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A155078 The representation of n=1,2,3,... in binary is a divisor of the same representation in another base. The sequence is the first such base. 3
3, 4, 5, 4, 3, 3, 4, 4, 5, 3, 13, 3, 15, 4, 8, 4, 8, 5, 21, 8, 4, 22, 25, 6, 27, 26, 5, 4, 31, 3, 4, 4, 8, 8, 32, 3, 39, 38, 38, 8, 43, 4, 45, 22, 8, 23, 19, 6, 9, 50, 8, 26, 24, 5, 46, 4, 5, 29, 18, 3, 63, 4, 5, 4, 7, 6, 69, 8, 25, 30, 47, 6, 4, 74, 17, 38, 79, 12, 60, 8, 79, 82, 85, 4, 8, 43 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

The pattern of solutions for each binary representation is notable. For 1001= decimal 9, the bases as solutions are 5,8,11,14,... whereas the pattern for 111=decimal 7 is 4,9,11,16,18,....

The binary representation of n corresponds to the unique polynomial p_n(x) with coefficients in {0,1} such that p(2) = n. a(n) is the least x >= 3 such that p_n(x) == 0 mod n. Thus 3 <= a(n) <= n + 2. - Robert Israel, Dec 15 2014

From Rémy Sigrist, Mar 15 2017: (Start)

If n is even then a(n) <= max(4, n).

If n is odd then a(n) <= n + 2.

If n is odd then n and a(n) are coprime.

If a(n)=4 then n belongs to A062846.

(End)

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

FORMULA

It is simply a matter of converting a binary number to another base to see if the resulting number is a multiple of n. The lowest other base is listed.

EXAMPLE

The n-th term is solved by converting the decimal n to binary then asking to what other base is this representation a multiple of n. For the 5th term, the binary representation is 101; if this is converted to base 3, 101 = 9+0+1 = 10, a multiple of 5. The base 3 is the first base producing a multiple of n: the 5th term is therefore 3.

MAPLE

A155078 := proc(n) local bdgs, b ; bdgs := convert(n, base, 2) ; for b from 3 do add(op(i, bdgs)*b^(i-1), i=1..nops(bdgs)) ; if mod n = 0 then RETURN(b); fi; od: end: seq(A155078(n), n=1..100) ; # R. J. Mathar, Mar 14 2009

# second Maple program:

f:= proc(n) local b, L, r, sols;

     L:= convert(n, base, 2);

     r:= add(L[i]*b^(i-1), i=1..nops(L));

     sols:= subs(0=n, 1=n+1, 2=n+2, map(t -> rhs(op(t)), {msolve(r, n)})) ;

     min(sols);

end proc:

3, seq(f(n), n=2..100); # Robert Israel, Dec 15 2014

CROSSREFS

Cf. A062846.

Sequence in context: A014238 A275719 A014250 * A332420 A115051 A094634

Adjacent sequences:  A155075 A155076 A155077 * A155079 A155080 A155081

KEYWORD

base,easy,nonn

AUTHOR

J. M. Bergot, Jan 19 2009

EXTENSIONS

Corrected and extended by R. J. Mathar, Mar 14 2009

STATUS

approved

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Last modified November 26 12:09 EST 2022. Contains 358359 sequences. (Running on oeis4.)