This sequence looks superficially like a simple continued-fraction expansion of some constant, but it's not. There are no numbers (I conjecture) ending with the digit 5. How did I arrive at my sequence? They are the magic multipliers (the final number on each line) in the following:
n = 1: (37+1)/2 = 19*1
n = 2: (73+1)/2 = 37*1
n = 3: (113+1)/2 = 19*3
n = 4: (149-1)/2 = 37*2
n = 5: (157+1)/2 = 79*1
n = 6: (193+1)/2 = 97*1
n = 7: (269-1)/2 = 67*2
n = 8: (277+1)/2 = 139*1
n = 9: (313+1)/2 = 157*1
n = 10: (353+1)/2 = 59*3
n = 11: (389-1)/2 = 97*2
n = 12: (397+1)/2 = 199*1
n = 13: (457+1)/2 = 229*1
n = 14: (557+1)/2 = 31*9
n = 2: (73+1)/2 = 37*1
n = 3: (113+1)/2 = 19*3
n = 4: (149-1)/2 = 37*2
n = 5: (157+1)/2 = 79*1
n = 6: (193+1)/2 = 97*1
n = 7: (269-1)/2 = 67*2
n = 8: (277+1)/2 = 139*1
n = 9: (313+1)/2 = 157*1
n = 10: (353+1)/2 = 59*3
n = 11: (389-1)/2 = 97*2
n = 12: (397+1)/2 = 199*1
n = 13: (457+1)/2 = 229*1
n = 14: (557+1)/2 = 31*9
etc.
The initial number on each line is A135952(n). Note that if the magic multiplier is odd, we add 1 before dividing by 2; if it is even, we subtract 1 before dividing by 2. The number immediately after each equal sign is the prime p where A135952(n) divides composite Fibonacci(p). In my sequence of magic multipliers, the first occurrences of the positive integers are at indices:
1, 4, 3, 157, 0, 24, 91, 71, 14, 78, 81, 802, 124, 149, 0, 720, 436, 292, 82, 347, 128, 389, 598, 113, 0, 1245, 454, 1728, 270, 39, 258, ...
