4, 65, 279, 6, 8, 10, 14, 427, 21, 33, 185, 20, 16, 22, 9, 77, 12, 51, 39, 32, 64, 28, 40, 49, 35, 18, 44, 56, 27, 69, 341, 15, 111, 94, 81, 24, 78, 123, 25, 177, 50, 134, … Après remplacement on a : (même suite des premiers 2–>53 suivie de) 5, 41, 2, 18, 46, 5, 2, 7, 9, 68, 10, 14, 42, 7, 2, 13, 3, 18, 5, 20, 16, 2, 2, 9, 7, 7, 12, 5, 13, 9, 3, 26, 42, 8, 40, 49, 3, 5, 62, 7, 69, ... où l’on voit que la suite des chiffres de E est la même que celle des chiffres « remplacés » (modulo mes erreurs de la nuit !-)

Il y a sûrement d’autres variantes à proposer sur ce thème (celle-ci par exemple, merci à M.), mais ce sera pour une autre fois ! 

à+ É.

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Hello M.

say that k(n) is the sum of the distinct prime factors of a(n); we would then have, for instance:

k(15) = 8 (= sum 3 + 5 as 15 = 3*5) k(16) = 2 (as 16 = 2*2*2*2) k(17) = 17 (as 17 is the only prime factor of 17) k(18) = 5 (as 18 = 2*3*3); etc. 

a) a first sequence that could be submitted is the k(n) of n :

k(n) = 0, 2, 3, 2, 5, 5, 7, 2, 3, 7, 11, 5, 13, 9, 8, 2, 17, 5, 19,  7,...

[this seq is already in the OEIS, there – wrote M. to me in private a few hours ago].

b) we see above that some terms are repeated (2 for instance, or 3, or 5, or 7). We might then submit those sequences:

 — integers having 2 as k(n) ; this sequence is already in the OEIS (powers of 2); we might just add a note there about the k(n) "concept":  2, 4, 8, 16, 32, 64,… 

 — integers having 3 as k(n) ; this sequence is also in the OEIS (powers of 3); a similar note could be added there:  3, 9, 27, 81, 243, 729, … — integers having 5 as k(n) ; this is new, I guees (and a nightmare to compute by hand): 5, 6, 12, 18, 24, 25, … — integers having 7 as k(n) ; this is new too [as are all the similar sequences: integers having 8 as k(n), integers having 9 as k(n), integers having 10 as k(n) etc.] Here too, a nightmare to compute by hand. We might submit those sequences up to the integer having 20 as k(n).

c) the _infinite_ lexicographically earliest sequence of distinct terms > 0 such that a(n) is the k(n) of a(n+1) :  2, 8, 15, 26, 69, 134, 393, 8043, 553173, …

Caveat: it is not obvious at all to find the smallest a(n+1)… The word _infinite_ above is important — else the sequence would stop immediately: 2, 4. The above infinite sequence "explodes" quickly…

d) everything started (for me) when I wanted to form the lexicographically earliest sequence E of distinct terms > 0 with the hereunder constraint:

« Substitute each term a(n) of E by its k(n); the concatenation of the a(n)s is equal to the concatenation of the k(n)s ». 

This is fun to compute by hand — because the lexico-earliest (etc.) constraint requires one to be (very) subtle! I get:

E = (put here the sequence of prime numbers from 2 to 53, folowed by) 54, 1218, 4, 65, 279, 6, 8, 10, 14, 427, 21, 33, 185, 20, 16, 22, 9, 77, 12, 51, 39, 32, 64, 28, 40, 49, 35, 18, 44, 56, 27, 69, 341, 15, 111, 94, 81, 24, 78, 123, 25, 177, 50, 134,… After the substitution we get: (same sequence of prime numbers from 2–>53, followed by) 5, 41, 2, 18, 46, 5, 2, 7, 9, 68, 10, 14, 42, 7, 2, 13, 3, 18, 5, 20, 16, 2, 2, 9, 7, 7, 12, 5, 13, 9, 3, 26, 42, 8, 40, 49, 3, 5, 62, 7, 69,... and we see that the successive digits of E are the same successive digits of the "substituted" terms sequence (modulo my usual hand mistakes ;-)

There must exist for sure a few interesting variants dealing with this idea (this seq is an example, thank you M.) — but ... enough for today!

Best,

É.

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A few approved seqs that are not yet published in the OEIS:

NAME https://oeis.org/draft/A364274

Lexicographically earliest sequence of distinct positive terms such that the cumulative sum Q(n) of the first n terms of the sequence has distinct digits. 

DATA 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 10, 13, 14, 15, 16, 17, 19, 18, 20, 21, 22, 23, 25, 24, 26, 27, 28, 29, 30, 31, 32, 33, 35, 34, 40, 36, 37, 38, 39, 41, 42, 43, 77, 44, 136, 45, 46, 48, 47, 49, 51, 50, 53, 55, 56, 57, 60, 52, 54, 58, 59, 61, 63, 134, 64, 65, 66, 67, 68,... 

