A096098 -id:A096098 - cp's OEIS frontend

A240588 a(1) = 1, a(2) = 2; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

1, 2, 3, 41, 7, 9, 137131, 61, 2023244487101, 13, 19, 11, 143, 142733, 21, 17, 193, 37, 3907, 1290366811360047359, 1805030483980039, 3803623, 123, 369, 27, 23, 58271, 47609, 523, 79, 307, 179, 73, 57, 18032419296851, 29, 31, 3281881401611107, 69, 171, 60244474373, 197, 97

Offset: 1

Paolo P. Lava, Apr 29 2014

From Scott R. Shannon, Dec 19 2019: (Start)

The next unknown term a(131) requires the factorization of a 517-digit composite number 46297...2963. (End)

			a(1)=1 and a(2)=2. a(1) U a(2) = 12 and its divisors are 1, 2, 3, 4, 6, 12. Therefore 3 is the least number not yet present in the sequence which divides 12. Again, a(1) U a(2) U a(3) = 123 and its divisors are 1, 3, 41, 123. Therefore a(4)=41. Etc.

Scott R. Shannon, Table of n, a(n) for n = 1..130.

Cf. A096097, A096098, A241811.

with(numtheory):
T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a,b,c,k,n; b:=12; print(1); print(2); c:=[1,2];
for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do
if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;
fi; od; od; end: P(19);

a = {1, 2}; While[Length[a] < 22,
  n = ToExpression[StringJoin[ToString /@ a]];
  AppendTo[a, SelectFirst[Sort[Divisors[n]], FreeQ[a, #] &]]
]; a

a(20)-a(40) from Alois P. Heinz, May 08 2014
a(22) corrected by Ryan Hitchman, Sep 14 2017
a(23)-a(25) from Robert Price, May 16 2019
a(23)-a(25) corrected, and a(26)-a(43) added by Scott R. Shannon, Dec 10 2019

A241811 a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

Original entry on oeis.org

1, 0, 2, 3, 11, 71, 29, 9, 683, 67, 7, 743, 739, 1933, 23, 161, 21, 37, 19, 17, 119, 49, 332534262883, 13, 39, 13739483941387, 83, 111, 79853560395691, 5431567, 70610371, 69, 51, 4112497, 28384496881337963, 353, 77, 1531, 42787, 63, 27, 41, 709, 33, 81, 487, 139697

Offset: 1

Paolo P. Lava, Apr 29 2014

			a(1)=1 and a(2)=0. a(1) U a(2) = 10 and its divisors are 1, 2, 5, 10. Therefore 2 is the least number not yet present in the sequence which divides 10. Again, a(1) U a(2) U a(3) = 102 and its divisors are 1, 2, 3, 6, 17, 34, 51, 102. Therefore a(4)=3, etc.

Chai Wah Wu, Table of n, a(n) for n = 1..68

Cf. A096097, A096098, A240588.

with(numtheory):
T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a,b,c,k,n; b:=10; print(1); print(0); c:=[0,1];
for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do
if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;
fi; od; od; end: P(30);

a(23)-a(28) from Zak Seidov, May 08 2014

a(29)-a(47) from Giovanni Resta, Aug 15 2019

A096097 a(1) = 2, a(2) = 1; for n >= 3, a(n) = least prime not included earlier that divides the concatenation of all previous terms.

Original entry on oeis.org

2, 1, 3, 71, 7, 10177, 2100001, 101770000001, 4603, 13, 107, 4013, 23, 3097349301044927552199565217412468305904367, 1847, 37, 367767021959, 54371, 3229, 17, 520063, 29, 389, 8059, 732713, 11, 7123120001, 137, 294563, 1656881076199062425029, 313583, 4817, 277

Offset: 1

Amarnath Murthy, Jun 24 2004

Conjecture:(1) Every concatenation is squarefree. (2) This is a rearrangement of the noncomposite numbers other than 5.

Conjecture (1) is false. 3^2 divides the concatenation for a(22) and a(30). - Sean A. Irvine, Nov 25 2009

			a(4) = 71 as 213 = 3*71.

Cf. A096098.

More terms from Sean A. Irvine, Nov 25 2009

a(30)-a(33) from Chai Wah Wu, Nov 29 2019

A382445 Lexicographically least increasing sequence of distinct positive integers such that for any n > 1, a(n) does not divide the concatenation of the earlier terms.

Original entry on oeis.org

1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70

Offset: 1

Rémy Sigrist, Mar 25 2025

			a(1) = 1.
a(2) must not divide 1; we can take a(2) = 2.
a(3) must not divide 12; we can take a(3) = 5.

Rémy Sigrist, PARI program

Cf. A007978, A096098, A382441.

PARI
```
\\ See Links section.
```

from itertools import count, islice
def agen(): # generator of terms
    an = t = 1
    while True:
        yield an
        an = next(k for k in count(an+1) if t%k != 0)
        t = t*10**len(str(an)) + an
print(list(islice(agen(), 54))) # Michael S. Branicky, Mar 26 2025

A240229 a(n) is the shortest concatenation of the Fibonacci numbers F(1), F(2), ..., divisible by F(n) = A000045(n), n >= 1. a(n) = 0 if there is no such concatenation.

Original entry on oeis.org

1, 1, 112, 11235, 11235, 112, 1123581321345589144233, 11235, 11235813213455891442333776109871597258, 11235813213455891442333776109871597258441816765, 1123581321345589144233377610987159725844181676, 1123581321345589144233377610987159725844181676510946177112865746368

Offset: 1

Wolfdieter Lang, May 10 2014

The corresponding numbers a(n)/F(n) are 1, 1, 56, 3745, 2247, 14, 86429332411199164941, 535, 330465094513408571833346356172694037, 204287512971925298951523201997665404698942123, 12624509228602125216105366415586064335327884, 7802648064899924612731788965188609207251261642437126229950456572, ...
The author's opinion is that this is an example of a not-so-interesting sequence. I call this a WOTS (waste of time sequence). But because I had to write a program to test similar proposed sequences I thought I would apply it to this prominent example.
The next entry a(13) has 324 digits for the divisibility by F(13) = 233 with a(13)/F(13) a 321 digit composite. The given a(n) are all nonprimes.
Question: is there an n with a(n) = 0?

			a(3) = 112 because neither 1 nor 11 are divisible by F(3) = 2, but 112, the concatenation of F(1), F(2) and F(3) is.

Cf. A096098, A096097, A240588.

See the name.

cp's OEIS Frontend

A240588 a(1) = 1, a(2) = 2; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

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A241811 a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

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A096097 a(1) = 2, a(2) = 1; for n >= 3, a(n) = least prime not included earlier that divides the concatenation of all previous terms.

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Extensions

A382445 Lexicographically least increasing sequence of distinct positive integers such that for any n > 1, a(n) does not divide the concatenation of the earlier terms.

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Programs

PARI

Python

A240229 a(n) is the shortest concatenation of the Fibonacci numbers F(1), F(2), ..., divisible by F(n) = A000045(n), n >= 1. a(n) = 0 if there is no such concatenation.

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