A190217 -id:A190217 - cp's OEIS frontend

A093696 Numbers n such that all divisors of n have an odd number of 1's in their binary expansions.

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 22, 26, 28, 31, 32, 37, 38, 41, 44, 47, 49, 52, 56, 59, 61, 62, 64, 67, 73, 74, 76, 79, 82, 88, 91, 94, 97, 98, 103, 104, 107, 109, 112, 118, 121, 122, 124, 127, 128, 131, 133, 134, 137, 143, 146, 148, 151, 152, 157, 158, 164, 167, 173

Offset: 1

Jason Earls, May 16 2004

Subsequence of A000069. - Michel Marcus, Feb 09 2014

Numbers all of whose divisors are odious. - Bernard Schott, Jul 22 2022

			14 is in the sequence because its divisors are [1, 2, 7, 14] and in binary: 1, 10, 111 and 1110, all have an odd number of 1's.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000069, A001969, A227872, A330289, A355968, A355969.
Similar sequences: A062687, A190217, A337741, A337941, A355596.
A000079 is a subsequence.

isA001969 := proc(n)
    if wt(n) mod 2 = 0 then
        true;
    else
        false;
    end if;
end proc:
isA093696 := proc(n)
    for d in numtheory[divisors](n) do
        if isA001969(d) then
            return false;
        end if;
    end do;
    true;
end proc:
for n from 1 to 200 do
    if isA093696(n) then
        printf("%d,",n);
    end if;
end do: # R. J. Mathar, Feb 13 2014

odiousQ[n_] := OddQ @ DigitCount[n, 2][[1]]; Select[Range[200], AllTrue[ Divisors[#], odiousQ ] &] (* Amiram Eldar, Dec 09 2019 *)

is(n)=fordiv(n,d,if(hammingweight(d)%2==0, return(0))); 1 \\ Charles R Greathouse IV, Mar 29 2013

from sympy import divisors, isprime
def c(n): return bin(n).count("1")&1
def ok(n): return n > 0 and all(c(d) for d in divisors(n, generator=True))
print([k for k in range(174) if ok(k)]) # Michael S. Branicky, Jul 24 2022

{n: A356018(n) =0 }. - R. J. Mathar, Aug 07 2022

A340548 Integers whose number of repdigit divisors sets a new record.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 66, 132, 264, 792, 3960, 14652, 26664, 29304, 79992, 146520, 399960, 1025640, 2666664, 7999992, 13333320, 39999960, 269333064, 807999192, 1346665320, 4039995960, 28279971720, 7999999999992, 8080799919192, 13333333333320, 13467999865320, 39999999999960, 40403999595960

Offset: 1

Bernard Schott, Jan 11 2021

The first 10 terms are the same as A093036, then A093036(11) = 1848 while a(11) = 3960, because from a(1) to a(10), all palindromic divisors are also repdigits, and then 616 is a non-repdigit palindromic divisor of 1848.
Number of repdigit divisors: 1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 17, 18, ...
Indices of repdigits: 1, 2, 3, 4, 7, ...

			132 has 12 divisors: {1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132} of which 10 are repdigits: {1, 2, 3, 4, 6, 11, 22, 33, 44, 66}. No positive integer smaller than 132 has as many as ten repdigit divisors; hence 132 is a term.

David A. Corneth, Table of n, a(n) for n = 1..53 (terms <= 10^30).

Cf. A010785, A190217.

Similar for: A053624 (odd), A181808 (even), A093036 (palindromes), A340549 (repunits).

repQ[n_] := Length @ Union @ IntegerDigits[n] == 1; s[n_] := DivisorSum[n, 1 &, repQ[#] &]; smax =  0; seq = {}; Do[s1 = s[n]; If[s1 > smax, smax = s1; AppendTo[seq, n]], {n, 1, 10^5}]; seq (* Amiram Eldar, Jan 11 2021 *)

isrd(n) = {1 == #Set(digits(n))}; \\ A010785
f(n) = sumdiv(n, d, isrd(d));
lista(nn) = {my(m = 0); for (n=1, nn, my(x = f(n)); if (x > m, print1(n, ", "); m = x););} \\ Michel Marcus, Jan 11 2021

a(16)-a(20) from Michel Marcus, Jan 11 2021
a(21)-a(26) from Amiram Eldar, Jan 12 2021
a(27) from Chai Wah Wu, Jan 14 2021
More terms from David A. Corneth, Jan 15 2021

A308851 Numbers >= 2 all of whose divisors > 1 are Brazilian.

