A034838 -id:A034838 - cp's OEIS frontend

A337184 Numbers divisible by their first digit and their last digit.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99, 101, 102, 104, 105, 111, 112, 115, 121, 122, 123, 124, 125, 126, 128, 131, 132, 135, 141, 142, 144, 145, 147, 151, 152, 153, 155, 156, 161, 162, 164, 165, 168, 171, 172, 175, 181, 182

Offset: 1

Bernard Schott, Jan 29 2021

The first 23 terms are the same first 23 terms of A034838 then a(24) = 101 while A034838(24) = 111.
Terms of A034709 beginning with 1 and terms of A034837 ending with 1 are terms.
All positive repdigits (A010785) are terms.
There are infinitely many terms m for any of the 53 pairs (first digit, last digit) of m described below: when m begins with {1, 3, 7, 9} then m ends with any digit from 1 to 9; when m begins with {2, 4, 6, 8}, then m must also end with {2, 4, 6, 8}; to finish, when m begins with 5, m must only end with 5. - Metin Sariyar, Jan 29 2021

David A. Corneth, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.

Intersection of A034709 and A034837.
Subsequences: A010785\{0}, A034838, A043037, A043040, A208259, A066622.
Cf. A139138.

Select[Range[175], Mod[#, 10] > 0 && And @@ Divisible[#, IntegerDigits[#][[{1, -1}]]] &] (* Amiram Eldar, Jan 29 2021 *)

is(n) = n%10>0 && n%(n%10)==0 && n % (n\10^logint(n,10)) == 0 \\ David A. Corneth, Jan 29 2021

def ok(n): s = str(n); return s[-1] != '0' and n%int(s[0])+n%int(s[-1]) == 0
print([m for m in range(180) if ok(m)]) # Michael S. Branicky, Jan 29 2021

(10n-9)/9 <= a(n) < 45n. (I believe the liminf of a(n)/n is 3.18... and the limsup is 6.18....) - Charles R Greathouse IV, Nov 26 2024

Conjecture: 3n < a(n) < 7n for n > 75. - Charles R Greathouse IV, Dec 02 2024

A342445 Numbers that are divisible by their nonzero digits but are not divisible by the product of their nonzero digits.

Original entry on oeis.org

22, 33, 44, 48, 55, 66, 77, 88, 99, 122, 124, 126, 155, 162, 168, 184, 202, 204, 222, 244, 248, 264, 280, 288, 303, 324, 330, 333, 336, 366, 396, 404, 408, 412, 420, 424, 440, 444, 448, 488, 505, 515, 555, 606, 636, 648, 660, 666, 707, 728, 770, 777, 784, 808, 824, 840

Offset: 1

Bernard Schott, Mar 20 2021

Numbers that are divisible by the product of their nonzero digits (A055471) are trivially divisible by each of their nonzero digits (A002796), but the converse is false. This sequence = A002796 \ A055471 and consists of these counterexamples.

This sequence differs from A337163: the first sixteen terms are the same but a(17) = 202 while A337163(17) = 222.

			204 is divisible by 2 and 4 but 204 is not divisible by 2*4 = 8, hence 204 is a term.
248 is divisible by 2, by 4 and by 8 but 248 is not divisible by 2*4*8 = 64, hence 248 is a term.

Harvey P. Dale, Table of n, a(n) for n = 1..2000

Equals A002796 \ A055471.

Cf. A337163 = A034838 \ A007602 (subsequence of zeroless numbers).

q[n_] := AllTrue[(d = Select[IntegerDigits[n], # > 0 &]), Divisible[n, #] &] && ! Divisible[n, Times @@ d]; Select[Range[840], q] (* Amiram Eldar, Mar 21 2021 *)
dnzQ[n_]:=With[{c=DeleteCases[IntegerDigits[n],0]},Union[Boole[Divisible[n,c]]]=={1}&&!Divisible[n,Times@@c]]; Select[ Range[ 1000],dnzQ] (* Harvey P. Dale, Jan 16 2025 *)

isok(m) = my(d=select(x->(x != 0), digits(m))); (m % vecprod(d)) && (sum(k=1, #d, m % d[k]) == 0); \\ Michel Marcus, Mar 22 2021

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.

A018298 Divisors of 135.

