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Numbers k for which A000120(k) = A000120(k'), where k' is arithmetic derivative of k (A003415).
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%I #15 Jan 27 2022 08:51:20

%S 0,2,4,6,9,16,20,22,27,38,40,42,45,49,52,60,65,68,69,78,86,88,96,99,

%T 100,104,105,108,110,114,116,117,118,133,134,144,148,154,164,166,174,

%U 175,184,192,204,209,212,214,216,224,225,230,232,244,245,256,259,262

%N Numbers k for which A000120(k) = A000120(k'), where k' is arithmetic derivative of k (A003415).

%C The sequence is infinite because (p^p)' = p^p, for any p prime number (A051674).

%C If m = 2^k (A000079) then 2^m is a term. Indeed, 2^m has a single 1 in binary expansion and (2^m)' = m*2^(m-1) = 2^(2^k + k - 1) also has a single 1 in its binary expansion.

%C If p is a prime number of the form p = 4*k + 1 (A002144), then 4*p is a term. Indeed, if p is in A002144, then its binary expansion is of the form b_1_b_2...b_k01, and for 4*p it is b_1_b_2...b_k0100. Because (4*p)' = 4*(p + 1) = 4*p + 4 the binary expansion of the number 4*p + 4 is b_1_b_2....b_k0100 + 100 = b_1_b_2....b_k1000, so with the same number of digits 1.

%e 0' = 0, so 0 is a term.

%e 2 = 10_2 and 2' = 1 = 1_2, so 2 is a term.

%e 20 = 10100_2 and 20' = 24 = 11000_2, so 20 is a term.

%p d:= n-> n*add(i[2]/i[1], i=ifactors(n)[2]):

%p q:= n-> (f-> is(f(n)=f(d(n))))(t-> add(i, i=Bits[Split](t))):

%p select(q, [$0..300])[]; # _Alois P. Heinz_, Jan 26 2022

%t d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); Select[Range[0, 300], Equal @@ DigitCount[{#, d[#]}, 2, 1] &] (* _Amiram Eldar_, Jan 25 2022 *)

%o (Magma) f:=func<n |n le 1 select 0 else n*(&+[Factorisation(n)[i][2] / Factorisation(n)[i][1]: i in [1..#Factorisation(n)]])>; [n: n in [0..265]|&+Intseq(n,2) eq &+Intseq(Floor(f(n)),2)];

%o (PARI) d(n) = vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415

%o isok(k) = hammingweight(k) == hammingweight(d(k)); \\ _Michel Marcus_, Jan 25 2022

%Y Cf. A000079, A000120, A002144, A003415, A051674.

%K nonn,base

%O 1,2

%A _Marius A. Burtea_, Jan 24 2022