A068953 -id:A068953 - cp's OEIS frontend

A059972 a(n) is the least positive integer k such that all digits of k are 0 or 1 in exactly n different bases B, where 2 <= B <= k; i.e., such that A068953(k)=n.

2, 3, 4, 9, 30, 81, 4096, 531441, 16777216

Offset: 1

Naohiro Nomoto, Mar 28 2002

Is every term except 30 a power of either 2 or 3?

			For n=4: 9 written in bases 2 through 9 is 1001, 100, 21, 14, 13, 12, 11, 10. In 4 bases, namely 2, 3, 8 and 9, all digits are 0 or 1.

Cf. A068953.

f[1]=0; f[k_] := Length[Select[Rest[Union[Divisors[k], Divisors[k-1]]], Max@@IntegerDigits[k, # ]==1&]]; a[n_] := For[k=1, True, k++, If[f[k]==n, Return[k]]]

Edited by Dean Hickerson, Mar 31 2002

A043306 Sum of all digits in all base-b representations for n, for 2 <= b <= n.

Original entry on oeis.org

1, 3, 4, 8, 10, 16, 17, 21, 25, 35, 34, 46, 52, 60, 58, 74, 73, 91, 92, 104, 114, 136, 128, 144, 156, 168, 171, 199, 193, 223, 221, 241, 257, 281, 261, 297, 315, 339, 333, 373, 367, 409, 416, 430, 452, 498, 472, 508, 515, 547, 556, 608, 598, 638, 634, 670, 698, 756, 717, 777

Offset: 2

Clark Kimberling

			5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 2 + 3 + 2 + 1 = 8.

Alois P. Heinz, Table of n, a(n) for n = 2..1000
Robin Fissum, Digit sums and the number of prime factors of the factorial n!=1.2...n, Bachelor's project in BMAT, Norwegian University of Science and Technology, 2020.
Jan-Christoph Puchta and Jürgen Spilker, Altes und Neues zur Quersumme, Math. Semesterber., Vol. 49, No. 2 (2002), pp. 209-226; alternative link.
Vladimir Shevelev, Compact integers and factorials, Acta Arith., Vol. 126, No. 3 (2007), pp. 195-236 (see p. 205).

Cf. A014837, A043000, A068953, A191251, A191322, A191350, A309531.

Row sums of A240236.

Table[Sum[Total[First[RealDigits[n, i]]], {i, 2, n}], {n, 2, 80}] (* Carl Najafi, Aug 16 2011 *)

a(n) = sum(i=2, n, vecsum(digits(n, i))); \\ Michel Marcus, Jan 03 2017

a(n) = sum(b=2, n, sumdigits(n, b)); \\ Michel Marcus, Aug 18 2017

from sympy.ntheory.digits import digits
def a(n): return sum(sum(digits(n, b)[1:]) for b in range(2, n+1))
print([a(n) for n in range(2, 62)]) # Michael S. Branicky, Apr 04 2022

From Vladimir Shevelev, Jun 03 2011: (Start)
a(n) = (n-1)*n - Sum_{i=2..n} (i-1)*Sum_{r>=1} floor(n/i^r).
a(n) <= (n-1)^2*log(n+1)/log(n).
Problem: find a better upper estimate. (End)
From Amiram Eldar, Apr 16 2021: (Start)
a(n) = A014837(n) + 1.
a(n) ~ (1-Pi^2/12)*n^2 + O(n^(3/2)) (Fissum, 2020). (End)

A043000 Number of digits in all base-b representations of n, for 2 <= b <= n.

Original entry on oeis.org

2, 4, 7, 9, 11, 13, 16, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 54, 56, 59, 61, 63, 65, 67, 70, 72, 74, 76, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126

Offset: 2

Clark Kimberling

From A.H.M. Smeets, Dec 14 2019: (Start)
a(n)-a(n-1) >= 2 due to the fact that n = 10_n, so there is an increment of at least 2. If n can be written as a perfect power m^s, an additional +1 comes to it for the representation of n in each base m.
For instance, for n = 729 we have 729 = 3^6 = 9^3 = 27^2, so there is an additional increment of 3. For n = 1296 we have 1296 = 6^4 = 36^2, so there is an additional increment of 2. For n = 4096 we have 4096 = 2^12 = 4^6 = 8^4 = 16^3= 64^2, so there is an additional increment of 5. (End)

			5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 3+2+2+2 = 9.

A.H.M. Smeets, Table of n, a(n) for n = 2..20000
Vaclav Kotesovec, Plot of a(n)/(2*n) for n = 2..1000000

Cf. A001597, A043306, A068953, A191322, A253642, A255165.