OFFSET 1,2 

COMMENTS The sequence is finite, as Q(n) cannot be > 9876543210. What is the last term of the sequence? 

EXAMPLE a(9) = 9 and Q(9) = 45; a(10) = 11 and Q(10) = 56: a(10) cannot = 10 as Q(10) would = 55; a(11) = 12 and Q(11) = 68: a(11) cannot = 10 as Q(11) would = 66; a(12) = 10 and Q(12) = 78; a(13) = 13 and Q(13) = 91; etc. 

NAME https://oeis.org/draft/A364275

Lexicographically earliest sequence of distinct positive terms such that the cumulative sum Q(n) of the first n terms of the sequence has at least one duplicated digit.

DATA

11, 22, 33, 34, 1, 9, 2, 3, 4, 12, 10, 14, 6, 5, 15, 7, 23, 13, 8, 20, 25, 26, 19, 16, 17, 18, 27, 24, 21, 29, 31, 28, 32, 30, 38, 35, 39, 37, 43, 41, 40, 51, 36, 42, 44, 47, 45, 46, 48, 50, 53, 58, 52, 54, 56, 49, 57, 55, 59, 61, 60, 64, 65, 62, 69, 63, 66, 72, 68, 74

OFFSET

1,1

COMMENTS

The sequence is infinite. Does a term a(n) exist such that from a(n) on a(n+1) is always a(n) + 1?

EXAMPLE

a(1) = 11 and Q(1) = 11 (with two 1s);

a(2) = 22 and Q(2) = 33 (with two 3s);

a(3) = 33 and Q(3) = 66 (with two 6s);

a(4) = 34 and Q(4) = 100 (with two 0s);

a(5) = 1 and Q(5) = 101 (with two 1s);

a(6) = 9 and Q(6) = 110 (with two 1s); etc.

NAME https://oeis.org/draft/A364325

Underline the k-th digit of a(n), k being the leftmost digit of a(n). This is the lexicographically earliest sequence of distinct terms > 0 such that the succession of underlined digits is the succession of the sequence's digits themselves.

DATA

1, 10, 20, 22, 200, 220, 221, 222, 201, 202, 223, 224, 203, 225, 226, 11, 227, 228, 229, 302, 204, 12, 312, 205, 322, 332, 342, 23, 352, 362, 24, 372, 206, 230, 382, 392, 25, 2200, 2201, 26, 13, 14, 2202, 2203, 27, 2204, 2205, 28, 2206, 2207, 29, 231, 207, 2208, 2209, 208, 240, 15, 2210, 232, 16, 2211

OFFSET

1,2

EXAMPLE

The leftmost digit of a(1) = 1 is 1: this digit underlines the 1st digit of a(1) which is (1);

The leftmost digit of a(2) = 10 is 1: this digit underlines the 1st digit of a(2) which is (1);

The leftmost digit of a(3) = 20 is 2: this digit underlines the 2nd digit of a(3) which is (0);

The leftmost digit of a(4) = 22 is 2: this digit underlines the 2nd digit of a(4) which is (2);

The leftmost digit of a(5) = 200 is 2: this digit underlines the 2nd digit of a(5) which is (0);

The leftmost digit of a(6) = 220 is 2: this digit underlines the 2nd digit of a(6) which is (2); etc.

We see that the parenthesized digits at the end of each line reproduce the succession of the original digits

NAME https://oeis.org/draft/A364326

Underline the k-th digit of a(n), k being the rightmost digit of a(n). This is the lexicographically earliest sequence of distinct terms > 0 such that the succession of the underlined digit is the succession of the sequence's digits themselves.

DATA

1, 11, 101, 111, 102, 112, 121, 131, 141, 151, 202, 12, 161, 171, 21, 181, 22, 191, 212, 31, 312, 412, 41, 512, 612, 51, 712, 32, 302, 42, 812, 52, 912, 61, 1001, 1011, 71, 1013, 62, 1021, 1031, 81, 1041, 72, 82, 1051, 91, 1061, 92, 1071, 122, 103, 1081, 113, 1091, 201, 142, 1101, 211, 242

OFFSET

1,2

EXAMPLE

The rightmost digit of a(1) = 1 is 1: this digit underlines the 1st digit of a(1) which is (1);

The rightmost digit of a(2) = 11 is 1: this digit underlines the 1st digit of a(2) which is (1);

The rightmost digit of a(3) = 101 is 1: this digit underlines the 1st digit of a(3) which is (1);

The rightmost digit of a(4) = 111 is 1: this digit underlines the 1st digit of a(4) which is (1);

The rightmost digit of a(5) = 102 is 2: this digit underlines the 2nd digit of a(5) which is (0);

The rightmost digit of a(6) = 112 is 2: this digit underlines the 2nd digit of a(6) which is (1); etc.

We see that the parenthesized digits at the end of each line reproduce the succession of the original digits.