A146025
Numbers that can be written in bases 2, 3, 4, and 5 using only the digits 0 and 1.
Original entry on oeis.org
82000 = 10100000001010000 (2) = 11011111001 (3) = 110001100 (4) = 10111000 (5).
- Stuart A. Burrell and Han Yu, Digit expansions of numbers in different bases, arXiv:1905.00832 [math.NT], 2019.
- Daniel Glasscock, Joel Moreira, and Florian K. Richter, Additive transversality of fractal sets in the reals and the integers, arXiv:2007.05480 [math.NT], 2020. See p. 5.
- James Grime and Brady Haran, Why 82,000 is an extraordinary number, Numberphile video (2015)
- Alex P. Klinkhamer, Digits of 82000, search algorithm with code and analysis.
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f[n_] := Total[Total@ Drop[RotateRight[DigitCount[n, #]], 2] & /@ Range[3, 5]]; Select[Range[0, 100000], f@ # == 0 &] (* Michael De Vlieger, Aug 29 2015 *)
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is(n)=vecmax(digits(n,5))<2 && vecmax(digits(n,4))<2 && vecmax(digits(n,3))<2 \\ Charles R Greathouse IV, Aug 31 2015
Removed keywords "fini" and "full", since it is only a conjecture that there are no further terms. -
N. J. A. Sloane, Feb 06 2016
A258981
Numbers containing only 1's and 0's in their base-2, base-3, and base-4 representations.
Original entry on oeis.org
0, 1, 4, 81, 84, 85, 256, 273, 324, 325, 336, 337, 1089, 1092, 1093, 20496, 20497, 20736, 20737, 20740, 65620, 65856, 65857, 81921, 81984, 81985, 82000, 86032, 86277, 86292, 86293, 86356, 262468, 262480, 263169
Offset: 1
81 is 10000 in base 3 and 1101 in base 4 so 81 is a term.
273 is 101010 in base 3 and 10101 in base 4 so 273 is a term.
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N:= 20: # to get all terms < 2*4^(N-1)
g:= proc(n)
local L, j, m, a;
L:= convert(n,base, 2);
a:= add(4^(j-1)*L[j],j=1..nops(L));
if has(convert(a,base,3),2) then NULL else a fi
end proc:
map(g, [$0..2^N]); # Robert Israel, Jul 14 2015
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ok3[n_] := 1 == Max@ IntegerDigits[n, 3]; to4[n_] := FromDigits[ IntegerDigits[n, 2], 4]; Select[to4/@ Range[2^20], ok3] (* Giovanni Resta, Jun 16 2015 *)
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digitsb(m)=vecsort(concat(digits(m,3),digits(m,4)),,8)
is_ok(n)={my(v=digitsb(n),r=0, i);for(i=2,9,r = r || vecsearch(v,i));!r}
first(m)={ my(v=vector(m),i,k=0);for(i=1, m, while(!is_ok(k), k++); v[i] = k;k++); v;} /* Anders Hellström, Jul 19 2015 */
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isok(n) = (n==0) || ((vecmax(digits(n,3)) < 2) && (vecmax(digits(n,4)) < 2)); \\ Michel Marcus, Aug 05 2015
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print1(0);for(n=1,1e5,vecmax(digits(t=subst(Pol(binary(n)),'x,4),3))<2&&print1(","t)) \\ M. F. Hasler, Feb 01 2016
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\\ See links too.
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def digits(n, b=10): # digits of n in base 2 <= b <= 62
x, y = n, ''
while x >= b:
x, r = divmod(x,b)
y += str(r) if r < 10 else (chr(r+87) if r < 36 else chr(r+29))
y += str(x) if x < 10 else (chr(x+87) if x < 36 else chr(x+29))
return y[::-1]
A258981_list = [n for n in (int(format(d,'b'),4) for d in range(10**4)) if max(digits(n,3)) <= '1'] # Chai Wah Wu, Aug 13 2015
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[0]+[n for n in [1..1000000] if max(n.digits(base=3))==1 and max(n.digits(base=4))==1] # Tom Edgar, Jul 11 2015
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