a mathematical curiosity that maybe could help the puzzle:
all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity
divided by 3
plus the sum of +1 to the same number divided by 3, results in an integer, odd number.
target = 100
target_2 = 100+1= 101
t1= target//3 = 33.333333333333336
t2= target_2//3 = 33.666666666666664
r= t1+t2 = 67
import bitcoin
import random
import secp256k1 as ice
target= 100
print("pk:",target)
pkaddr = ice.scalar_multiplication(target)
pubkey= pkaddr.hex()
fh=ice.to_cpub(pubkey)
print("target_pub:",fh)
N = 3
h = bitcoin.divide(fh, N)
h2 = target/N
print("N:", N)
print("R:", h)
print("R_dec:", h2)
r_upub = ice.pub2upub(h)
pubkey_u= r_upub.hex()
print(" pub:",pubkey_u)
print("fase 2")
target2= target+1
print("pk:",target2)
pkaddr2 = ice.scalar_multiplication(target2)
pubkey2= pkaddr2.hex()
fh2=ice.to_cpub(pubkey2)
print("target_pub2:",fh2)
h_2 = bitcoin.divide(fh2, N)
h2_2 = target2/N
print("N:", N)
print("R2:", h_2)
print("R_dec2:", h2_2)
r_upub_2 = ice.pub2upub(h_2)
pubkey_u= r_upub_2.hex()
print(" pub:",pubkey_u)
b = ice.pub2upub(h)
b2= ice.pub2upub(h_2)
add_pub = ice.point_addition(b, b2)
res = add_pub.hex()
f=ice.to_cpub(res)
print("add_pub:", f)
print("add_pub_dec:", str(int(h2+h2_2)))