A054055 -id:A054055 - cp's OEIS frontend

A083895 Number of divisors of n with largest digit = 8 (base 10).

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 1, 2, 1, 1, 1, 2, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Reinhard Zumkeller, May 08 2003

			n=72, 2 of the 12 divisors of 72 have largest digit =8: {8,18}, therefore a(72)=2.

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

Cf. A054055, A000005, A083903, A083888, A083889, A083890, A083891, A083892, A083893, A083894, A083896, A283611.

f:= proc(n) nops(select(t -> max(convert(t, base, 10))=d, numtheory:-divisors(n))) end proc:
d:= 8:
map(f, [$1..200]); # Robert Israel, Oct 06 2019

With[{k = 8}, Array[DivisorSum[#, 1 &, And[#[[k]] > 0, Total@ #[[k + 1 ;; 9]] == 0] &@ DigitCount[#] &] &, 105]] (* Michael De Vlieger, Oct 06 2019 *)

a(n) = A000005(n) - A083888(n) - A083889(n) - A083890(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083896(n) = A083903(n) - A083902(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A283611(k) = 11.62909500243165896645... . - Amiram Eldar, Jan 04 2024

A083888 Number of divisors of n with largest digit = 1 (base 10).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1

Offset: 1

Reinhard Zumkeller, May 08 2003

			n=110, 4 of the divisors of 110 {1,2,5,10,11,22,55,110} have largest digit =1: {1,10,11,110}, therefore a(110)=4.

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

Cf. A054055, A000005, A007088, A083889, A083890, A083891, A083892, A083893, A083894, A083895, A083896.

[#[d:d in Divisors(n) | Max(Intseq(d)) eq 1]: n in [1..110]]; // Marius A. Burtea, Oct 06 2019

f:= proc(n) nops(select(t -> max(convert(t,base,10))=d, numtheory:-divisors(n))) end proc:
d:= 1:
map(f, [$1..200]); # Robert Israel, Oct 06 2019

With[{k = 1}, Array[DivisorSum[#, 1 &, And[#[[k]] > 0, Total@ #[[k + 1 ;; 9]] == 0] &@ DigitCount[#] &] &, 105]] (* Michael De Vlieger, Oct 06 2019 *)

a(n) = A000005(n) - A083889(n) - A083890(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083895(n) - A083896(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A007088(k) = 1.23840561530559480971... . - Amiram Eldar, Jan 04 2024

A083889 Number of divisors of n with largest digit = 2 (base 10).

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 0, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 2, 0, 2, 0, 2, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 3, 0, 1, 1, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 2, 0, 2, 0, 1, 1

Offset: 1

Reinhard Zumkeller, May 08 2003

			n=120, 4 of the 16 divisors of 120 have largest digit =2: {2,12,20,120}, therefore a(120)=4.

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

Cf. A054055, A000005, A083897, A083888, A083890, A083891, A083892, A083893, A083894, A083895, A083896, A277964.

[#[d:d in Divisors(n) | Max(Intseq(d)) eq 2]: n in [1..120]]; // Marius A. Burtea, Oct 06 2019

f:= proc(n) nops(select(t -> max(convert(t, base, 10))=d, numtheory:-divisors(n))) end proc:
d:= 2:
map(f, [$1..200]); # Robert Israel, Oct 06 2019

With[{k = 2}, Array[DivisorSum[#, 1 &, And[#[[k]] > 0, Total@ #[[k + 1 ;; 9]] == 0] &@ DigitCount[#] &] &, 105]] (* Michael De Vlieger, Oct 06 2019 *)

a(n) = A000005(n) - A083888(n) - A083890(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083895(n) - A083896(n) = A083897(n) - A083888(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A277964(k) = 0.85636382912390578285... . - Amiram Eldar, Jan 04 2024

A083896 Number of divisors of n with largest digit = 9 (base 10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0

Offset: 1

Reinhard Zumkeller, May 08 2003

			n = 117, 2 of the 6 divisors of 117 have largest digit = 9: {9,39}, therefore a(117) = 2.

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

Cf. A054055, A000005, A083888, A083889, A083890, A083891, A083892, A083893, A083894, A083895.

Cf. A001620, A082838.

f:= proc(n) nops(select(t -> max(convert(t, base, 10))=d, numtheory:-divisors(n))) end proc:
d:= 9:
map(f, [$1..200]); # Robert Israel, Oct 06 2019

Table[Count[Divisors[n],?(Max[IntegerDigits[#]]==9&)],{n,110}] (* _Harvey P. Dale, Mar 15 2018 *)

a(n) = A000005(n) - A083888(n) - A083889(n) - A083890(n) - A083891(n) - A083892(n) - A083893(n) - A083894(n) - A083895(n) = A000005(n) - A083903(n).

