A064438 Numbers which are divisible by the sum of their quaternary digits.

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

Offset: 1

Author

Len Smiley, Oct 01 2001

Keywords

• base
• easy
• nice
• nonn

Comments

A good "puzzle" sequence -- guess the rule given the first twenty or so terms.

Examples

			Quaternary representation of 28 is 130, 1 + 3 + 0 = 4 divides 28.
		

Links

• Harry J. Smith, Table of n, a(n) for n = 1..1000
• Paul Dalenberg and Tom Edgar, Consecutive factorial base Niven numbers, Fibonacci Quart. (2018) Vol. 56, No. 2, 163-166.

Crossrefs

Cf. A005349 (decimal), A049445 (binary), A064150 (ternary).

Programs

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;

Mathematica

Select[Range[200],Divisible[#,Total[IntegerDigits[#,4]]]&] (* Harvey P. Dale, Jun 09 2011 *)

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

Extensions

More terms from Matthew Conroy, Oct 02 2001

Offset changed from 0 to 1 by Harry J. Smith, Sep 14 2009