I know exactly what range puzzle number 130 is in. It will reduce the key to be scanned by 50%. I will sell it to anyone who wants to buy it. proof is available. If anyone wants to know the range of PUZZLE 130 UN, I can tell you for money.
if you use something like this:
import secp256k1 as ice
#1361129467683753853853498429727072845823
#100000000000000000000000000000000
target_public_key = "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
target = ice.pub2upub(target_public_key)
num = 13611294 # number of times.
sustract= 100000000000000000000000000000000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
h= (res[t*65:t*65+65]).hex()
hc= ice.to_cpub(h)
data = open("data-base.txt","a")
data.write(str(hc)+"\n")
data.close()
one of those pubs is in the range lower than puzzle
#107