Original entry on oeis.org

7, 13, 31, 43, 73, 91, 127, 157, 211, 217, 241, 301, 307, 403, 421, 463, 511, 559, 601, 757, 889, 949, 1093, 1099, 1123, 1333, 1477, 1483, 1651, 1687, 1723, 2041, 2149, 2263, 2551, 2743, 2801, 2821, 2947, 2971, 3133, 3139, 3241, 3307, 3541, 3907, 3913, 3937

Offset: 1

Bernard Schott, Jun 28 2019

nonn

The terms of this sequence are the Brazilian primes and the products of two or more distinct Brazilian primes.

There are no even numbers because 2 is not Brazilian.

			91 is a term because all divisors of 91 that are > 1: {7, 13, 91} are Brazilian numbers with 7 = 111_2, 13 = 111_3 and 91 = 77_12.

Amiram Eldar, Table of n, a(n) for n = 1..1000
Index entries for sequences related to Brazilian numbers

Cf. A085104 (subsequence), A125134.

Similar with even numbers: A000079, with odd numbers: A005408, with palindromes: A062687, with repdigits: A190217.

brazQ[n_] := Block[{k, b, ok}, If[FindInstance[k (1 + b) == n && 1 < b < n - 1 && 0 < k < b, {k, b}, Integers] != {}, True, b = 2; ok = False; While[1 + b + b^2 <= n && ! ok, ok = Length@ Union@ IntegerDigits[n, b++] == 1]; ok]]; Select[ Range[3, 4000, 2], AllTrue[ Rest@ Divisors@ #, brazQ] &] (* Giovanni Resta, Jun 29 2019 *)
max = 5000; fQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; A125134 = Select[Range[4, max], fQ]; Select[Range[2, max], Intersection[A125134, Rest[Divisors[#]]] == Rest[Divisors[#]] &] (* Vaclav Kotesovec, Jun 29 2019, using a subroutine from T. D. Noe *)

isb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1)));
isok(n) = {fordiv(n, d, if ((d>1) && ! isb(d), return (0));); return (1);} \\ Michel Marcus, Jun 29 2019

A337741 Numbers all of whose divisors are Niven numbers (A005349).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 36, 40, 54, 63, 72, 81, 108, 162, 216, 243, 324, 486, 648, 972, 1944

Offset: 1

Amiram Eldar, Sep 17 2020

Since the only prime Niven numbers are the single-digit primes 2, 3, 5 and 7, all the terms are 7-smooth numbers (A002473).

If k is a term, all the divisors of k are also terms. Since all the terms are 7-smooth, every term is of the form p * k, where p is in {2, 3, 5, 7} and k is a smaller term. Thus it is easy to verify that there are only 31 terms in this sequence, and 1944 being the last term.

			6 is a term since all the divisors of 6, i.e., 1, 2, 3 and 6, are Niven numbers.

Subsequence of A002473 and A005349.
Cf. A332268, A335708, A335709.
Similar sequences: A062687, A190217, A329419.

nivenQ[n_] := Divisible[n, Plus @@ IntegerDigits[n]]; allQ[n_] := AllTrue[Divisors[n], nivenQ]; p = {1, 2, 3, 5, 7}; s = {1}; n = 0; While[Length[s] != n, n = Length[s]; s = Select[Union @ Flatten @ Outer[Times, s, p], allQ]]; s

A355699 a(n) is the smallest number that has exactly n repdigit divisors.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 72, 66, 666, 132, 1332, 264, 2664, 792, 13320, 3960, 14652, 26664, 48840, 29304, 79992, 341880, 146520, 399960, 1333332, 1025640, 2799720, 8879112, 2666664, 18666648, 7999992, 44395560, 13333320, 93333240, 39999960, 279999720, 269333064

Offset: 1

Bernard Schott, Jul 14 2022

			72 has 12 divisors: {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72}, only {1, 2, 3, 4, 6, 8, 9} are repdigits; no positive integer smaller than 72 has seven repdigit divisors, hence a(7) = 72.

David A. Corneth, Table of n, a(n) for n = 1..135
David A. Corneth, PARI program

Cf. A010785, A087990, A190217, A340548, A355698.

