A334955 Number of 1's in n-th string when A000002 is expressed as 1, 2, 2, 11, 21, 221, 22112, 11221211, 21221121121, 2212211212212112, ....

0, 2, 1, 1, 2, 5, 6, 7, 12, 19, 29, 43, 62, 95, 148, 216, 322, 482, 727, 1083, 1627, 2457, 3660, 5518, 8250, 12403, 18594, 27914, 41863, 62781, 94158, 141205, 211797, 317782, 476743, 714948, 1072526, 1608732, 2413223, 5428986

Offset: 3

Author

Rakesh Khanna A, May 24 2020

Keywords

• nonn

Comments

Representing A000002 as a tree. Each branch of this tree is a string. Starting from n=3, each 1 in n-th string generates either 1 or 2 in (n+1)-th string and each 2 in n-th string generates either 11 or 22 in (n+1)-st string based on the previously generated term of either 2 or 1. Hence number of terms in (n+1)-st string is sum of all terms in n-th string.

Crossrefs

Cf. A000002.

Programs

Python

MAX_NUM = 10000 # Number of terms in Kolakoski Sequence
K = [1,2,2]
previous = 2
# Generate Kolakoski Sequence
for index1 in range(2,MAX_NUM):
    generator = K[index1]
    if (generator == 1 and previous == 1):
        K.append(2)
        previous = 2
    elif(generator == 2 and previous == 1):
        K.append(2)
        K.append(2)
        previous = 2
    elif(generator == 1 and previous == 2):
        K.append(1)
        previous = 1
    elif(generator == 2 and previous == 2):
        K.append(1)
        K.append(1)
        previous = 1
branch_sum = 1
index2 = 2
cntr1 = 1
while(index2 < MAX_NUM):
    ones_cntr = 0
    # This for loop extracts strings from Kolakoski Sequence
    start_index = index2
    end_index = index2 + branch_sum
    branch_sum = 0
    for index3 in range(start_index , end_index):
        branch_sum = branch_sum + K[index3]
        ones_cntr = ones_cntr + (K[index3]%2)
    index2 = end_index
    print(str(cntr1)+"   "+str(ones_cntr)+"\n")
    cntr1 = cntr1 + 1