A014950 -id:A014950 - cp's OEIS frontend

A225782 Numbers such that every permutation of digits of n is divisible by sum of digits of n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90, 100, 102, 108, 111, 117, 120, 126, 135, 144, 153, 162, 171, 180, 200, 201, 204, 207, 210, 216, 222, 225, 234, 240, 243, 252, 261, 270, 288

Offset: 1

Jayanta Basu, May 15 2013

Subsets of both A005349 and A225780. First member of A225780 missing here is 209. Next one is 308.
From Robert Israel, May 11 2017: (Start)
Numbers n such that n is divisible by A007953(n) and 9*d (mod A007953(n)) are all equal for all digits d of n.
If n is in the intersection of this sequence and A011540, then so is 10*n.  In particular, the sequence is infinite.
If n is in the sequence and A007953(n) > 81, then n = d*A002275(r) where 1 <= d <= 9 and r is in A014950. (End)

			126 is a member since 126, 162, 216, 261, 612 and 621 are all divisible by (1+2+6)=9. 209 is not a member since 29 is not divisible by (2+9)=11.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A002275, A005349, A007953, A011540, A014950, A225780.

filter:= proc(n) local s,L;
     L:= convert(n,base,10);
     s:= convert(L,`+`);
     n mod s = 0 and nops({seq(9*d mod s, d = L)}) = 1
end proc:
select(filter, [$1..1000]); # Robert Israel, May 11 2017

d[n_]:=IntegerDigits[n]; sod[n_]:=Total[d[n]]; t={}; Do[t1=Table[FromDigits[k],{k,Permutations[d[n]]}]; If[Select[t1,Mod[#,sod[n]]!=0 &]=={},AppendTo[t,n]],{n,288}]; t

A273906 Primes equal to the concatenation of two nonzero palindromic numbers.

Original entry on oeis.org

11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 113, 199, 211, 223, 227, 229, 233, 277, 311, 331, 337, 433, 443, 449, 499, 557, 577, 599, 661, 677, 733, 773, 811, 877, 881, 883, 887, 911, 977, 991, 997, 1013, 1019, 1117, 1151, 1171, 1181

Offset: 1

Giovanni Teofilatto, Jun 03 2016

The only palindrome in this sequence below 10^9 is 11 (per request of Giovanni Teofilatto). A004022 is a subsequence. - David A. Corneth, Jun 10 2016

If we have a concatenation of two palindromes A = A', B = B' which is palindromic, concat(A,B) =: A.B = (A.B)' = B'.A' = B.A, then A*(10^LB-1) = B*(10^LA-1) (LX = length of X) <=> A*R(LB) = B*R(LA), where R(n) = (10^n-1)/9. To have A.B prime we also must have gcd(A,B) = 1, thus A | R(LA) and B | R(LB). Such numbers are listed in A249647 (not A014950), the only palindromes there are of the form 1...1, 3...3 or 9...9. Thus the only palindromic terms in this sequence A273906 are the repunit primes A004022. - M. F. Hasler, Jun 10 2016

			The prime 1013 is a term since 101 and 3 are palindromic.
The prime 101 is not a term, since it is not a concatenation of two nonzero palindromic numbers.
The prime 131 is not a term because it is not a concatenation of two nonzero palindromic numbers.

David A. Corneth, Table of n, a(n) for n = 1..25912
David A. Corneth, PARI program

Cf. A000040, A002113, A004022, A014950, A096489, A249647, A273906.

Take[#, 62] &@ Select[Sort@ Map[FromDigits@ Flatten@ IntegerDigits@ # &, Tuples[#, 2]], PrimeQ] &@ Select[Range[10^3], Reverse@ # == # &@ IntegerDigits@ # &] (* Michael De Vlieger, Jun 03 2016 *)
nxtPal[n_]:=With[{c=Join[{2},Flatten[Table[{10*10^d,11*10^d},{d,0,10}]]]},SelectFirst[n+c,PalindromeQ]]; Take[Join[{11},Select[ #[[1]]*10^IntegerLength[ #[[2]]]+#[[2]]&/@ Flatten[{#,Reverse[#]}&/@Subsets[Join[Range[8],NestList[nxtPal,9,100]],{2}],1],PrimeQ]//Union],60] (* Harvey P. Dale, Dec 08 2024 *)

