A sequence $(a_n)$ is defined by means of the recursion
\[a_1 = 1, a_{n+1} = \frac{1 + 4a_n +\sqrt{1+ 24a_n}}{16}.\]
Find an explicit formula for $a_n.$
1. Given the sequence \((a_n)\) defined by the recursion:
   \[
   a_1 = 1, \quad a_{n+1} = \frac{1 + 4a_n + \sqrt{1 + 24a_n}}{16}
   \]
   we aim to find an explicit formula for \(a_n\).

2. Start by manipulating the recursion formula. Multiply both sides by 16 to clear the denominator:
   \[
   16a_{n+1} = 1 + 4a_n + \sqrt{1 + 24a_n}
   \]

3. Isolate the square root term:
   \[
   \sqrt{1 + 24a_n} = 16a_{n+1} - 1 - 4a_n
   \]

4. Square both sides to eliminate the square root:
   \[
   1 + 24a_n = (16a_{n+1} - 1 - 4a_n)^2
   \]

5. Expand the right-hand side:
   \[
   1 + 24a_n = (16a_{n+1} - 1 - 4a_n)^2 = (16a_{n+1} - 1 - 4a_n)(16a_{n+1} - 1 - 4a_n)
   \]
   \[
   1 + 24a_n = (16a_{n+1} - 1)^2 - 2(16a_{n+1} - 1)(4a_n) + (4a_n)^2
   \]
   \[
   1 + 24a_n = 256a_{n+1}^2 - 32a_{n+1} + 1 - 128a_{n+1}a_n + 8a_n + 16a_n^2
   \]

6. Simplify the equation:
   \[
   1 + 24a_n = 256a_{n+1}^2 - 32a_{n+1} + 1 - 128a_{n+1}a_n + 8a_n + 16a_n^2
   \]

7. Notice that the left-hand side is linear in \(a_n\), while the right-hand side is quadratic. This suggests a simpler approach might be needed. Let's introduce a new variable \(b_n\) such that:
   \[
   b_n = \sqrt{24a_n + 1}
   \]

8. Substitute \(b_n\) into the recursion:
   \[
   b_{n+1} = \sqrt{24a_{n+1} + 1}
   \]

9. From the recursion formula, we have:
   \[
   24a_{n+1} + 1 = \left(\frac{\sqrt{24a_n + 1} + 3}{2}\right)^2
   \]
   \[
   24a_{n+1} + 1 = \left(\frac{b_n + 3}{2}\right)^2
   \]

10. Simplify the right-hand side:
    \[
    24a_{n+1} + 1 = \frac{(b_n + 3)^2}{4}
    \]
    \[
    24a_{n+1} + 1 = \frac{b_n^2 + 6b_n + 9}{4}
    \]
    \[
    24a_{n+1} + 1 = \frac{24a_n + 1 + 6b_n + 9}{4}
    \]
    \[
    24a_{n+1} + 1 = \frac{24a_n + 10 + 6b_n}{4}
    \]

11. Multiply both sides by 4:
    \[
    96a_{n+1} + 4 = 24a_n + 10 + 6b_n
    \]
    \[
    96a_{n+1} = 24a_n + 6b_n + 6
    \]
    \[
    16a_{n+1} = a_n + \frac{b_n + 1}{4}
    \]

12. Since \(b_n = \sqrt{24a_n + 1}\), we have:
    \[
    b_{n+1} = \frac{b_n + 3}{2}
    \]

13. This is a linear recurrence relation for \(b_n\). Solving it, we get:
    \[
    b_n = 3 + 2^{2-n}
    \]

14. Substitute back to find \(a_n\):
    \[
    b_n = \sqrt{24a_n + 1} = 3 + 2^{2-n}
    \]
    \[
    24a_n + 1 = (3 + 2^{2-n})^2
    \]
    \[
    24a_n + 1 = 9 + 6 \cdot 2^{2-n} + 2^{2(2-n)}
    \]
    \[
    24a_n = 8 + 6 \cdot 2^{2-n} + 2^{2(2-n)}
    \]
    \[
    a_n = \frac{8 + 6 \cdot 2^{2-n} + 2^{2(2-n)}}{24}
    \]
    \[
    a_n = \frac{(3 + 2^{2-n})^2 - 1}{24}
    \]

Conclusion:
\[
\boxed{a_n = \frac{(3 + 2^{2-n})^2 - 1}{24}}
\]