A355302 -id:A355302 - cp's OEIS frontend

A355303 a(n) is the smallest integer that has n normal undulating divisors.

1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 126, 60, 320, 144, 168, 120, 252, 180, 560, 240, 630, 420, 780, 360, 1890, 960, 1920, 720, 1560, 1080, 1260, 1440, 1680, 4368, 2160, 3240, 3120, 3360, 4320, 2520, 6300, 6120, 8640, 6240, 13104, 5040, 12480, 9360, 12240, 7560

Offset: 1

Bernard Schott, Jun 29 2022

Normal undulating numbers are in A355301.

The first ten terms are the same as A005179, then A005179(11) = 1024 while a(11) = 126 (see example); also, a(n) = A005179(n) for n = 12, 16, 18, 20, 24 (up to n = 50).

			16 has 5 divisors: {1, 2, 4, 8, 16}, all of which are normal undulating integers; no positive integer smaller than 16 has five normal undulating divisors, hence a(5) = 16.
126 has 12 divisors: {1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126}; only 126 is not normal undulating; no positive integer smaller than 126 has eleven normal undulating divisors, hence a(11) = 126.

David A. Corneth, Table of n, a(n) for n = 1..402

Cf. A005179, A355301, A355302, A355304.

nuQ[n_] := AllTrue[(s = Sign[Differences[IntegerDigits[n]]]), # != 0 &] && AllTrue[Differences[s], # != 0 &]; f[n_] := DivisorSum[n, 1 &, nuQ[#] &]; 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[50, 10^5] (* Amiram Eldar, Jun 29 2022 *)

isok(m) = if (m<10, return(1)); my(d=digits(m), dd = vector(#d-1, k, sign(d[k+1]-d[k]))); if (#select(x->(x==0), dd), return(0)); my(pdd = vector(#dd-1, k, dd[k+1]*dd[k])); #select(x->(x>0), pdd) == 0; \\ A355301
a(n) = my(k=1); while (sumdiv(k, d, isok(d)) != n, k++); k; \\ Michel Marcus, Jun 30 2022

Terms a(11) and beyond from Amiram Eldar, Jun 29 2022

A356018 a(n) is the number of evil divisors (A001969) of n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 0, 2, 2, 0, 3, 0, 0, 3, 0, 1, 4, 0, 3, 1, 0, 1, 4, 1, 0, 3, 0, 1, 6, 0, 0, 2, 2, 1, 6, 0, 0, 2, 4, 0, 2, 1, 0, 5, 2, 0, 5, 0, 2, 3, 0, 1, 6, 1, 0, 2, 2, 0, 9, 0, 0, 3, 0, 2, 4, 0, 3, 2, 2, 1, 8, 0, 0, 4, 0, 1, 4, 0, 5, 3, 0, 1, 3, 3, 2

Offset: 1

Bernard Schott, Jul 23 2022

a(n) = 0 iff n is in A093696.

			12 has 6 divisors: {1, 2, 3, 4, 6, 12} of which three {3, 6, 12} have an even number of 1's in their binary expansion with 11, 110 and 11100 respectively; hence a(12) = 3.

Cf. A000005, A001969, A093688, A093696 (location of 0s), A227872, A356019, A356020.

Similar sequences: A083230, A087990, A087991, A332268, A355302.

A356018 := proc(n)
    local a,d ;
    a := 0 ;
    for d in numtheory[divisors](n) do
        if isA001969(d) then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A356018(n),n=1..200) ;  # R. J. Mathar, Aug 07 2022

a[n_] := DivisorSum[n, 1 &, EvenQ[DigitCount[#, 2, 1]] &]; Array[a, 100] (* Amiram Eldar, Jul 23 2022 *)

a(n) = my(v = valuation(n, 2)); n>>=v; d=divisors(n); sum(i=1, #d, bitand(hammingweight(d[i]), 1) == 0) * (v+1) \\ David A. Corneth, Jul 23 2022

from sympy import divisors
def c(n): return bin(n).count("1")&1 == 0
def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Jul 23 2022

a(n) = A000005(n) - A227872(n).

More terms from David A. Corneth, Jul 23 2022

A355593 a(n) is the number of alternating integers that divide n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 1, 6, 1, 4, 3, 5, 1, 6, 1, 5, 4, 2, 2, 7, 3, 2, 4, 5, 2, 7, 1, 6, 2, 3, 3, 9, 1, 3, 2, 6, 2, 7, 2, 3, 5, 3, 2, 8, 3, 6, 2, 4, 1, 8, 2, 7, 2, 4, 1, 9, 2, 2, 6, 6, 3, 4, 2, 4, 4, 7, 1, 11, 1, 3, 4, 5, 2, 5, 1, 7, 5, 3, 2, 9, 3, 3, 4, 4, 2, 11, 2, 5, 2, 4, 2, 10, 1, 6, 3, 7

Offset: 1

Bernard Schott, Jul 08 2022

This sequence first differs from A355302 at index 13, where a(13) = 1 while A355302(13) = 2.

