A000866 -id:A000866 - cp's OEIS frontend

A004644 Duplicate of A000866.

1, 2, 4, 13, 31, 112, 224, 1003, 2011, 4022, 13044, 31143, 112341, 230232, 1011014

Offset: 0

dead

A368866 The smallest positive number such that 2^a(n) when written in base n contains adjacent equal digits.

2, 2, 4, 5, 6, 3, 6, 12, 16, 14, 11, 15, 8, 4, 8, 23, 16, 14, 16, 21, 9, 17, 20, 14, 30, 27, 16, 15, 10, 5, 10, 29, 48, 14, 46, 19, 18, 15, 32, 36, 27, 36, 18, 12, 56, 41, 37, 24, 58, 22, 26, 46, 58, 40, 29, 24, 36, 14, 20, 18, 12, 6, 12, 60, 62, 50, 49, 50, 20, 35, 36, 55, 61, 52, 53, 77

Offset: 2

Scott R. Shannon, Jan 08 2024

In the first 10000 terms the largest value is a(9031) = 1924, with a corresponding power of 2 of approximately 1.52*10^579.

			a(2) = 2 as 2^2 = 4 written in base 2 = 100_2 which contains adjacent 0's.
a(6) = 6 as 2^6 = 64 written in base 6 = 144_6 which contains adjacent 4's.
a(10) = 16 as 2^16 = 65536 written in base 10 = 65536_10 which contains adjacent 5's.

Scott R. Shannon, Table of n, a(n) for n = 2..10000

Cf. A000079, A171901, A011557, A004642, A004643, A000866.

f:= proc(n) local k,L;
  for k from 1 do
    L:= convert(2^k,base,n);
    if member(0, L[2..-1]-L[1..-2]) then return k fi
  od
end proc:
map(f, [$2..100]); # Robert Israel, Jan 09 2024

from itertools import count
from sympy.ntheory.factor_ import digits
def A368866(n):
    k = 1
    for m in count(1):
        k <<= 1
        s = digits(k,n)[1:]
        if any(s[i]==s[i+1] for i in range(len(s)-1)):
            return m # Chai Wah Wu, Jan 08 2024

A055223 One-fourth the digital sum of base 5 representations of 2^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 3, 3, 3, 2, 3, 4, 5, 5, 5, 4, 3, 5, 6, 7, 5, 8, 8, 6, 8, 10, 10, 8, 6, 7, 8, 8, 10, 7, 9, 9, 10, 11, 10, 9, 10, 9, 11, 11, 11, 11, 12, 13, 13, 12, 14, 10, 14, 17, 15, 13, 13, 12, 15, 14, 16, 15, 12, 14, 15, 15, 16, 15, 15, 15, 16, 13, 12, 16, 17, 14, 20, 20, 20

Offset: 2

Lekraj Beedassy, Jun 20 2000

			a(19) = 4 because 2^20 = 1048576 = 232023301 (to scale 5) and (2+3+2+0+2+3+3+0+1)/4 = 4.

Cf. A000866.

a[n_] := 1/4* Module[{aux = IntegerDigits[2^n, 5]},Sum[aux[[i]], {i, 1, Length[aux]}]]; Table[a[n], {n, 2, 83}] (* José María Grau Ribas, Feb 13 2010 *)

More terms from James Sellers, Jul 04 2000

A055476 Powers of ten written in base 5.

Original entry on oeis.org

1, 20, 400, 13000, 310000, 11200000, 224000000, 10030000000, 201100000000, 4022000000000, 130440000000000, 3114300000000000, 112341000000000000, 2302320000000000000, 101101400000000000000, 2022033000000000000000

Offset: 0

Henry Bottomley, Jun 27 2000

The leading numbers free of the trailing end 0's in the entries of sequence a(n) are the corresponding powers of 2 written in base 5, i.e., A000866(n). - Lekraj Beedassy, Oct 26 2010

The first formula follows from the fact that the quinary representation of 10^n - 1 is equal to the concatenation of the quinary representation of 2^n - 1 with four times the n-th repunit; so the successor 10^n is the concatenation of 2^n with n zeros. See the Regan link. - Washington Bomfim, Dec 24 2010

Rick Regan, Nines in quinary

Cf. A000468, A011557.

