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Consider a number of k digits n = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + … + d_(2)*10 + d_(1). Sequence lists the numbers n such that Sum_{i=1..k-1}{phi(Sum_{j=1..i}{d_(j)*10^(j-1)})} = Sum_{i=1..k-1}{sigma(Sum_{j=1..i}{d_(k-j+1)*10^(i-j)})} (see example below).
3

%I #18 Jan 03 2024 10:02:08

%S 11,21,53,75,83,95,211,506,523,708,908,932,955,1008,5086,6535,7272,

%T 7557,9126,20534,31165,51301,52695,71665,73713,85173,90902,93026,

%U 93565,210021,313370,330173,406945,423775,521427,633190,728687,850123,926281

%N Consider a number of k digits n = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + … + d_(2)*10 + d_(1). Sequence lists the numbers n such that Sum_{i=1..k-1}{phi(Sum_{j=1..i}{d_(j)*10^(j-1)})} = Sum_{i=1..k-1}{sigma(Sum_{j=1..i}{d_(k-j+1)*10^(i-j)})} (see example below).

%C No single-digit terms are permitted. - _Harvey P. Dale_, Mar 08 2015

%e If n = 423775, starting from the least significant digit, let us cut the number into the set {5, 75, 775, 3775, 23775}. We have:

%e sigma(5) = 6;

%e sigma(75) = 124;

%e sigma(775) = 992;

%e sigma(3775) = 4712;

%e sigma(23775) = 39432.

%e Then, starting from the most significant digit, let us cut the number into the set {4, 42, 423, 4237, 42377}. We have:

%e phi(4) = 2;

%e phi(42) = 12;

%e phi(423) = 276;

%e phi(4237) = 3996;

%e phi(42377) = 40980.

%e Finally, 6 + 124 + 992 + 4712 + 39432 = 2 + 12 + 276 + 3996 + 40980 = 45266.

%p with(numtheory); P:=proc(q) local a, b, k, n; for n from 10 to q do

%p a:=0; k:=1; while trunc(n/10^k)>0 do a:=a+phi(trunc(n/10^k)); k:=k+1; od;

%p b:=0; k:=1; while (n mod 10^k)<n do b:=b+sigma(n mod 10^k); k:=k+1; od;

%p if a=b then print(n); fi; od; end: P(10^9);

%t dsepQ[n_]:=Module[{idn=IntegerDigits[n],len},len=Length[idn]-1; Total[ DivisorSigma[1,#]&/@(FromDigits/@Table[Take[idn,-k],{k,If[Last[idn] == 0,2,1],len}])]==Total[EulerPhi/@(FromDigits/@Table[Take[idn,i],{i, len}])]]; Select[Range[10,10^6],dsepQ] (* _Harvey P. Dale_, Mar 08 2015 *)

%Y Cf. A000010, A000203, A240894-A240903, A241207, A241502, A241503, A244068.

%K nonn,base

%O 1,1

%A _Paolo P. Lava_, Jun 19 2014

%E Corrected (a(18) added) by _Harvey P. Dale_, Mar 08 2015