A357063 Lengths of the B blocks associated with A091787.

1, 4, 13, 42, 127, 382, 1149, 3448, 10345, 31044, 93133, 279400, 838203, 2514610, 7543831, 22631496, 67894489, 203683468, 611050413, 1833151240, 5499453721, 16498361166, 49495083499, 148485250498, 445455751497, 1336367254492, 4009101763477, 12027305290463, 36081915871390, 108245747614173, 324737242842520, 974211728527561, 2922635185582686, 8767905556748059, 26303716670244178, 78911150010732543, 236733450032197630, 710200350096592891

Offset: 1

Author

Levi van de Pol, Sep 10 2022

Keywords

• nonn

Comments

The B blocks are explained in the article "A slow-growing sequence defined by an unusual occurrence". They have superscript (2).

Examples

			The third B-block of order 2 is B_3^{(2)}=2223222322233. Therefore, a(3)=13.
		

Links

• F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), Article 07.1.2.

Crossrefs

Cf. A091787, A357068.

Programs

Python

number_of_terms=38
def Cn(X):
    l=len(X)
    cn=1
    for i in range(1,int(l/2)+1):
        j=i
        while(X[l-j-1]==X[l-j-1+i]):
            j=j+1
            if j>=l:
                break
        candidate=int(j/i)
        if candidate>cn:
            cn=candidate
    return cn
# This algorithm generates a prefix of the level-3 Gijswijt sequence
def Generate_A3(number):
  glue_lengths=[]
  A3=[3]
  S=[3]
  i=0
  while(True):
      c=Cn(A3)
      if c<3:
        glue_lengths.append(len(S))
        i=i+1
        if i==number:
            break
        S=[]
      A3.append(max(c,3))
      S.append(max(c,3))
  return glue_lengths
glue_lengths=Generate_A3(number_of_terms-1)
beta_lengths=[1]
beta_length=1
for l in glue_lengths:
    beta_length=3*beta_length+l
    beta_lengths.append(beta_length)
print(beta_lengths)

Extensions

By special permission, more than the usual number of terms are shown. - N. J. A. Sloane, Oct 23 2022