X n=0 bnxn= 1 − x + 2 ∞ X k=0 x2k(1 − x2k) Qk j=0 1 − x2j

Problem 11685

(American Mathematical Monthly, Vol.120, January 2013)

Proposed by Donald Knuth (USA).

Prove that

∞

Y

k=0



1 + 1 2^2k− 1



= 1 2+

∞

X

k=0

1 Qk−1

j=0 2^2j− 1
.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will show that F (x) = G(x) for |x| < 2/3 where

F (x) =

∞

X

n=0

anxⁿ=

∞

Y

k=0

 1

1 − x^2k



and G(x) =

∞

X

n=0

bnxⁿ= 1 − x + 2

∞

X

k=0

x^2k(1 − x^2k) Qk

j=0 1 − x^2j
. The required equation is equivalent to F (1/2) = G(1/2).

Note that

F (x²) =

∞

Y

k=0

 1

1 − x^2k+1



=

∞

Y

k=1

 1

1 − x^2k



= (1 − x)F (x)

which implies that a_2n+1= a_2n, a_2n= a_2n−1+ a_n for n ≥ 1.

Moreover

G(x²) = 1 − x²+ 2

∞

X

k=0

x^2k+1(1 − x^2k+1) Qk

j=0 1 − x^2j+1
= 1 − x²+ 2

∞

X

k=0

x^2k+1(1 − x^2k+1) Qk+1

j=1 1 − x^2j

= 1 − x²+ 2(1 − x)

∞

X

k=0

x^2k+1(1 − x^2k+1) Qk+1

j=0 1 − x^2j
= 1 − x²+ 2(1 − x)

∞

X

k=1

x^2k(1 − x^2k) Qk

j=0 1 − x^2j

= (1 − x) 1 + x + 2

∞

X

k=0

x^2k(1 − x^2k) Qk

j=0 1 − x^2j
− x

!!

= (1 − x)G(x).

Therefore b2n+1= b2n, b2n= b2n−1+ bn for n ≥ 1.

Since a0 = F (0) = 1 = G(0) = b0, it follows that an = bn for all n ≥ 0, and therefore, the power series F and G coincide in (−R, R) where R = 1/ lim sup_n→∞(an)^1/n. It is easy to verify by induction that 1 ≤ an≤ (3/2)ⁿ for all n ≥ 0, so R ≥ 2/3.