[&+[Floor(Log(i,i*n)):k in [2..n]]:n in [1..70]]; // Marius A. Burtea, Nov 13 2019

A043000 := proc(n) add( nops(convert(n,base,b)),b=2..n) ; end proc: # R. J. Mathar, Jun 04 2011

Table[Total[IntegerLength[n,Range[2,n]]],{n,2,60}] (* Harvey P. Dale, Apr 23 2019 *)

a(n)=sum(b=2,n,#digits(n,b)) \\ Jeppe Stig Nielsen, Dec 14 2019

a(n)= n-1 +sum(b=2,n,logint(n,b)) \\ Jeppe Stig Nielsen, Dec 14 2019

a(n) = {2*n-2+sum(i=2, logint(n, 2), sqrtnint(n, i)-1)} \\ David A. Corneth, Dec 31 2019

first(n) = my(res = vector(n)); res[1] = 2; for(i = 2, n, inc = numdiv(gcd(factor(i+1)[,2]))+1; res[i] = res[i-1]+inc); res \\ David A. Corneth, Dec 31 2019

def count(n,b):
    c = 0
    while n > 0:
        n, c = n//b, c+1
    return c
n = 0
while n < 50:
    n = n+1
    a, b = 0, 1
    while b < n:
        b = b+1
        a = a + count(n,b)
    print(n,a) # A.H.M. Smeets, Dec 14 2019

a(n) = Sum_{i=2..n} floor(log_i(i*n)); a(n) ~ 2*n. - Vladimir Shevelev, Jun 03 2011 [corrected by Vaclav Kotesovec, Apr 05 2021]
a(n) = A070939(n) + A081604(n) + A110591(n) + ... + 1. - R. J. Mathar, Jun 04 2011
From Ridouane Oudra, Nov 13 2019: (Start)
a(n) = Sum_{i=1..n-1} floor(n^(1/i));
a(n) = n - 1 + Sum_{i=1..floor(log_2(n))} floor(n^(1/i) - 1);
a(n) = n - 1 + A255165(n). (End)
If n is in A001597 then a(A001597(m)) - a(A001597(m)-1) = 2 + A253642(m), otherwise a(n) - a(n-1) = 2. - A.H.M. Smeets, Dec 14 2019

A088323 Number of numbers b>1 such that n is a repunit in base b representation.

Original entry on oeis.org

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

Offset: 2

Reinhard Zumkeller, Nov 06 2003

is a(n) < 4 ?;
n>2: a(n) > 0 as n = (n-1)^1 + (n-1)^0.
a(A119598(n)) > 3; a(A053696(n)) > 2; a(A085104(n)) > 2. - Reinhard Zumkeller, Jan 22 2014

			a(31)=3: 31 = 2^4+2^3+2^2+2^1+2^0 = 5^2+5^1+5^0 = 30^1+30^0.

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000
Eric Weisstein's World of Mathematics, Repunit

Cf. A000225, A003462, A002275, A068953.

a088323 n = sum $ map (f n) [2 .. n-1] where
   f x b = if x == 0 then 1 else if d /= 1 then 0 else f x' b
                                 where (x',d) = divMod x b
-- Reinhard Zumkeller, Jan 22 2014

Example corrected by Reinhard Zumkeller, Jan 22 2014

A341434 a(n) is the number of bases 1 < b < n in which n is divisible by its product of digits.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 3, 2, 2, 1, 5, 2, 3, 4, 6, 1, 5, 1, 5, 4, 4, 1, 9, 2, 2, 4, 5, 1, 7, 3, 9, 4, 2, 3, 12, 1, 2, 3, 10, 1, 7, 2, 7, 7, 2, 1, 15, 2, 5, 3, 6, 1, 10, 3, 10, 4, 3, 1, 14, 1, 2, 7, 14, 3, 8, 1, 6, 3, 6, 1, 20, 2, 3, 8, 7, 3, 7, 1, 16, 7, 2, 1, 14

Offset: 1

Amiram Eldar, Feb 11 2021

			a(3) = 1 since 3 is divisible by its product of digits only in base 2: 3 = 11_2 and 1*1 | 3.
a(6) = 2 since 6 is divisible by its product of digits in 2 bases: in base 4, 6 = 12_4 and 1*2 | 6, and in base 5, 6 = 11_5 and 1*1 | 6.

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

Cf. A007602, A068953, A080221, A085104, A088323, A119598.

q[n_, b_] := (p = Times @@ IntegerDigits[n, b]) > 0 && Divisible[n, p]; a[n_] := Count[Range[2, n], _?(q[n, #] &)]; Array[a, 100]

a(n) = sum(b=2, n-1, my(x=vecprod(digits(n, b))); x && !(n%x)); \\ Michel Marcus, Feb 12 2021

a(n) > 0 for all numbers n > 2 since n in base b = n-1 is 11.
a(n) > 1 for all even numbers > 4 since n in base b = n-2 is 12. Similarly, a(n) > 1 for all composite numbers > 4 since if n = k*m, then n is divisible by its product of digits in bases n-m and n-k.
a(p) > 1 for primes p in A085104.
a(p) > 2 for primes p in A119598 (i.e., 31, 8191, ...).
a(n) >= A088323(n), with equality if n = 4 or if n is a prime.

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A059972 a(n) is the least positive integer k such that all digits of k are 0 or 1 in exactly n different bases B, where 2 <= B <= k; i.e., such that A068953(k)=n.

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A043306 Sum of all digits in all base-b representations for n, for 2 <= b <= n.

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A043000 Number of digits in all base-b representations of n, for 2 <= b <= n.

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A088323 Number of numbers b>1 such that n is a repunit in base b representation.

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A341434 a(n) is the number of bases 1 < b < n in which n is divisible by its product of digits.

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