Surely this hyperbolic honeycomb is possible as a mere alternation of x5x3x5x only. Accordingly its edges (within an euclidean setting) would have size x(4) = q = sqrt(2) = 1.414214, x(6) = h = sqrt(3) = 1.732051, and x(10) = sqrt[(5+sqrt(5))/2] = 1.902113 respectively.

Incidence matrix according to Dynkin symbol

s5s3s5s   (N → ∞)

demi( . . . . )   | 30N |   2   1   4   2 |   2   1   6   6 | 2  2   4
------------------+-----+-----------------+-----------------+---------
      s 2 s .   & |   2 | 30N   *   *   * |   0   0   2   2 | 1  1   2  x(4)
      s 2 . s     |   2 |   * 15N   *   * |   0   0   0   4 | 0  2   2  x(4)
sefa( s5s . . ) & |   2 |   *   * 60N   * |   1   0   1   1 | 1  1   1  x(10)
sefa( . s3s . )   |   2 |   *   *   * 30N |   0   1   2   0 | 2  0   1  x(6)
------------------+-----+-----------------+-----------------+---------
      s5s . .   & |   5 |   0   0   5   0 | 12N   *   *   * | 1  1   0
      . s3s .     |   3 |   0   0   0   3 |   * 10N   *   * | 2  0   0
sefa( s5s3s . ) & |   3 |   1   0   1   1 |   *   * 60N   * | 1  0   1
sefa( s5s 2 s ) & |   3 |   1   1   1   0 |   *   *   * 60N | 0  1   1
------------------+-----+-----------------+-----------------+---------
      s5s3s .   & |  60 |  30   0  60  60 |  12  20  60   0 | N  *   *  snid
      s5s 2 s   & |  10 |   5   5  10   0 |   2   0   0  10 | * 6N   *  pap
sefa( s5s3s5s )   |   4 |   2   1   2   1 |   0   0   2   2 | *  * 30N  verf(x5x3x5x)