This sequence first differs from A332268 at index 14, where a(14) = 4 while A332268(14) = 3.

			40 has 8 divisors: {1, 2, 4, 5, 8, 10, 20, 40} of which 2 are not alternating integers: {20, 40}, hence a(40) = 8 - 2 = 6.

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

Cf. A030141 (alternating integers), A355594, A355595, A355596.

Similar to A332268 (with Niven numbers) and A355302 (with undulating integers).

Alt:= [$1..9, seq(seq(10*i+r - (i mod 2), r=[1,3,5,7,9]),i=1..9)]:
V:= Vector(100):
for t in Alt do J:= [seq(i,i=t..100,t)]; V[J]:= V[J] +~ 1 od:
convert(V,list); # Robert Israel, Nov 26 2023

q[n_] := !MemberQ[Differences[Mod[IntegerDigits[n], 2]], 0]; a[n_] := DivisorSum[n, 1 &, q[#] &]; Array[a, 120] (* Amiram Eldar, Jul 08 2022 *)

alternate(n,d=digits(n))=for(i=2,#d, if((d[i]-d[i-1])%2==0, return(0))); 1
a(n)=sumdiv(n,d,alternate(d)) \\ Charles R Greathouse IV, Jul 08 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 a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Jul 08 2022

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=2} 1/A030141(n) = 5.1... (the sums up to 10^10, 10^11 and 10^12 are 5.1704..., 5.1727... and 5.1738..., respectively). - Amiram Eldar, Jan 06 2024

A355301 Normal undulating numbers where "undulating" means that the alternate digits go up and down (or down and up) and "normal" means that the absolute differences between two adjacent digits may differ.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 101, 102, 103, 104, 105, 106, 107, 108, 109, 120, 121, 130, 131, 132, 140, 141, 142, 143, 150

Offset: 1

Bernard Schott, Jun 27 2022

This definition comes from Patrick De Geest's link.
Other definitions for undulating are present in the OEIS (e.g., A033619, A046075).
When the absolute differences between two adjacent digits are always equal (e.g., 85858), these numbers are called smoothly undulating numbers and form a subsequence (A046075).
The definition includes the trivial 1- and 2-digit undulating numbers.
Subsequence of A043096 where the first different term is A043096(103) = 123 while a(103) = 130.
This sequence first differs from A010784 at a(92) = 101, A010784(92) = 102.
The sequence differs from A160542 (which contains 100). - R. J. Mathar, Aug 05 2022

			111 is not a term here, but A033619(102) = 111.
a(93) = 102, but 102 is not a term of A046075.
Some terms: 5276, 918230, 1053837, 263915847, 3636363636363636.
Are not terms: 1331, 594571652, 824327182.

Patrick De Geest, Smoothly Undulating Palindromic Primes, World of Numbers.

Cf. A033619, A043096, A046075.
Cf. A059168 (subsequence of primes).
Differs from A010784, A241157, A241158.
Cf. A355302, A355303, A355304.

isA355301 := proc(n)
    local dgs,i,back,forw ;
    dgs := convert(n,base,10) ;
    if nops(dgs) < 2 then
        return true;
    end if;
    for i from 2 to nops(dgs)-1 do
        back := op(i,dgs) -op(i-1,dgs) ;
        forw := op(i+1,dgs) -op(i,dgs) ;
        if back*forw >= 0 then
            return false;
        end if ;
    end do:
    back := op(-1,dgs) -op(-2,dgs) ;
    if back = 0 then
        return false;
    end if ;
    return true ;
end proc:
A355301 := proc(n)
    option remember ;
    if n = 1 then
        0;
    else
        for a from procname(n-1)+1 do
            if isA355301(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A355301(n),n=1..110) ; # R. J. Mathar, Aug 05 2022

q[n_] := AllTrue[(s = Sign[Differences[IntegerDigits[n]]]), # != 0 &] && AllTrue[Differences[s], # != 0 &]; Select[Range[0, 100], q] (* Amiram Eldar, Jun 28 2022 *)

isok(m) = if (m<10, return(1)); my(d=digits(m), dd = vector(#d-1, k, sign(d[k+1]-d[k]))); if (#select(x->(x==0), dd), return(0)); my(pdd = vector(#dd-1, k, dd[k+1]*dd[k])); #select(x->(x>0), pdd) == 0; \\ Michel Marcus, Jun 30 2022

