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 A061282 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 3. A stopping problem: begin with n and at each stage if a multiple of 3 divide by 3, otherwise subtract 1. 4
 0, 1, 2, 2, 3, 4, 3, 4, 5, 3, 4, 5, 4, 5, 6, 5, 6, 7, 4, 5, 6, 5, 6, 7, 6, 7, 8, 4, 5, 6, 5, 6, 7, 6, 7, 8, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 7, 8, 9, 8, 9, 10, 9, 10, 11, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 7, 8 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS n > 0 occurs A001590(n+2) times in this sequence. - Peter Kagey, Jul 19 2015 a(n) gives the number of iterations of A260316 to reach 0. - Peter Kagey, Jul 22 2015 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..10000 FORMULA a(n) = A062153(n) + A053735(n) = (number of base 3 digits of n) + (sum of base 3 digits of n)-1. a(3n) = a(n)+1, a(3n+1) = a(n)+2, a(3n+2) = a(n)+3; a(0)=0, a(1)=1, a(2)=2. EXAMPLE a(25)=7 since 25=((0+1+1)*3+1+1)*3+1. MAPLE a:= n-> (l-> nops(l)+add(i, i=l)-1)(convert(n, base, 3)): seq(a(n), n=0..105);  # Alois P. Heinz, Jul 16 2015 PROG (PARI) a(n)=sumdigits(n, 3)+#digits(n, 3)-1 \\ Charles R Greathouse IV, Jul 16 2015 (Haskell) c i = if i `mod` 3 == 0 then i `div` 3 else i - 1 b 0 foldCount = foldCount b sheetCount foldCount = b (c sheetCount) (foldCount + 1) a061282 n = b n 0 -- Peter Kagey, Sep 02 2015 CROSSREFS Cf. A001590, A007089, A053735, A056792, A062153, A260316. Analogous sequences with a different multiplier k: A056792 (k=2), A260112 (k=4). Sequence in context: A199408 A285325 A135529 * A244040 A064514 A112342 Adjacent sequences:  A061279 A061280 A061281 * A061283 A061284 A061285 KEYWORD nonn,easy AUTHOR Henry Bottomley, Jun 06 2001 STATUS approved

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Last modified September 21 13:37 EDT 2019. Contains 327253 sequences. (Running on oeis4.)