\\ See program link from David A. Corneth, Jun 10 2016.

a(n) = A096489(n+1), n=1..21. - R. J. Mathar, Jun 12 2016. (This is a pure accident, I think, since A096489 is a finite sequence. - N. J. A. Sloane, Jun 12 2016)

More terms from Michael De Vlieger, Jun 03 2016

A343682 Zuckerman numbers which when divided by the product of their digits, give a quotient which is a Niven (Harshad) number.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 24, 36, 111, 128, 135, 144, 175, 216, 315, 384, 432, 672, 735, 1296, 1575, 2916, 11115, 11232, 11664, 12132, 12288, 12312, 13212, 13824, 14112, 16416, 22176, 23112, 23328, 26112, 27216, 31212, 32832, 34272, 34992, 42624, 72128, 77175

Offset: 1

Bernard Schott, Apr 26 2021

Repunit R(k) is a term iff k divides R(k) (A014950).

			36 is a Zuckerman number as 36/(3*6) = 2, 2/2 = 1 that is a Niven number, and 36 is a term.
315 is a Zuckerman number as 315/(3*1*5) = 21, 21/(2+1) = 7 that is a Niven number, and 315 is a term.

Giovanni Resta, Zuckerman numbers, Numbers Aplenty.

Cf. A005349, A007602, A235507, A288069, A343680, A343681.

nivenQ[n_] := IntegerQ[n] && (sum = Plus @@ IntegerDigits[n]) > 0 && Divisible[n, sum]; Select[Range[10^5], (prod = Times @@ IntegerDigits[#]) > 0 && nivenQ[# / prod] &] (* Amiram Eldar, Apr 26 2021 *)

isn(n) = !(n%sumdigits(n)); \\ A005349
isz(n) = my(p=vecprod(digits(n))); p && !(n % p); \\ A007602
isok(n) = isz(n) && isn(n/vecprod(digits(n))); \\ Michel Marcus, Apr 26 2021

More terms from Michel Marcus, Apr 26 2021

A375590 Numbers m such that there exists a nonnegative integer k for which the concatenation of m, m-1, ..., m-k is an m-digit number.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 150, 153, 156, 159, 162, 165, 168

Offset: 1

Nicholas M. R. Frieler, Aug 19 2024

Let d be the number of digits of m (in base 10). One can show that the number m is a member of this sequence if and only if:
   (a) m is divisible by d and m >= 10^(d-1) * d/(d+1)
    or
   (b) 10^(d-1) == 1 (mod d-1) (i.e., d-1 is a term of A014950) and m < 10^d * d/(d+1).

			13 is a term because the concatenation of 13, 12, ..., 5 is a 13-digit number.
100 is not a term because the concatenation of 100, 99, ..., 52 is a 99-digit number and concatenating this number with 51 yields a 101-digit number.

Cf. A375461 (self-consecutive).

SelfDownwardConsecutiveQ[n_] :=
 Module[{len = Length@IntegerDigits[n], num, c = 1, numDigits = 0},
  numDigits = len*Ceiling[n + 1 - 10^(len - 1)];
If[numDigits >= n, Return[Mod[n, len] == 0]];
num = n - Ceiling[n + 1 - 10^(len - 1)];
While[numDigits < n + 1,
  If[(len - c)*Ceiling[num + 1 - 10^(len - c - 1)] >= n - numDigits,
   Return[Mod[n - numDigits, len - c] == 0],
   numDigits += (len - c)*Ceiling[num + 1 - 10^(len - c - 1)]
    ];
  num -= Ceiling[num + 1 - 10^(len - c - 1)];
  c++;
   ]
 ]
Select[Range[1000], SelfDownwardConsecutiveQ]

A334929 Positive integers k such that there exists a positive integer m consisting of k identical digits and such that m is a multiple of k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 22, 24, 27, 36, 42, 44, 45, 54, 63, 66, 72, 78, 81, 84, 88, 108, 111, 126, 132, 135, 156, 162, 168, 189, 198, 205, 216, 222, 234, 242, 243, 252, 264, 294, 312, 324, 333, 342, 378, 396, 404, 405, 444, 462, 465, 468, 484, 486

Offset: 1

Reiner Moewald, May 16 2020

For k=3^t, t>=1 you can always find numbers m.