Original entry on oeis.org

1, 3, 5, 9, 15, 27, 45, 135

Offset: 1

N. J. A. Sloane

Notice that the base-10 digits of 135 occur as the first three divisors of this number, in order. Hence 135 is in A034838, numbers divisible by each of their digits. - Alonso del Arte, Sep 17 2017

Index entries for sequences related to divisors of numbers

Divisors[135] (* Alonso del Arte, Sep 17 2017 *)

divisors(135) \\ Charles R Greathouse IV, Jun 22 2017

A116957 Lynch-Bell numbers n such that 5 is a digit of n.

Original entry on oeis.org

5, 15, 135, 175, 315, 735, 1395, 1935, 3195, 3915, 9135, 9315

Offset: 1

Walter Kehowski, Apr 03 2006

A Lynch-Bell number is a positive integer n with distinct nonzero digits such that each of its digits divides the number: n mod d = 0 if d is a digit of n.

			a(3)=135 since 135 is the third Lynch-Bell number that contains a 5.

Cf. A115569, A034838, A034709, A063527.

lbn5Q[n_]:=Module[{idn=IntegerDigits[n]},MemberQ[idn,5]&&FreeQ[idn,0]&&Max[DigitCount[n]]==1&&AllTrue[n/idn,IntegerQ]]; Select[Range[ 10000],lbn5Q] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Aug 19 2019 *)

A116960 Lynch-Bell numbers k such that 1 is not a digit of k.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 24, 36, 48, 248, 264, 324, 384, 396, 432, 624, 648, 672, 728, 735, 784, 824, 864, 936, 2364, 2436, 3264, 3276, 3492, 3624, 3648, 3864, 3924, 4236, 4368, 4392, 4632, 4872, 4896, 4932, 4968, 6324, 6384, 6432, 6984, 8496, 8736, 9324, 9432

Offset: 1

Walter Kehowski, Apr 03 2006

The Lynch-Bell numbers are those positive integers k with distinct nonzero digits such that each digit divides k: k mod d = 0 if d is a digit of k.

			a(9)=24 since it is the 9th Lynch-Bell number that does not contain a 1.

Cf. A115569, A034838, A034709, A063527.

A376003 Positive integers k such that each digit of k^2 is a factor of k.

Original entry on oeis.org

1, 6, 12, 36, 54, 108, 156, 168, 192, 204, 288, 306, 408, 432, 486, 696, 804, 1104, 1146, 1188, 1488, 1512, 1632, 1764, 1806, 1932, 2232, 2904, 3114, 3408, 3456, 3528, 4014, 4104, 4392, 4596, 4608, 4704, 4788, 4872, 4932, 4944, 5208, 5304, 5868, 6012, 6696, 6792

Offset: 1

Sam N. Harrison, Sep 28 2024

0 is never a factor so k^2 must be zeroless and this sequence is a subset of A052040.
The first term > 1 that is not divisible by 6 is 47768.
From Andrew Howroyd, Sep 28 2024: (Start)
Except for the first term, all terms are even since all squares with at least 2 digits contain an even digit. This implies k^2 cannot contain the digit 5.
All numbers of the form (100*1000^k-1)/3+3 are terms. These are the numbers 36, 33336, 33333336, 33333333336, etc. This shows that the sequence is infinite. (End)

			k = 12 is a term since k^2 = 144 has digits 1 and 4 and both are factors of k.
k = 2 is not a term since k^2 = 4 has a digit 4 which is not a factor of k.

Cf. A052040, A034838.

q:= n-> andmap(x-> x>0 and irem(n, x)=0, convert(n^2, base, 10)):
select(q, [$1..10000])[];  # Alois P. Heinz, Sep 28 2024

isok(k) = my(d=Set(digits(k^2))); if(!vecmin(d), return(0)); for (i=1, #d, if (k % d[i], return(0))); return(1); \\ Michel Marcus, Sep 28 2024

def is_valid_k(k):
    k_squared = k ** 2
    for digit in str(k_squared):
        d = int(digit)
        if d == 0 or k % d != 0:
            return False
    return True
def find_valid_k(max_k):
    valid_k = []
    for k in range(1, max_k + 1):
        if is_valid_k(k):
            valid_k.append(k)
    return valid_k
max_k = 10000
result = find_valid_k(max_k)
print(result)

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A337184 Numbers divisible by their first digit and their last digit.

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A342445 Numbers that are divisible by their nonzero digits but are not divisible by the product of their nonzero digits.

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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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A018298 Divisors of 135.

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A116957 Lynch-Bell numbers n such that 5 is a digit of n.

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Mathematica

A116960 Lynch-Bell numbers k such that 1 is not a digit of k.

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A376003 Positive integers k such that each digit of k^2 is a factor of k.

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Maple

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Python

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.