Sum_{k=1..n} a(k) ~ n * (log(n) + c), where c = 2*A001620 - 1 - A082838 = -22.766245... . - Amiram Eldar, Apr 17 2025

A091049 a(n) = first term which reduces to an unchanging value in n steps via repeated interpretation of a(n) as a base b+1 number where b is the largest digit of a(n).

Original entry on oeis.org

1, 10, 15, 17, 18, 58, 72, 80, 88, 507, 683, 838, 1384, 1807, 3417, 12651, 18316, 41841, 80852, 132815, 388315, 1182482, 2202048, 6408851, 15438855, 34630248, 72141683, 332386516, 764388521, 1867287828, 5451218338, 24187765577, 68380483575, 215445843883, 677083325011

Offset: 0

Chuck Seggelin, Dec 15 2003, Jul 09 2008

There is no maximum number of steps and for any value of n, there MUST be a term a(n) that reduces in n steps. This is demonstrable as follows: take any term in the above sequence and convert it to base 2. The resulting value, if interpreted as a base 10 value will require one additional step to reduce. The resulting value may not be the FIRST value to resolve in that many steps, however, so it may not belong in this sequence.

			a(0) = 1 because 1 is the first term that reduces to an unchanging value in zero steps (i.e. 1 is already fully reduced.) a(1) = 10 because 10 reduces in one step (10 in base 2 is 2, 2 does not reduce further.) a(8) = 88 because 88 reduces in 8 steps: 88 --> 80 --> 72 --> 58 --> 53 --> 33 --> 15 --> 11 --> 3.

Bert Dobbelaere, Table of n, a(n) for n = 0..100
Bert Dobbelaere, Backtracking program (Python)
Chuck Seggelin, Interesting Base Conversions.

Cf. A054055 (largest digit of n) A068505 (n as base b+1 number where b=largest digit of n) A091047 (a(n) = the final value of n reached through repeated interpretation of n as a base b+1 number where b is the largest digit of n) A091048 (number of times n must be interpreted as a base b+1 number where b is the largest digit of n until an unchanging value is reached).

def A091049(n):
    k = 1
    while True:
        m1 = k
        for i in range(n+1):
            m2 = int(str(m1),1+max(int(d) for d in str(m1)))
            if m1 == m2:
                if i == n:
                    return k
                else:
                    break
            m1 = m2
        k += 1 # Chai Wah Wu, Jan 07 2015

a(30)-a(31) from Chai Wah Wu, Jan 14 2015

A277964 Numbers whose largest decimal digit is 2.

Original entry on oeis.org

2, 12, 20, 21, 22, 102, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1002, 1012, 1020, 1021, 1022, 1102, 1112, 1120, 1121, 1122, 1200, 1201, 1202, 1210, 1211, 1212, 1220, 1221, 1222, 2000, 2001, 2002, 2010, 2011, 2012, 2020, 2021, 2022

Offset: 1

Colin Barker, Nov 06 2016

Number of terms less than 10^n is 3^n-2^n, i.e., A001047(n). - Chai Wah Wu, Nov 06 2016 [extended by Felix Fröhlich, Nov 07 2016]

Numbers n such that A054055(n) = 2. - Felix Fröhlich, Nov 07 2016

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)

Cf. A007088, A277965, A277966.

Filtered([1..2100],n->Maximum(ListOfDigits(n))=2); # Muniru A Asiru, Mar 01 2019

N:= 6: # to get all terms of at most N digits
R:= 2: B:= {1}: C:= {1,2}:
for  d from 2 to N do B:= map(t -> (10*t,10*t+1),B);
C:= map(t -> (10*t,10*t+1,10*t+2),C);
R:= R, op(sort(convert(C minus B,list)))
od:
R; # Robert Israel, Nov 07 2016

A277964Q = Max[IntegerDigits[#]] == 2 &; Select[Range[2000], A277964Q] (* JungHwan Min, Nov 06 2016 *)

L=List(); for(n=1, 10000, if(vecmax(digits(n))==2, listput(L, n))); Vec(L)

A156979 Primes p such that 1 = abs(largest digit of p - sum of all the other digits of p).

