A117846 Given n, define a(n) as follows: let a sequence b(k) be defined by b(k+1)=b(k)+b(k)mod k; b(1)=2n-1. (Here b(k)mod k denotes the least nonnegative residue of b(k) modulo k). Let a(n) be the common value of b(k+1)-b(k) for all large k if such exists; otherwise let a(n) be 0.

97, 1, 2, 2, 316, 2, 3, 3, 3, 4, 12, 4, 4, 12, 11, 11, 316, 11, 316, 316, 6, 316, 316, 316, 316, 97, 316, 316, 13, 316, 13, 13, 13, 13, 8, 13, 13, 12, 13, 13, 13, 13, 13, 13, 14, 14, 316, 14, 316, 316, 316, 97, 9, 97, 97, 13, 10, 10, 11, 10, 14, 11, 12, 12, 97, 12, 97, 132

Offset

1,1

Comments

Putting b(1)=2n gives essentially the same sequence as putting b(1)=2n-1. It is a plausible conjecture or at least an interesting open problem that a(n) is never zero; that is all the sequences b(k) are arithmetic progressions from some point on. Sequence A073117 is the sequence b(k) with b(1)=1. Do the values a(n) include all positive numbers?

Links

Table of n, a(n) for n=1..68.

Example

n=4: b(1)=7 and the sequence b(k) continues 7,8,10,12,14...with b(k+1)-b(k)=2 for all k>3, so a(4)=2.

Crossrefs

Cf. A073117.

Sequence in context: A133402 A249922 A163494 * A058286 A051330 A106429

Adjacent sequences: A117843 A117844 A117845 * A117847 A117848 A117849

Keyword

nonn

Author

Alex Abercrombie, Mar 22 2007

Status

approved