A062853 - OEIS

%I #40 Mar 07 2023 07:40:43

%S 0,1,2,53,91,182,194,273,546,582,948,1092,1236,2184,2527,9373,19238,

%T 28119,57714,84357,173142,185640,452807,21774372,48833136,65323116,

%U 1145127998,3435383994,4804366457,11296002941,14224061544,18500792316,28413081060,33888008823

%N When expressed in base 3 and then interpreted in base 4, is a multiple of the original number.

%C From _Jon E. Schoenfield_, Mar 06 2023: (Start)

%C Let u(k) be the result of expressing an integer k in base 3 and interpreting the result as a base-4 number, and define the ratio r(k) = u(k)/k. Then (after the initial term 0) the sequence consists of the integers k > 0 such that r(k) is an integer.

%C Note that, among all numbers k in any interval [m*3^j, (m+1)*3^j - 1] where m > 0, r(k) is maximized at k = m*3^j and minimized at (m+1)*3^j - 1. Consequently, there cannot be any terms in that interval unless there is at least one integer in the interval [r((m+1)*3^j - 1), r(m*3^j)]. (This observation is implemented in the Magma program below, which, when run on the Magma Calculator, computes the first 34 terms in about 0.5 seconds.) (End)

%C Numbers k such that A023717(k) is a multiple of k. - _Michel Marcus_, Mar 07 2023

%H Jon E. Schoenfield, <a href="/A062853/b062853.txt">Table of n, a(n) for n = 1..55</a> (all terms < 3^48).

%e 53 = 1222_3; 1222_4 = 106 = 2*53.

%t fQ[n_] := Mod[ FromDigits[ IntegerDigits[n, 3], 4], n] == 0;

%t k = 1; lst = {};

%t While[k < 10^10/8, If[ fQ@k, AppendTo[ lst, k]; Print@k]; k++ ];

%t lst (* _Robert G. Wilson v_, Feb 24 2010 *)

%o (Magma)

%o N := 34; // max # of terms

%o A := [0];

%o D := [1]; // base-3 dgts (reversed) at curr srch point

%o j := 1; // pointer (at ones place)

%o while #A lt N do

%o    if j eq 1 then // test a single integer (k)

%o       k := Seqint(D, 3);

%o       if Seqint(D, 4) mod k eq 0 then

%o          A[#A+1] := k;

%o       end if;

%o       D[j] +:= 1;

%o    else // test the interval [k0, k1]

%o       k0 := Seqint(D, 3);

%o       k1 := k0 + 3^(j - 1) - 1;

%o       u0 := Seqint(D, 4);

%o       u1 := Seqint(Intseq(k1, 3), 4);

%o       if u0 div k0 gt (u1 - 1) div k1 then

%o          // at least 1 integer in interval [u1/k1, u0/k0]

%o          j -:= 1; // test its 3 subintervals

%o       else

%o          D[j] +:= 1;

%o       end if;

%o    end if;

%o    while D[j] eq 3 do // all 3 subintervals tested

%o       D[j] := 0; // reset

%o       j +:= 1; // move up to larger interval

%o       if j gt #D then

%o          D[j] := 1; // add a digit

%o          break;

%o       end if;

%o       D[j] +:= 1;

%o    end while;

%o end while;

%o A; // _Jon E. Schoenfield_, Mar 05 2023

%Y Cf. A023717.

%K base,nonn

%O 1,3

%A _Erich Friedman_, Jul 21 2001

%E a(21)-a(27) from _Robert G. Wilson v_, Feb 24 2010

%E Offset changed to 1 and a(28), a(29) from _Georg Fischer_, Mar 03 2023

%E a(30)-a(34) from _Jon E. Schoenfield_, Mar 05 2023