A189202 Let s_k(n) denote the sum of digits of n in base k. Then a(n) is the smallest m>0 such that both s_2(m*(n-1)) and s_n(2*m*(n-1))/(n-1) are even, or a(n)=0, if such m does not exist.

3, 5, 5, 3, 13, 4, 9, 5, 11, 6, 19, 20, 15, 47, 17, 9, 19, 10, 21, 32, 23, 12, 37, 13, 40, 41, 29, 15, 46, 16, 33, 17, 35, 18, 37, 56, 39, 20, 41, 21, 85, 22, 45, 68, 47, 72, 73, 25, 51, 26, 79, 80, 109, 28, 57, 87, 59, 30, 91, 153, 63, 191, 65, 33, 67, 34, 69

Offset: 2

Author

Vladimir Shevelev, May 02 2011

Keywords

• nonn

Comments

Conjecture: For all n>=2, a(n)>0.

For a general problem, see SeqFan link.

Links

• Alois P. Heinz, Table of n, a(n) for n = 2..10000
• Vladimir Shevelev, "A new digital problem", SeqFan Discussion, Apr 2011.

Crossrefs

Cf. A026741, A145051.

Programs

Maple

s:= proc(n, b) local m, t;
      t:= 0; m:= n;
      while m<>0 do t:= t+ irem(m, b, 'm') od; t
    end:
a:= proc(n) local m;
      for m while irem(s(m*(n-1), 2), 2)<>0 or
                  irem(s(2*m*(n-1), n)/(n-1), 2)<>0 do od; m
    end:
seq(a(n), n=2..100);  # Alois P. Heinz, May 02 2011

Mathematica

s[n_, b_] := Module[{m, t}, t = 0; m = n; While[m != 0 , t = t + Mod[m, b]; m = Quotient[m, b]]; t];
a[n_] := Module[{m}, For[m = 1, Mod[s[m*(n-1), 2], 2] != 0 || Mod[s[2*m*(n-1), n]/(n-1), 2] != 0, m++]; m];
Table[a[n], {n, 2, 100}](* Jean-François Alcover, May 20 2025, after Alois P. Heinz *)