Similar sequences: A087997, A333456, A355303, A355695.

f[n_] := DivisorSum[n, 1 &, Length[Union[IntegerDigits[#]]] == 1 &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[24, 10^6] (* Amiram Eldar, Jul 15 2022 *)

isrep(n) = 1==#Set(digits(n)); \\ A010785
a(n) = my(k=1); while (sumdiv(k, d, isrep(d)) != n, k++); k; \\ Michel Marcus, Jul 15 2022

\\ See PARI link. - David A. Corneth, Jul 26 2022

from sympy import divisors
from itertools import count, islice
def c(n): return len(set(str(n))) == 1
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
    n, adict = 1, dict()
    for k in count(1):
        fk = f(k)
        if fk not in adict: adict[fk] = k
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 21))) # Michael S. Branicky, Jul 26 2022

a(9)-a(35) from Michael S. Branicky, Jul 14 2022

a(36)-a(37) from Michael S. Branicky, Jul 15 2022

A355773 Numbers all of whose divisors are members of A333369.

Original entry on oeis.org

1, 3, 5, 7, 9, 13, 15, 17, 19, 31, 35, 37, 39, 51, 53, 57, 59, 71, 73, 79, 91, 93, 95, 97, 111, 137, 139, 153, 157, 159, 173, 179, 193, 197, 221, 223, 227, 229, 317, 333, 359, 371, 379, 395, 397, 443, 449, 519, 537, 571, 579, 591, 593, 661, 663, 669, 719, 739

Offset: 1

Bernard Schott, Jul 18 2022

All terms are necessarily odd because 2 is not in A333369

			111 is a term since all the divisors of 111, i.e., 1, 3, 37 and 111, are in A333369.

Michael S. Branicky, Table of n, a(n) for n = 1..10000
Project Euler, Problem 520: Simbers.

Cf. A333369, A353735, A355770, A355771, A355772.
Similar sequences: A062687, A190217, A329419, A337741
.
Subsequences: A155045, A355853.

simQ[n_] := AllTrue[Tally @ IntegerDigits[n], EvenQ[Plus @@ #] &]; Select[Range[1000], AllTrue[Divisors[#], simQ] &] (* Amiram Eldar, Jul 19 2022 *)

issimber(m) = my(d=digits(m), s=Set(d)); for (i=1, #s, if (#select(x->(x==s[i]), d) % 2 != (s[i] % 2), return (0))); return (1); \\ A333369
isok(k) = fordiv(k, d, if (!issimber(d), return(0))); return(1); \\ Michel Marcus, Jul 19 2022

from sympy import divisors, isprime
def c(n): s = str(n); return all(s.count(d)%2 == int(d)%2 for d in set(s))
def ok(n): return n > 0 and all(c(d) for d in divisors(n, generator=True))
print([k for k in range(740) if ok(k)]) # Michael S. Branicky, Jul 24 2022

A337941 Numbers whose divisors are all Zuckerman numbers (A007602).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 24, 1111111111111111111, 11111111111111111111111

Offset: 1

Bernard Schott, Oct 01 2020

Inspired by A337741.
Zuckerman numbers are numbers that are divisible by the product of their digits (see link).
The next term is the repunit prime R_317 which is too large to include in the data.
Primes in this sequence are 2, 3, 5, 7 and all the repunit primes (see A004023).
This sequence is infinite if and only if there are infinitely many repunit primes.

			6 is a term since all the divisors of 6, i.e., 1, 2, 3 and 6, are Zuckerman numbers.

Giovanni Resta, Zuckerman numbers, Numbers Aplenty

Subsequence of A007602.
Similar sequences: A062687, A190217, A308851, A329419, A337741.
Cf. A004023, A335037, A335038.
Cf. A004022 (subsequence of prime repunits).

zuckQ[n_] := (prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod]; Select[Range[24], AllTrue[Divisors[#], zuckQ] &] (* Amiram Eldar, Oct 01 2020 *)

isok(m) = {fordiv(m, d, my(p=vecprod(digits(d))); if (!p || (d % p), return (0))); return (1);} \\ Michel Marcus, Oct 05 2020

A190220 Numbers all of whose divisors are numbers whose decimal digits are in nonincreasing order.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 21, 22, 31, 33, 40, 41, 43, 44, 53, 55, 61, 62, 63, 66, 71, 73, 77, 82, 83, 86, 88, 93, 97, 99, 110, 211, 220, 311, 331, 421, 422, 431, 433, 440, 443, 511, 521, 541, 622, 631, 633, 641, 643, 653, 661, 662, 733, 743, 751

Offset: 1

Jaroslav Krizek, May 06 2011

Subset of A009996. Superset of A028867, A190219 and A190217.

