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

3,2

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.

Links

Table of n, a(n) for n=3..42.

Prog

(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

Crossrefs

Cf. A000002.

Sequence in context: A172483 A216396 A273488 * A117848 A025242 A356696

Adjacent sequences: A334952 A334953 A334954 * A334956 A334957 A334958

Keyword

nonn

Author

Rakesh Khanna A, May 24 2020

Status

approved