A087142 Numbers divisible by their individual digits, but not by the sum of their digits (counted with multiplicity).
11, 15, 22, 33, 44, 55, 66, 77, 88, 99, 115, 122, 124, 128, 155, 168, 175, 184, 212, 244, 248, 366, 384, 412, 424, 488, 515, 636, 672, 728, 784, 816, 824, 848, 1111, 1112, 1113, 1115, 1124, 1131, 1144, 1155, 1176, 1184, 1197, 1222, 1244, 1248, 1266, 1288, 1311
Offset: 1
Examples
488 is in the sequence as its divisible by its individual digits but not by the sum of its digits counted with multiplicity. That is 488 is divisible by 4 and 8 but not by 4 + 8 + 8 = 20. - _David A. Corneth_, Jan 28 2021
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harvey P. Dale)
Programs
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Mathematica
didQ[n_]:=Module[{idn=IntegerDigits[n]},FreeQ[idn,0]&&AllTrue[n/idn, IntegerQ] && !Divisible[n,Total[idn]]]; Select[Range[1300], didQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Apr 18 2016 *) -
PARI
is(n) = { my(d = digits(n), sd = vecsum(d), s = Set(d)); if(n == 0 || s[1] == 0, return(0)); if(n % sd != 0, for(i = 1, #s, if(n % s[i] != 0, return(0) ) ); return(1) ); 0 } \\ David A. Corneth, Jan 28 2021 -
Python
def ok(n): d = list(map(int, str(n))) return 0 not in d and n%sum(d) and all(n%di == 0 for di in set(d)) print([k for k in range(1312) if ok(k)]) # Michael S. Branicky, Nov 15 2021
Comments