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.
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
Examples
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.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 1..10000
- 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.
Crossrefs
Programs
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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
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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
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PARI
a(n) = {v = Vecrev(digits(n, 4)); for (i=1, #v, if (v[i] != 1, return (i-1));); return(#v);}
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
Extensions
Edited by N. J. A. Sloane, Oct 31 2015 and Nov 06 2015.
Comments