			12 is a term since 444444444444 = 12*37037037037.

Cf. A014950.

ok[n_] := AnyTrue[(10^n - 1)/9 Range@9, Mod[#, n] == 0 &]; Select[ Range[486], ok] (* Giovanni Resta, May 24 2020 *)

t = "1"
list = [1]
for i in range(1, 1000):
    t = "1" + t
    m = int(t)
    weiter = 0
    for k in range(1, 10):
        if k * m % (i + 1) == 0:
            weiter = 1
    if weiter == 1:
        list.append(i + 1)
print(list)

A345467 Ratios R(k)/k for which R(k) / k is an integer, where R(k) = A002275(k) is a repunit.

Original entry on oeis.org

1, 37, 12345679, 4115226337448559670781893, 1371742112482853223593964334705075445816186556927297668038408779149519890260631

Offset: 1

Thomas T. Burgess, Jun 20 2021

This is the sequence where fractions of repunits (A002275) and their number of digits in base 10 R(k) / k are integers, where k is A014950(n). This happens for all k of the form k=3^m; this is true because R(3k) / R(k) = 10^(2k) + 10^n*k + 1, which is divisible by 3. Therefore R(3^m) is divisible by 3^m by induction on m. There are additional solutions in A014950.

			For n = 2, a(2) = 111/3 = 37. For n = 3, a(3) = 111111111/9 = 12345679.

Cf. A002275, A014950.

s = Join[{1}, Select[Range[3, 81, 6], PowerMod[10, #, #] == 1 &]]; Table[(10^n - 1)/(9*n), {n, s}] (* Amiram Eldar, Jun 20 2021 *)

[(10**n-1)//(9*n) for n in range(1, 300) if not (10**n-1)//9 % n]

a(n) = A002275(A014950(n))/A014950(n).

A377228 Repdigits which are also Harshad numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 111, 222, 333, 444, 555, 666, 777, 888, 999, 111111111, 222222222, 333333333, 444444444, 555555555, 666666666, 777777777, 888888888, 999999999, 111111111111111111111111111, 222222222222222222222222222, 333333333333333333333333333, 444444444444444444444444444

Offset: 1

John Bibby, Oct 20 2024

			999 is in the sequence as it is a repdigit (all digits are equal (in base 10)) and 999 is divisible by its sum of digits; 9 + 9 + 9 = 27 is a divisors of 999. - _David A. Corneth_, Oct 20 2024

Intersection of A005349 and A010785.

Cf. A014950.

Select[Union[FromDigits/@Flatten[Table[PadRight[{}, i, n], {n, 9}, {i, 27}], 1]],Divisible[#,DigitSum[#]]&] (* James C. McMahon, Jan 07 2025 *)

repd(n) = 10^((n+8)\9)\9*((n-1)%9+1); \\ A010785
lista(nn) = select(x->(x%sumdigits(x)==0), vector(nn, k, repd(k))); \\ Michel Marcus, Jan 07 2025

More terms from David A. Corneth, Oct 20 2024

cp's OEIS Frontend

A225782 Numbers such that every permutation of digits of n is divisible by sum of digits of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A273906 Primes equal to the concatenation of two nonzero palindromic numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A343682 Zuckerman numbers which when divided by the product of their digits, give a quotient which is a Niven (Harshad) number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A375590 Numbers m such that there exists a nonnegative integer k for which the concatenation of m, m-1, ..., m-k is an m-digit number.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A334929 Positive integers k such that there exists a positive integer m consisting of k identical digits and such that m is a multiple of k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

A345467 Ratios R(k)/k for which R(k) / k is an integer, where R(k) = A002275(k) is a repunit.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

A377228 Repdigits which are also Harshad numbers.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

PARI

Extensions