A202147
Numbers k such that the sum of digits^4 of k equals Sum_{d|k, 1
Original entry on oeis.org
1005, 5405, 89195, 92029, 107707, 149851, 323723, 524371, 610171, 999643, 1119253, 1134227, 1728787, 1900523, 2045171, 2170451, 2668381, 3351833, 3361717, 3611227, 5364059, 6571483, 7710883, 7865659, 8938691, 9286331, 9362051, 9593833, 10841387, 11507813
Offset: 1
1005 is in the sequence because 1^4 + 0^4 + 0^4 + 5^4 = 626, and the sum of the divisors 1< d<1005 is 3 + 5 +15 + 67 + 201+ 335 = 626.
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Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^4]]; Select[Range[2, 10^7], Q]
A202279
Numbers k such that the sum of digits^3 of k equals Sum_{d|k, 1
Original entry on oeis.org
142, 160, 1375, 6127, 12643, 51703, 86833, 103039, 104647, 112093, 137317, 218269, 261883, 266923, 449881, 505891, 617569, 907873
Offset: 1
160 is in the sequence because 1^3 + 6^3 + 0^3 = 217, and the sum of the divisors 1< d<160 is 2 + 4 + 5 + 8 + 10 + 16 + 20 + 32 + 40 + 80 = 217.
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A055012 := proc(n)
add(d^3,d=convert(n,base,10)) ;
end proc:
A048050 := proc(n)
if n > 1 then
numtheory[sigma](n)-1-n ;
else
0;
end if;
end proc:
isA202279 := proc(n)
A055012(n) = A048050(n) ;
end proc:
for n from 1 do
if isA202279(n) then
printf("%d,\n",n);
end if;
end do; # R. J. Mathar, Dec 15 2011
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Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^3]]; Select[Range[2, 5*10^7], Q]
Select[Range[1000000],DivisorSigma[1,#]-#-1==Total[IntegerDigits[#]^3]&] (* Harvey P. Dale, Jul 19 2014 *)
A202285
Numbers k such that the sum of digits^5 of k equals Sum_{d|k, 1
Original entry on oeis.org
118678, 459385, 4150651, 4351003, 15033631, 20402671, 33224707, 35188159, 40460929, 42454261, 50067673, 54610051, 62004127, 77278261, 88720939, 106412347, 113660551, 113852653, 118203559, 121732873, 125252137, 128083639, 162748279, 163869049, 164863987
Offset: 1
k=118678 is in the sequence because 1^5 + 1^5 + 8^5 + 6^5 + 7^5 + 8^5 = 90121, and the sum of the divisors 1 < d < k = sigma(k) - k - 1 = 90121.
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Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^5]]; Select[Range[2, 10^7], Q]
a(11)-a(25) and keywords fini and full added by
Giovanni Resta, Oct 05 2018
Showing 1-3 of 3 results.
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