A355304 Integers whose number of normal undulating divisors sets a new record.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720, 1080, 1260, 1440, 1680, 2160, 2520, 5040, 7560, 10080, 15120, 21840, 28080, 32760, 56160, 65520, 98280, 131040, 196560, 393120, 589680, 786240, 1113840, 1670760, 2227680, 3341520, 6683040, 13366080, 20049120

Offset: 1

Bernard Schott, Jun 30 2022

Normal undulating integers are in A355301.
The first 14 terms are also the first 14 highly composite numbers in A002182, then A002182(15) = 840 while a(15) = 1080. Indeed, 840 is the smallest integer that has 32 divisors of which only 28 are normal undulating integers, while 1080 has also 32 divisors of which 30 are normal undulating integers.
Corresponding records of number of normal undulating divisors are 1, 2, 3, 4, 6, 8, 9, 10, 12, ...

			a(6) = 24 is in the sequence because A355302(24) is larger than any earlier value in A355302.

Cf. A002182, A355301, A355302, A355303.

Similar, but with divisors that are: A046952 (squares), A053624 (odd), A181808 (even), A093036 (palindromes), A340548 (repdigits), A340549 (repunits), A350756 (triangular).

nuQ[n_] := AllTrue[(s = Sign[Differences[IntegerDigits[n]]]), # != 0 &] && AllTrue[Differences[s], # != 0 &]; dm = -1; seq = {}; Do[If[(d = DivisorSum[n, 1 &, nuQ[#] &]) > dm, dm = d; AppendTo[seq, n]], {n, 1, 10^5}]; seq (* Amiram Eldar, Jun 30 2022 *)

More terms from Amiram Eldar, Jun 30 2022

A355770 a(n) is the number of terms of A333369 that divide n.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 3, 2, 1, 2, 2, 2, 4, 1, 2, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 2, 1, 4, 2, 1, 2, 2, 4, 3, 2, 2, 4, 2, 1, 3, 1, 3, 5, 1, 1, 2, 2, 2, 4, 2, 2, 3, 2, 2, 4, 1, 2, 4, 1, 2, 4, 1, 3, 4, 1, 2, 2, 4, 2, 3, 2, 2, 5, 2, 2, 4, 2, 2, 3, 1, 1, 3, 3, 1, 2

Offset: 1

Bernard Schott, Jul 16 2022

Cf. A333369, A353735, A355771, A355772, A355773.

Similar sequences: A083230, A087990, A087991, A332268, A355302.

q[n_] := AllTrue[Tally @ IntegerDigits[n], EvenQ[Plus @@ #] &]; a[n_] := DivisorSum[n, 1 &, q[#] &]; Array[a, 100] (* Amiram Eldar, Jul 16 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
a(n) = sumdiv(n, d, issimber(d)); \\ Michel Marcus, Jul 18 2022

from sympy import divisors
def c(n): s = str(n); return all(s.count(d)%2 == int(d)%2 for d in set(s))
def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Jul 16 2022

More terms from Michael S. Branicky, Jul 16 2022

A357171 a(n) is the number of divisors of n whose digits are in strictly increasing order (A009993).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 3, 1, 6, 2, 4, 4, 5, 2, 6, 2, 4, 3, 2, 2, 8, 3, 4, 4, 6, 2, 6, 1, 5, 2, 4, 4, 9, 2, 4, 4, 5, 1, 6, 1, 3, 6, 4, 2, 10, 3, 4, 3, 5, 1, 7, 2, 8, 4, 4, 2, 8, 1, 2, 4, 5, 3, 4, 2, 6, 4, 6, 1, 11, 1, 3, 5, 5, 2, 8, 2, 6, 4, 2, 1, 9, 3, 2, 3, 4, 2, 9, 3, 5, 2, 3, 3, 10, 1, 5, 3, 5

Offset: 1

Bernard Schott, Sep 16 2022

As A009993 is finite with 512 terms, a(n) is bounded with a(n) <= 511 and not 512, since A009993(1) = 0.

			22 has 4 divisors {1, 2, 11, 22} of which two have decimal digits that are not in strictly increasing order: {11, 22}, hence a(22) = 4-2 = 2.
52 has divisors {1, 2, 4, 13, 26, 52} and a(52) = 5 of them have decimal digits that are in strictly increasing order (all except 52 itself).

