A052423 -id:A052423 - cp's OEIS frontend

A069715 GCD of digits of n is 1.

1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 23, 25, 27, 29, 31, 32, 34, 35, 37, 38, 41, 43, 45, 47, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 65, 67, 71, 72, 73, 74, 75, 76, 78, 79, 81, 83, 85, 87, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 103, 104, 105, 106, 107, 108

Offset: 1

Labos Elemer, Apr 02 2002

			All numbers with at least one digit equal to 1 are here.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.

Cf. A011531 (subsequence), A240913 (subsequence).

a069715 n = a069715_list !! (n-1)
a069715_list = filter ((== 1) . a052423) [1..]
-- Reinhard Zumkeller, Apr 14 2014

Do[s=Apply[GCD, IntegerDigits[n]]; If[Equal[s, 1], Print[n]], {n, 1, 256}]

is(n)=gcd(digits(n))==1 \\ Charles R Greathouse IV, Nov 01 2014

A052423(a(n)) = 1. - Reinhard Zumkeller, Apr 14 2014

a(n) ~ n. In fact a(n) = n + O(n^(log 5/log 10)). - Charles R Greathouse IV, Nov 01 2014

A240913 Numbers m such that GCD of digits of m is 1 and no digit of m is 1.

Original entry on oeis.org

23, 25, 27, 29, 32, 34, 35, 37, 38, 43, 45, 47, 49, 52, 53, 54, 56, 57, 58, 59, 65, 67, 72, 73, 74, 75, 76, 78, 79, 83, 85, 87, 89, 92, 94, 95, 97, 98, 203, 205, 207, 209, 223, 225, 227, 229, 230, 232, 233, 234, 235, 236, 237, 238, 239, 243, 245, 247, 249

Offset: 1

Reinhard Zumkeller, Apr 14 2014

A052423(a(n)) = 1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Subsequence of A069715.

a240913 n = a240913_list !! (n-1)
a240913_list = filter (not . elem '1' . show) a069715_list

gcdQ[n_]:=Module[{idn=IntegerDigits[n]},!MemberQ[idn,1]&&GCD@@idn==1]; Select[ Range[300],gcdQ] (* Harvey P. Dale, Dec 17 2014 *)

is(n)=my(d=Set(digits(n))); setsearch(d,1)==0 && gcd(d)==1 \\ Charles R Greathouse IV, Nov 01 2014

A062078 LCM of the digits of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 2, 6, 4, 10, 6, 14, 8, 18, 0, 3, 6, 3, 12, 15, 6, 21, 24, 9, 0, 4, 4, 12, 4, 20, 12, 28, 8, 36, 0, 5, 10, 15, 20, 5, 30, 35, 40, 45, 0, 6, 6, 6, 12, 30, 6, 42, 24, 18, 0, 7, 14, 21, 28, 35, 42, 7, 56, 63

Offset: 1

Amarnath Murthy, Jun 15 2001

			a(25) = 10, a(24) = 4.

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)

Cf. A007954, A052423.

a:= n-> ilcm(convert(n, base, 10)[]):
seq(a(n), n=1..99);  # Alois P. Heinz, May 15 2024

Table[LCM@@IntegerDigits[n],{n,120}] (* Harvey P. Dale, Feb 19 2017 *)

a(n)={ lcm(digits(n)) } \\ Harry J. Smith, Jul 31 2009

a(n) <= 2520. - Sean A. Irvine, Mar 18 2023

A350494 Divide n by the greatest common divisor of the nonzero digits of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 10, 21, 11, 23, 12, 25, 13, 27, 14, 29, 10, 31, 32, 11, 34, 35, 12, 37, 38, 13, 10, 41, 21, 43, 11, 45, 23, 47, 12, 49, 10, 51, 52, 53, 54, 11, 56, 57, 58, 59, 10, 61, 31, 21, 32, 65, 11, 67

Offset: 1

Rémy Sigrist, Jan 01 2022

			a(42) = 42 / gcd(4, 2) = 42 / 2 = 21.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A052423, A069715.

PARI
```
a(n) = n / gcd(digits(n))
```

from math import gcd
def a(n): return n//gcd(*[int(d) for d in str(n)])
print([a(n) for n in range(1, 68)]) # Michael S. Branicky, Jan 01 2022

a(n) = n / A052423(n).
a(n) = n iff n belongs to A069715.
a(a(n)) = a(n).