			Number 110 is in sequence because all divisors of 110 (1, 2, 5, 10, 11, 22, 55, 110) are numbers whose decimal digits are in nonincreasing order.

Nathaniel Johnston, Table of n, a(n) for n = 1..500

with(numtheory): A190220 := proc(n) option remember: local d, dd, i, j, k, m, poten: if(n=1)then return 1: fi: for k from procname(n-1)+1 do d:=divisors(k): poten:=1: for i from 1 to nops(d) do m:=-1: dd:=convert(d[i], base, 10): for j from 1 to nops(dd) do if(m<=dd[j])then m:=dd[j]: else poten:=0: break: fi: od: if(poten=0)then break:fi: od: if(poten=1)then return k: fi: od: end: seq(A190220(n), n=1..64); # Nathaniel Johnston, May 14 2011

A355596 Numbers all of whose divisors are alternating numbers (A030141).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 21, 23, 25, 27, 29, 32, 36, 41, 43, 47, 49, 50, 54, 58, 61, 63, 67, 69, 81, 83, 87, 89, 94, 98, 101, 103, 107, 109, 123, 125, 127, 129, 141, 145, 147, 149, 161, 163, 167, 181, 183, 189, 214, 218, 250, 254, 290, 298

Offset: 1

Bernard Schott, Jul 12 2022

The smallest alternating number that is not a term is 30, because of 15.

			32 is a term since all the divisors of 32, i.e., 1, 2, 4, 8, 16 and 32, are alternating numbers

Subsequence of A030141.
Cf. A355593, A355594, A355595.
Similar sequences: A062687, A190217, A329419, A337941.

q[n_] := AllTrue[Divisors[n], !MemberQ[Differences[Mod[IntegerDigits[#], 2]], 0] &]; Select[Range[300], q] (* Amiram Eldar, Jul 12 2022 *)

isokd(n, d=digits(n))=for(i=2, #d, if((d[i]-d[i-1])%2==0, return(0))); 1; \\ A030141
isok(m) = sumdiv(m, d, isokd(d)) == numdiv(m); \\ Michel Marcus, Jul 12 2022

from sympy import divisors
def p(d): return 0 if d in "02468" else 1
def c(n):
    if n < 10: return True
    s = str(n)
    return all(p(s[i]) != p(s[i+1]) for i in range(len(s)-1))
def ok(n):
    return c(n) and all(c(d) for d in divisors(n, generator=True))
print([k for k in range(1, 200) if ok(k)]) # Michael S. Branicky, Jul 12 2022

a(51) and beyond from Michael S. Branicky, Jul 12 2022

A190221 Numbers all of whose divisors are numbers whose decimal digits are in nondecreasing order.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 125

Offset: 1

Jaroslav Krizek, May 06 2011

Subset of A009994. Superset of A028864, A190218 and A190217.

			Number 112 is in sequence because all divisors of 112 (1, 2, 4, 7, 8, 14, 16, 28, 56, 112) are numbers whose decimal digits are in nondecreasing order.

Nathaniel Johnston, Table of n, a(n) for n = 1..1000

with(numtheory): A190221 := proc(n) option remember: local d, dd, i, j, k, m, poten: if(n=1)then return 1: fi: for k from procname(n-1)+1 do d:=divisors(k): poten:=1: for i from 1 to nops(d) do m:=10: dd:=convert(d[i], base, 10): for j from 1 to nops(dd) do if(m>=dd[j])then m:=dd[j]: else poten:=0: break: fi: od: if(poten=0)then break:fi: od: if(poten=1)then return k: fi: od: end: seq(A190221(n), n=1..64); # Nathaniel Johnston, May 06 2011

ndoQ[n_]:=Min[Differences[IntegerDigits[n]]]>=0; Select[Range[ 200],AllTrue[ Divisors[#],ndoQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 02 2021 *)

cp's OEIS Frontend

A093696 Numbers n such that all divisors of n have an odd number of 1's in their binary expansions.

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A340548 Integers whose number of repdigit divisors sets a new record.

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A308851 Numbers >= 2 all of whose divisors > 1 are Brazilian.

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A337741 Numbers all of whose divisors are Niven numbers (A005349).

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A355699 a(n) is the smallest number that has exactly n repdigit divisors.

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A355773 Numbers all of whose divisors are members of A333369.

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Mathematica

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A337941 Numbers whose divisors are all Zuckerman numbers (A007602).

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