# Problem 588: Quintinomial coefficients
The coefficients in the expansion of \$(x+1)\^k\$ are called binomial
coefficients. Analoguously the coefficients in the expansion of
\$(x\^4+x\^3+x\^2+x+1)\^k\$ are called quintinomial coefficients.
(quintus= Latin for fifth). Consider the expansion of
\$(x\^4+x\^3+x\^2+x+1)\^3\$:
\$x\^{12}+3x\^{11}+6x\^{10}+10x\^9+15x\^8+18x\^7+19x\^6+18x\^5+15x\^4+10x\^3+6x\^2+3x+1\$
As we can see 7 out of the 13 quintinomial coefficients for \$k=3\$ are
odd. Let \$Q(k)\$ be the number of odd coefficients in the expansion of
\$(x\^4+x\^3+x\^2+x+1)\^k\$. So \$Q(3)=7\$. You are given \$Q(10)=17\$
and \$Q(100)=35\$. Find \$\\sum\_{k=1}\^{18}Q(10\^k) \$.