A285271 Numbers that are divisible by each of their digits but that are either not divisible by the sum of their digits or are not divisible by the product of their digits or both.
11, 15, 22, 33, 44, 48, 55, 66, 77, 88, 99, 115, 122, 124, 126, 128, 155, 162, 168, 175, 184, 212, 222, 244, 248, 264, 288, 324, 333, 336, 366, 384, 396, 412, 424, 444, 448, 488, 515, 555, 636, 648, 666, 672, 728, 777, 784, 816, 824, 848, 864, 888, 936, 999, 1111, 1112
Offset: 1
Examples
15 is divisible by its digits 1 and 5, and 15 is divisible by the product of its digits 1*5 = 5, but 15 is not divisible by the sum of its digits 1+5 = 6, hence 15 is a term. 48 is divisible by its digits 4 and 8, and 48 is divisible by the sum of its digits 4+8 = 12, but 48 is not divisible by the product of its digits 4*8 = 32, hence 48 is a term. 124 is divisible by its digits 1, 2 and 4, but 124 is not divisible by the product of its digits 1*2*4 = 8 and 124 is not divisible by the sum of its digits 1+2+4 = 7, hence 124 is a term. 24 is divisible by its digits 2 and 4, and 24 is divisible by the sum of its digits 2+4 = 6, and 24 is also divisible by the product of its digits 2*4 = 8, hence 24 is NOT a term.
Links
- Robert G. Wilson v, Table of n, a(n) for n = 1..1520
Programs
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Maple
filter:= proc(n) local F; F:= convert(n,base,10); andmap(t -> t > 0 and n mod t = 0, F) and not(n mod convert(F,`+`) = 0 and n mod convert(F,`*`) = 0) end proc: select(filter, [$11 .. 2000]); # Robert Israel, Jul 05 2017
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Mathematica
fQ[n_] := Block[{ind = IntegerDigits@ n}, Union[ IntegerQ@# & /@ (n/ind)] == {True} && (!IntegerQ[n/Plus @@ ind] || !IntegerQ[n/Times @@ ind])]; Select[Range@ 1112, fQ] (* Robert G. Wilson v, Jul 05 2017 *) nddQ[n_]:=With[{idn=IntegerDigits[n]},FreeQ[idn,0]&&AllTrue[n/idn,IntegerQ]&&(!IntegerQ[n/Times@@idn]||!IntegerQ[n/Total[idn]])]; Select[Range[1200],nddQ] (* Harvey P. Dale, May 04 2025 *)
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PARI
isok(n) = {d = digits(n); if (vecmin(d), for (k=1, #d, if (n % d[k], return (0));); return ((n % vecsum(d)) || (n % prod(k=1, #d, d[k])));); return (0);} \\ Michel Marcus, Jul 02 2017
Extensions
Definition clarified by Harvey P. Dale, May 04 2025
Comments