Cf. A009993, A357172, A357173, A160218.

Similar: A087990 (palindromic), A355302 (undulating), A355593 (alternating).

f:= proc(n) local d,L,i,t;
  t:= 0;
  for d in numtheory:-divisors(n) do
    L:= convert(d,base,10);
    if `and`(seq(L[i]>L[i+1],i=1..nops(L)-1)) then t:= t+1 fi
  od;
  t
end proc:
map(f, [$1..100]); # Robert Israel, Sep 16 2022

a[n_] := DivisorSum[n, 1 &, Less @@ IntegerDigits[#] &]; Array[a, 100] (* Amiram Eldar, Sep 16 2022 *)

isok(d) = Set(d=digits(d)) == d; \\ A009993
a(n) = sumdiv(n, d, isok(d)); \\ Michel Marcus, Sep 16 2022

from sympy import divisors
def c(n): s = str(n); return s == "".join(sorted(set(s)))
def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Sep 16 2022

G.f.: Sum_{n in A009993} x^n/(1-x^n). - Robert Israel, Sep 16 2022

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n=2..512} 1/A009993(n) = 4.47614714667538759358... (this is a rational number whose numerator and denominator have 1037 and 1036 digits, respectively). - Amiram Eldar, Jan 06 2024

A355698 a(n) is the number of repdigits divisors of n (A010785).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 3, 2, 5, 1, 3, 3, 4, 1, 5, 1, 4, 3, 4, 1, 6, 2, 2, 3, 4, 1, 5, 1, 4, 4, 2, 3, 6, 1, 2, 2, 5, 1, 5, 1, 6, 4, 2, 1, 6, 2, 3, 2, 3, 1, 5, 4, 5, 2, 2, 1, 6, 1, 2, 4, 4, 2, 8, 1, 3, 2, 4, 1, 7, 1, 2, 3, 3, 4, 4, 1, 5, 3, 2, 1, 6, 2, 2, 2, 8, 1, 6, 2, 3, 2, 2, 2, 6, 1, 3, 6, 4, 1, 4, 1, 4, 4

Offset: 1

Bernard Schott, Jul 14 2022

More than the usual number of terms are displayed in order to show the difference from A087990.
The first 100 terms are the same first 100 terms of A087990, then a(101) = 1 while A087990(101) = 2, because 101 is the smallest palindrome that is not repdigit; the next difference is 121.
Inequalities: 1 <= a(n) <= A087990(n).

			66 has 8 divisors: {1, 2, 3, 6, 11, 22, 33, 66} that are all repdigits, hence a(66) = 8.
121 has 3 divisors: {1, 11, 121} of which 2 are repdigits: {1, 11}, hence a(121) = 2.

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

Cf. A010785, A065444, A087990, A190217, A340548, A355699.

Similar sequences: A083230, A087990, A087991, A332268, A355302.

isrepdig:= proc(n) nops(convert(convert(n,base,10),set))=1 end proc:
f:= proc(n) nops(select(isrepdig, numtheory:-divisors(n))) end proc:
map(f, [$1..200]); # Robert Israel, Aug 07 2024

a[n_] := DivisorSum[n, 1 &, Length[Union[IntegerDigits[#]]] == 1 &]; Array[a, 100] (* Amiram Eldar, Jul 14 2022 *)

a(n) = my(ret=0,u=1); while(u<=n, ret+=sum(d=1,9, n%(u*d)==0); u=10*u+1); ret; \\ Kevin Ryde, Jul 14 2022

isrep(n) = {1==#Set(digits(n))}; \\ A010785
a(n) = sumdiv(n, d, isrep(d)); \\ Michel Marcus, Jul 15 2022

from sympy import divisors
def c(n): return len(set(str(n))) == 1
def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
print([a(n) for n in range(1, 105)]) # Michael S. Branicky, Jul 14 2022

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (7129/2520) * A065444 = 3.11446261209177581335... . - Amiram Eldar, Apr 17 2025

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A355303 a(n) is the smallest integer that has n normal undulating divisors.

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A356018 a(n) is the number of evil divisors (A001969) of n.

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A355593 a(n) is the number of alternating integers that divide n.

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A355301 Normal undulating numbers where "undulating" means that the alternate digits go up and down (or down and up) and "normal" means that the absolute differences between two adjacent digits may differ.

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A355304 Integers whose number of normal undulating divisors sets a new record.

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A355770 a(n) is the number of terms of A333369 that divide n.

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A357171 a(n) is the number of divisors of n whose digits are in strictly increasing order (A009993).

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