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.
-
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
A202240
a(n) is the smallest number k such that the sum of the n-th powers of the digits of k equals the sum of the divisors of k other than 1 and k.
Original entry on oeis.org
125, 142, 1005, 118678, 706862, 18481615, 122003411, 30330043, 5923078409, 22110133333, 120175787632, 5971473681952
Offset: 2
a(5) = 118678 because 1^5 + 1^5 + 8^5 + 6^5 + 7^5 + 8^5 = 90121, and sum of the divisors 1 < d < a(5) = sigma(118678) - 118678 - 1 = 90121.
Cf.
A070308 (n=2, "Canada perfect numbers").
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f(k, n) = my(d=digits(k)); sum(i=1, #d, d[i]^n);
a(n) = my(k=1); while(f(k, n) != sigma(k)-k-1, k++); k; \\ Michel Marcus, Sep 29 2018
Showing 1-3 of 3 results.
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