A293011 a(n) is the smallest positive k such that f(k) = n*g(k) where f = A007953 and g = A000120, or 0 if no such k exists.

1, 2, 6, 4, 32, 48, 16, 8, 288, 64, 128, 8196, 256, 2048, 16896, 278528, 2097664, 589824, 4096, 8192, 8388609, 16384, 536870944, 268435488, 65536, 32768, 268959744, 17179869440, 524288, 4294967298, 1048576, 8589934594, 8589934596

Offset: 1

Author

Altug Alkan, Sep 28 2017

Keywords

• nonn
• base
• look

Comments

Conjecture: a(n) > 0 for all n >= 1.

Numbers n such that a(n) is not a power of 2 are 3, 6, 9, 12, 15, 16, 17, 18, ...

a(21) = 8388609 = 2^23 + 1 is the second odd term of this sequence after a(1) = 1.

Smallest n such that a(n + 1) = a(n) + 2 is 32 and a(32) = 2*(2^32 + 1).

For n <= 170, A000120(a(n)) <= 2. - Robert Israel, Nov 22 2019

Examples

			a(9) = 288 = 2^8 + 2^5 because A007953(288) = 2 + 8 + 8 = 18, 18 / 2 = 9 and 288 is the least number with this property.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..170
• Robert Israel, Color-coded logarithmic plot

Crossrefs

Cf. A000120, A001370, A007953.

Programs

Maple

# This code returns a(n) if A000120(a(n)) <= 3 and it can prove that no
# smaller number with A000120 >= 4 can have A007953 large enough. If it
# can't prove that, it returns FAIL.
sdd:= n -> convert(convert(n,base,10),`+`):
g:= proc(n) local found, k1, k2, k3, x, y, m,bd;
  found:= false;
  for k1 from 1 while not found do
    for k2 from 0 to k1-1 do
      x:= 2^k1 + 2^k2;
      if sdd(x) = 2*n then found:= true; break fi
  od od;
  for k1 from 0 to ilog2(x) do
    if sdd(2^k1) = n then x:= 2^k1; break fi
  od;
  m:= ilog10(x);
  bd:= floor(x/10^m)+9*m;
  if bd <= 3*n then return x fi;
  found:= false;
  for k1 from 2 to ilog2(x) while not found do
    for k2 from 1 to k1-1 while not found do
      for k3 from 0 to k2-1 do
         y:= 2^k1 + 2^k2 + 2^k3;
         if y > x or sdd(y) = 3*n then found:= true; break fi;
  od od od;
  if found then x:= min(x,y) fi;
  bd:= floor(x/10^m)+9*m;
  if bd <= 4*n then x else FAIL fi;
end proc:
map(g, [$1..50]); # Robert Israel, Nov 22 2019

PARI

a(n) = {my(k=1); while ((hammingweight(k))*n != sumdigits(k), k++); k; }