A306824 Integer k such that digsum(k) = digsum (k/p(1)) = digsum (k/p(2)) = ... for all prime factors p(i) of k, where digsum(k) = A007953(k) is the digital sum of k.
1, 27, 54, 81, 108, 135, 162, 216, 243, 270, 324, 351, 361, 405, 432, 513, 540, 621, 702, 703, 810, 1026, 1053, 1080, 1215, 1242, 1458, 1620, 1728, 1944, 2071, 2079, 2106, 2133, 2160, 2187, 2403, 2413, 2592, 2700, 2701, 2916, 3024, 3051, 3105, 3267, 3321, 4023, 4033, 4050, 4158
Offset: 1
Examples
4158 = 2*3^3*7*11 is in the sequence because 4 + 1 + 5 + 8 = 18, and: 4158/2 = 2079 and digsum(2079) = 18; 4158/3 = 1386 and digsum(1386) = 18; 4158/7 = 594 and digsum(594) = 18; 4158/11 = 378 and digsum(378) = 18.
Programs
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Maple
with(numtheory):nn:=4200: for k from 1 to nn do: d:=factorset(k):n1:=nops(d):it:=0: b:=convert(k, base, 10):n2:=nops(b):s:=sum(‘b[i]’, ‘i’=1..n2): for i from 1 to n1 do: x:=k/d[i]:b1:=convert(x, base, 10):n3:=nops(b1): s1:=sum(‘b1[i]’, ‘i’=1..n3): if s1=s then it:=it+1: else fi: od: if it=n1 then printf(`%d, `,k): else fi: od: -
Mathematica
sod[n_] := Total@IntegerDigits@n; Select[Range[1, 5000], {sod[#]} == Union[sod /@ (#/First /@ FactorInteger[#])] &] (* Giovanni Resta, Mar 12 2019 *) -
PARI
isok(k) = {my(pf = factor(k)[,1]~, sd = sumdigits(k)); for (i=1, #pf, if (sumdigits(k/pf[i]) != sd, return (0));); return (1);} \\ Michel Marcus, Mar 12 2019
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