Original entry on oeis.org

23, 43, 67, 89, 113, 131, 157, 179, 197, 199, 223, 241, 263, 269, 311, 313, 331, 337, 353, 359, 373, 379, 397, 421, 449, 461, 463, 571, 593, 607, 641, 643, 661, 683, 719, 733, 739, 751, 809, 827, 829, 863, 881, 919, 937, 953, 971, 991, 1013, 1031, 1033

Offset: 1

Juri-Stepan Gerasimov, Feb 20 2009

			If prime=197(1<7<9) and 1=abs(9-(1+7)), then 197=a(10). If prime=199(1<9=9) and 1=abs(9-(9+1)), then 199=a(11). If prime=223(2=2<3) and 1=abs(3-(2+2)), then 223=a(12), etc.

Harvey P. Dale, Table of n, a(n) for n = 1..5000

Cf. A000040, A040997.

From R. J. Mathar, Mar 18 2010: (Start)
A007953 := proc(n) local d ; add(d,d= convert(n,base,10)) ; end proc:
A054055 := proc(n) local d ; max(op(convert(n,base,10))) ; end proc:
isA156979 := proc(n) isprime(n) and abs(A007953(n)-2*A054055(n)) = 1 ; end proc:
for n from 1 to 1050 do if isA156979(n) then printf("%d,",n); end if; end do: (End)

ldodQ[n_]:=Module[{sidn=Sort[IntegerDigits[n]]},Abs[Total[Most[ sidn]]- Last[ sidn]] == 1]; Select[Prime[Range[200]],ldodQ] (* Harvey P. Dale, Nov 13 2013 *)

A209928 Largest digit of all divisors of n.

Original entry on oeis.org

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

Offset: 1

Jaroslav Krizek, Mar 20 2012

Also largest digit of concatenation of all divisors of n (A037278, A176558).
a(n) = 9 for almost all n. - Charles R Greathouse IV, Mar 20 2012
With an offset of 1 rather than 0, A016186 tells us how many integers among the first 10^n have 9s among their digits, and those numbers are therefore guaranteed to index a 9 in this sequence. More interesting of course are those numbers that don't have a 9 in their own digits but do have a 9 among the digits of their nontrivial divisors. - Alonso del Arte, Mar 23 2012

			a(12) = 6 because digit 6 is largest digit of all divisors of 12: (1, 2, 3, 4, 6, 12).

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A054055 (largest digit of n).

Flatten[Table[Take[Sort[Flatten[IntegerDigits[Divisors[n]]]], -1], {n, 100}]] (* Alonso del Arte, Mar 23 2012 *)

a(n)=my(t);fordiv(n, d, t=max(t, vecmax(eval(Vec(Str(d))))); if(t>8, return(t)));t \\Charles R Greathouse IV, Mar 20 2012

from sympy import divisors
def a(n): return int(max("".join(map(str, divisors(n)))))
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Feb 22 2021

A060418 Largest decimal digit in n-th prime.

Original entry on oeis.org

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

Offset: 1

Labos Elemer, Apr 05 2001

a(n) = A054055(A000040(n)) = A262410(A000040(n)). - Reinhard Zumkeller, Sep 25 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..78498, all primes < 10^6.

Cf. A054055.

Cf. A262401.

a060418 = a054055 . a000040  -- Reinhard Zumkeller, Sep 25 2015

Table[Max[IntegerDigits[Prime[w]]], {w, 1, 1000}]

a(n) = vecmax(digits(prime(n))); \\ Michel Marcus, Dec 26 2013

Offset corrected to 1 by Michel Marcus, Dec 26 2013

A083898 Number of divisors of n with largest digit <= 3 (base 10).

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, May 08 2003

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A054055, A000005, A007090, A083890, A083891, A083897, A083899.

Table[Count[Divisors[n],?(Total[Most[Drop[DigitCount[#],3]]]==0&)],{n,120}] (* _Harvey P. Dale, Dec 21 2018 *)

a(n) = A083897(n) + A083890(n) = A083899(n) - A083891(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A007090(k) = 2.93694402167748963905... . - Amiram Eldar, Jan 04 2024

cp's OEIS Frontend

A083895 Number of divisors of n with largest digit = 8 (base 10).

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A083888 Number of divisors of n with largest digit = 1 (base 10).

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Magma

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A083889 Number of divisors of n with largest digit = 2 (base 10).

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A083896 Number of divisors of n with largest digit = 9 (base 10).

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A091049 a(n) = first term which reduces to an unchanging value in n steps via repeated interpretation of a(n) as a base b+1 number where b is the largest digit of a(n).

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A277964 Numbers whose largest decimal digit is 2.

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A156979 Primes p such that 1 = abs(largest digit of p - sum of all the other digits of p).

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