FromDigits[IntegerDigits[#,5]]&/@(10^Range[0,20]) (* Harvey P. Dale, Feb 03 2019 *)

a(n) = A000866(n) followed by n zeros.

More terms from James Sellers, Jul 04 2000

A364049 a(n) is the least k such that the base-n digits of 2^k are not all distinct.

Original entry on oeis.org

2, 2, 4, 5, 6, 3, 6, 11, 16, 14, 11, 12, 8, 4, 8, 15, 16, 12, 16, 18, 9, 17, 15, 14, 24, 13, 16, 15, 10, 5, 10, 19, 24, 14, 21, 15, 18, 15, 19, 17, 17, 28, 18, 12, 24, 23, 31, 24, 31, 20, 26, 44, 35, 33, 25, 18, 36, 14, 14, 18, 12, 6, 12, 23, 45, 37, 38, 24, 20, 35, 36, 26, 51, 31, 33, 47, 34, 34

Offset: 2

Robert Israel, Jul 03 2023

			a(10) = 16 because 2^16 = 65536 does not have all distinct digits in base 10, while 2^k does have all distinct digits for 1 <= k <= 15.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A004642, A004643, A000866, A004645, A004646, A004647, A001357, A000079, A011557, A364052, A364089.

f:= proc(n) local k,L;
  for k from 2 do
    L:= convert(2^k,base,n);
    if nops(L) <> nops(convert(L,set)) then return k fi
  od;
end proc:
map(f, [$2..100]);

from itertools import count
from sympy.ntheory import digits
def a(n): return next(k for k in count(2) if len(set(d:=digits(1<Michael S. Branicky, Jul 05 2023

A364089 a(n) is the greatest k such that the base-n digits of 2^k are all distinct.

Original entry on oeis.org

1, 1, 3, 4, 5, 8, 5, 10, 29, 19, 19, 19, 16, 18, 7, 43, 41, 37, 45, 39, 55, 33, 43, 60, 35, 61, 56, 50, 44, 69, 9, 64, 44, 80, 43, 88, 53, 71, 56, 68, 59, 78, 76, 74, 95, 109, 111, 81, 86, 136, 117, 75, 98, 83, 84, 99, 104, 116, 95, 118, 60, 81, 11, 119, 119, 172, 140, 97, 105, 113, 93, 122, 92

Offset: 2

Robert Israel, Jul 04 2023

a(n) <= log_2(A062813(n)).

			a(10) = 29 because all decimal digits of 2^29 = 536870912 are distinct.

Cf. A004642, A004643, A000866, A004645, A004646, A004647, A001357, A000079, A364049.

f:= proc(b) local M,k,L;
  M:= b^b - (b^b-b)/(b-1)^2;
  for k from ilog2(M) to 1 by -1 do
    L:= convert(2^k,base,b);
    if nops(L) = nops(convert(L,set)) then return k fi
  od
end proc:
map(f, [$2..100]);

from sympy.ntheory.factor_ import digits
def A364089(n):
    m = 1<<(l:=((r:=n**n)-(r-n)//(n-1)**2).bit_length()-1)
    while len(d:=digits(m,n)[1:]) > len(set(d)):
        l -= 1
        m >>= 1
    return l # Chai Wah Wu, Jul 07 2023

cp's OEIS Frontend

A004644 Duplicate of A000866.

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A368866 The smallest positive number such that 2^a(n) when written in base n contains adjacent equal digits.

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A055223 One-fourth the digital sum of base 5 representations of 2^n.

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A055476 Powers of ten written in base 5.

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A364049 a(n) is the least k such that the base-n digits of 2^k are not all distinct.

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Maple

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A364089 a(n) is the greatest k such that the base-n digits of 2^k are all distinct.

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Author

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Comments

Examples

Crossrefs

Programs

Maple

Python