[seqfan] Unexpected help by A052515
r.rosenthal at web.de
Thu Feb 11 14:12:23 CET 2010
A friend of mine asked for numbers z such that
S^2 = P holds, where S = sum of digits and
P = product of digits.
Don't be afraid that this becomes a digit-related
posting. After having found lots of solutions I
asked myself what the solutions would look like
if one confined ones search to ternary digits.
Besides the trivial solutions z=0 and z=1 I was
able to find z=22222222 and z=1111111111112222222222
which can be written as [0,8] and [12,10], where
[a,b] means a solution with a 1-digits and b 2-digits.
As sum of digits=S=a+2b and product of digits=P=2^b,
we are looking for integer solutions of S^2=P, i.e.
(a+2b)^2 = 2^b
After some experiments I found more solutions:
[0,8], [12,10], [40,12], [100,14], [224,16]
and thus became interested in the sequence 1,12,40,100,224,...
And, alas! there it was:
Now I know (without proof) that integer solutions [a,b]
for (a+2b)^2 = 2^b are given by
a = 2*A052515(n)
b = 2*n+2
for n > 2.
Examples: n=3: a=2*A052515(3)= 0, b=2*3+2 = 8
n=4: a=2*A052515(4)= 12, b=2*4+2 = 10
n=5: a=2*A052515(5)= 40, b=2*5+2 = 12
I checked all available members of A052515 an they all worked
I would like to put my experimental findings as another
comment into the OEIS.
The most serious comment would be that a=2*A052515(n) and
b=2*n+2 provide integer solutions for (a+2b)^2 = 2^b.
A more playful remark could refer to the original problem:
Finding ternary numbers with S and P as their sum resp. product
of digits and satisfying S^2 = P. The number of digits 1 and 2
is then a and b respectively: a=2*A052515(n), b=2*n+2.
I am quite happy about my findings and like to share them
with SeqFan. I will wait with my submission of the comment
since I am hoping for some helpful advice, how to present
the findings appropriately.
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