A364135 Let d_r d_{r-1} ... d_1 d_0 be the decimal expansion of n; a(n) is the number of nonnegative integer solutions x_r ... x_0 to the Diophantine equation d_rx_r + ... + d_0x_0 = n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 7, 5, 4, 4, 3, 3, 3, 3, 1, 11, 12, 4, 7, 3, 5, 2, 4, 2, 1, 11, 6, 12, 3, 3, 7, 2, 2, 5, 1, 11, 11, 4, 12, 3, 4, 2, 7, 2, 1, 11, 6, 4, 3, 12, 2, 2, 2, 2, 1, 11, 11, 11, 6, 3, 12, 2, 3, 4, 1, 11, 6, 4, 3, 3, 2, 12, 2, 2, 1, 11, 11

Offset: 1

Ctibor O. Zizka, Jul 10 2023

For a formula for a(n), please see the Samsonadze article in Links section. a(n) = P(b) if n = b AND the nonzero digits of b are the coefficients a_i (in the article).

a(n) is the number of partitions of n into parts that are nonzero digits of n. - Stefano Spezia, Feb 17 2024

			For n = 10: 1*x_1 + 0*x_0 = 10, the solution is x_1 = 10, thus a(10) = 1.
For n = 22: 2*x_1 + 2*x_0 = 22, the solutions are (0,11), (2,10), ..., (11,0), thus a(22) = 12.

Robert P. P. McKone, Table of n, a(n) for n = 1..636
Eteri Samsonadze, On the number of integer non-negative solutions of a linear Diophantine equation, arXiv:2108.04756 [math.NT], 2021.

Cf. A007954, A034838, A052423, A055642.

a[n_Integer] := Module[{ds = IntegerDigits[n], p, t, v}, p = Table[If[d == 0, {0}, Range[0, Quotient[n, d]]], {d, ds}]; t = Tuples[p]; v = Select[t, ds . # == n &]; Length[v]]; Table[a[n], {n, 1, 82}] (* Robert P. P. McKone, Aug 25 2023 *)
a[n_]:=Length[IntegerPartitions[n, All,DeleteCases[ IntegerDigits[n],0]]]; Array[a,82] (* Stefano Spezia, Feb 17 2024 *)

a(n) = 1 for n = (d 0 ... 0), the digit d >= 1, the number of zeros >= 0.

a(n) = (1 ... 1) + 1 for n = (d ... d), the digit d >= 1, n >= 10.

A378513 a(1) = 1, a(n) = a(n-1) + n if all the digits of a(n-1) do not share a divisor greater than 1. Otherwise, a(n) = a(n-1) divided by the gcd of all its individual digits.

Original entry on oeis.org

1, 3, 1, 5, 1, 7, 1, 9, 1, 11, 22, 11, 24, 12, 27, 43, 60, 10, 29, 49, 70, 10, 33, 11, 36, 12, 39, 13, 42, 21, 52, 84, 21, 55, 11, 47, 84, 21, 60, 10, 51, 93, 31, 75, 120, 166, 213, 261, 310, 360, 120, 172, 225, 279, 334, 390, 130, 188, 247, 307, 368, 430

Offset: 1

Stuart Coe, Nov 29 2024

It can be shown that this sequence will grow indefinitely: Once it has reached a given number of digits in length, it cannot drop below that number of digits. Regardless of the number of times the number is reduced by dividing by GCDs, n will inevitably be great enough to increase a(n) to a larger and larger number of digits in length.

			a(37) = 47 + 37, because the digits of 47 do not share a factor greater than 1.
a(38) = 84 / 4 = 21, because the gcd of 8 and 4 is 4.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A052423 (gcd of digits).

a:= proc(n) option remember; `if`(n=1, 1, (t-> (g->
      `if`(g=1, t+n, t/g))(igcd(convert(t, base, 10)[])))(a(n-1)))
    end:
seq(a(n), n=1..62);  # Alois P. Heinz, Nov 29 2024

Module[{n = 1, g}, NestList[If[n++; (g = GCD @@ IntegerDigits[#]) == 1, # + n, #/g] &, 1, 100]] (* Paolo Xausa, Dec 16 2024 *)

lista(nn) = my(v=vector(nn)); v[1] = 1; for (n=2, nn, my(g=gcd(digits(v[n-1]))); if (g == 1, v[n] = v[n-1]+n, v[n] = v[n-1]/g);); v; \\ Michel Marcus, Dec 09 2024

cp's OEIS Frontend

A069715 GCD of digits of n is 1.

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A240913 Numbers m such that GCD of digits of m is 1 and no digit of m is 1.

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A062078 LCM of the digits of n.

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A350494 Divide n by the greatest common divisor of the nonzero digits of n.

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A364135 Let d_r d_{r-1} ... d_1 d_0 be the decimal expansion of n; a(n) is the number of nonnegative integer solutions x_r ... x_0 to the Diophantine equation d_r*x_r + ... + d_0*x_0 = n.

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A378513 a(1) = 1, a(n) = a(n-1) + n if all the digits of a(n-1) do not share a divisor greater than 1. Otherwise, a(n) = a(n-1) divided by the gcd of all its individual digits.

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Programs

Maple

Mathematica

PARI

A364135 Let d_r d_{r-1} ... d_1 d_0 be the decimal expansion of n; a(n) is the number of nonnegative integer solutions x_r ... x_0 to the Diophantine equation d_rx_r + ... + d_0x_0 = n.