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A258059
Let n = Sum_{i=0..k} d_i*4^i be the base-4 expansion of n, with 0 <= d_i < 4. Then a(n) = minimal i such that d_i is not 1, or k+1 if there is no such i.
2
1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1
OFFSET
1,5
COMMENTS
This is the "General Ruler Sequence Base 4 Focused at 1" of Webster (2015).
LINKS
Richard C. Webster, One Sequence to Rule Them All: The Ruler Sequence and Its Relation to Odd Perfect Numbers and Multiplicative Order, MS Thesis, California State Polytechnic University, Pomona, CA, 2015.
FORMULA
Recurrence: a(1)=1; thereafter a(4*n+1) = a(n)+1, a(4*n+j) = 0 for j = 0,2,3. G.f. g(x) = Sum_{k>=0} k * x^((4^k-1)/3) * (1 + x^(2*4^k) + x^(3*4^k))/(1 - x^(4*4^k)) satisfies g(x) = x*g(x^4) + x/(1-x^4). - Robert Israel, Jun 08 2015
EXAMPLE
1 = 0*4+1, so a(1)=1.
7 = 1*4+3, so a(7)=0.
21 = 0*4^3+1*4^2+1*4+1, so a(21)=3.
523 base 10 is 20023 in base 4, so a(523)=0.
1365 base 10 is 111111 in base 4, so a(1365)=6.
MAPLE
f:= proc(n)
if n mod 4 = 1 then procname((n-1)/4) + 1 else 0 fi
end proc:
map(f, [$1..1000]); # Robert Israel, Jun 08 2015
PROG
(PARI) a(n) = {v = Vecrev(digits(n, 4)); for (i=1, #v, if (v[i] != 1, return (i-1)); ); return(#v); }
(Haskell)
a258059 = f 0 . a030386_row where
f i [] = i
f i (t:ts) = if t == 1 then f (i + 1) ts else i
-- Reinhard Zumkeller, Nov 08 2015
CROSSREFS
The nonzero terms give A263845.
This sequence and A263845 are analogs of the pair of ruler sequences A007814 and A001511.
Cf. A030386.
Sequence in context: A276935 A235127 A358345 * A093956 A160383 A330023
KEYWORD
nonn,easy,base
AUTHOR
Richard C. Webster, May 17 2015
EXTENSIONS
Edited by N. J. A. Sloane, Oct 31 2015 and Nov 06 2015.
STATUS
approved