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 A085407 Runs of zeros in binomial(3k+2,k+1)/(3k+2) modulo 2 (A085405). 2
 1, 1, 3, 1, 5, 1, 1, 11, 1, 1, 3, 1, 21, 1, 1, 3, 1, 5, 1, 1, 43, 1, 1, 3, 1, 5, 1, 1, 11, 1, 1, 3, 1, 85, 1, 1, 3, 1, 5, 1, 1, 11, 1, 1, 3, 1, 21, 1, 1, 3, 1, 5, 1, 1, 171, 1, 1, 3, 1, 5, 1, 1, 11, 1, 1, 3, 1, 21, 1, 1, 3, 1, 5, 1, 1, 43, 1, 1, 3, 1, 5, 1, 1, 11, 1, 1, 3, 1, 341, 1, 1, 3, 1, 5, 1, 1, 11 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Construction: start with strings S(1)={1}, S(2)={1,3}, S(3)={1,5}; to obtain S(k) for k>3, concatenate all previous strings excluding S(k-1), then replace the last number L in this resulting string with {4L-(-1)^k}. This sequence is the limit of S(k) as k grows, generating strings with Fibonacci growth: {1,1,11}, {1,1,3,1,21}, {1,1,3,1,5,1,1,43}, {1,1,3,1,5,1,1,11,1,1,3,1,85}, ... LINKS Table of n, a(n) for n=1..97. FORMULA For n>0, a(F(n))=A001045(n), where F(n) is the n-th Fibonacci number with F(1)=1, F(2)=2. EXAMPLE To generate string S(4) at k=4: concatenate {S(1),S(2)} = {1, 1,3}, then replace the last number L=3 by 4*L-1=11 to obtain S(4)={1,1,11}. At k=5: concatenate {S(1),S(2),S(3)} = {1, 1,3, 1,5}, then replace the last number L=5 by 4*L+1=21 to obtain S(5)={1,1,3,1,21}. CROSSREFS Cf. A006013, A001045, A085405, A022340 (positions of ones). Sequence in context: A326454 A227507 A134700 * A325523 A352483 A016475 Adjacent sequences: A085404 A085405 A085406 * A085408 A085409 A085410 KEYWORD nonn AUTHOR Paul D. Hanna, Jun 29 2003 STATUS approved

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Last modified August 4 17:39 EDT 2024. Contains 374923 sequences. (Running on oeis4.)