A254606 - OEIS

A254606

The minimum absolute difference between k*p1 and p2 (p1<p2), where p1*p2 is the n-th term of A087112.

0, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 3, 0, 1, 1, 2, 1, 2, 0, 1, 1, 2, 3, 5, 4, 0, 1, 1, 1, 2, 3, 6, 2, 0, 1, 1, 2, 2, 1, 3, 6, 4, 0, 1, 1, 1, 1, 4, 3, 5, 9, 6, 0, 1, 1, 1, 3, 2, 5, 3, 7, 8, 2, 0, 1, 1, 2, 2, 4, 2, 3, 1, 9, 8, 6, 0, 1, 1, 1, 1, 3, 2, 7, 3, 5

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OFFSET

1,9

COMMENTS

k is an integer that minimizes |k*p1-p2|. It is trivial that if j is the integer part of p2/p1, k is either j or j+1.

LINKS

Lei Zhou, Table of n, a(n) for n = 1..10000

EXAMPLE

A087112(1)=4=2*2, 2-2=0, so a(1)=0;

A087112(2)=6=2*3, 3-2=2*2-3=1, so a(2)=1;

...

A087112(9)=35=5*7, 7-5=2, and 2*5-7=3, the smaller is 2. So a(9)=2.

MATHEMATICA

NumDiff[n1_, n2_] := Module[{c1 = n1, c2 = n2}, If[c1 < c2, c1 = c1 + c2; c2 = c1 - c2; c1 = c1 - c2]; k = Floor[c1/c2]; a1 = c1 - k*c2; If[a1 == 0, a2 = 0, a2 = (k + 1) c2 - c1]; Return[Min[a1, a2]]];

p1 = 2; p2 = 1; Table[p2 = NextPrime[p2]; If[p2 > p1, p1 = p2; p2 = 2]; NumDiff[p1, p2], {n, 1, 100}]

CROSSREFS

Cf. A087112, A254605.

Sequence in context: A333467 A120648 A215401 * A175358 A235330 A029394

Adjacent sequences: A254603 A254604 A254605 * A254607 A254608 A254609

KEYWORD

nonn,easy

AUTHOR

Lei Zhou, Feb 02 2015

STATUS

approved