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A064438
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Numbers which are divisible by the sum of their quaternary digits.
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22
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1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 20, 21, 24, 28, 30, 32, 33, 35, 36, 40, 42, 48, 50, 52, 54, 60, 63, 64, 66, 68, 69, 72, 76, 78, 80, 81, 84, 88, 90, 91, 96, 100, 102, 108, 112, 114, 120, 126, 128, 129, 132, 136, 138, 140, 144, 148, 150, 154, 156, 160, 162, 168, 171, 180
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OFFSET
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1,2
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COMMENTS
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A good "puzzle" sequence -- guess the rule given the first twenty or so terms.
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LINKS
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EXAMPLE
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Quaternary representation of 28 is 130, 1 + 3 + 0 = 4 divides 28.
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MATHEMATICA
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Select[Range[200], Divisible[#, Total[IntegerDigits[#, 4]]]&] (* Harvey P. Dale, Jun 09 2011 *)
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PROG
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(ARIBAS): maxarg := 190; for n := 1 to maxarg do if n mod sum(quaternarray(n)) = 0 then write(n, " "); end; end; function quaternarray(n: integer): array; var k: integer; stk: stack; begin while n > 0 do k := n mod 4; stack_push(stk, k); n := (n - k) div 4; end; return stack2array(stk); end; .
(PARI)
SumD(x)= { local(s); s=0; while (x>9, s+=x-10*(x\10); x\=10); return(s + x) }
baseE(x, b)= { local(d, e, f); e=0; f=1; while (x>0, d=x-b*(x\b); x\=b; e+=d*f; f*=10); return(e) }
{ n=0; for (m=1, 10^9, if (m%(SumD(baseE(m, 4)))==0, write("b064438.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Sep 14 2009
(PARI) isok(n) = !(n % sumdigits(n, 4)); \\ Michel Marcus, Jun 24 2018
(Python)
from sympy.ntheory.factor_ import digits
print([n for n in range(1, 201) if n%sum(digits(n, 4)[1:]) == 0]) # Indranil Ghosh, Apr 24 2017
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CROSSREFS
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KEYWORD
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base,easy,nice,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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