[seqfan] Re: 4-analogue of Ulam numbers?

[Alois Heinz]

Nice problem!

The 4-analogue of Ulam numbers:

1, 2, 3, 4, 10, 16, 17, 18, 19, 22, 64, 65, 66, 68, 69, 128, 132, 188,
190, 191, 194, 252, 253, 255, 313, 314, 318, 374, 376, 377, 436, 441,
496, 497, 499, 500, 502, 560, 561, 563, 621, 622, 626, 682, 684, 685,
687, 745, 746, 805, 811, 865, 866, 867, 869, 870, 871, 930, 932, 990

For every k>=2 there is a k-analogue of Ulam numbers:

1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, 38, 47, ...
1, 2, 3, 6, 9, 10, 11, 12, 28, 29, 30, 53, 56, 57, 80, ...
1, 2, 3, 4, 10, 16, 17, 18, 19, 22, 64, 65, 66, 68, 69, ...
1, 2, 3, 4, 5, 15, 25, 26, 27, 28, 29, 35, 43, 45, 165, ...
1, 2, 3, 4, 5, 6, 21, 36, 37, 38, 39, 40, 41, 51, 61, ...
1, 2, 3, 4, 5, 6, 7, 28, 49, 50, 51, 52, 53, 54, 55, ...
1, 2, 3, 4, 5, 6, 7, 8, 36, 64, 65, 66, 67, 68, 69, ...
1, 2, 3, 4, 5, 6, 7, 8, 9, 45, 81, 82, 83, 84, 85, ...
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 55, 100, 101, 102, 103, ...

Best regards, Alois

Am 01.01.2011 19:02, schrieb Jonathan Post:
> Much is known now about
> A002858 Ulam numbers: a(1) = 1; a(2) = 2; for n>2, a(n) = least
> number>  a(n-1) which is a unique sum of two distinct earlier terms.
>
> I've just had a comment approved of a 3-analogue of Ulam numbers:
> A007086  Next term is uniquely the sum of 3 earlier terms.
> COMMENTS
> a(1)=1, a(2)=2, a(3)=3, for n>3, a(n) = least number which is a unique
> sum of three distinct earlier terms.  Written this way, we see that
> this is to 3 as Ulam number A002858 is to 2. - Jonathan Vos Post
> (jvospost3(AT)gmail.com)
>
> The next sequence in the supersequence appears, based on quick work by
> hand, to be:
>
> 4-analogue of Ulam numbers: a(1) = 1; a(2) = 2; a(3) = 3, a(4) = 4,
> for n>4, a(n) = least number>  a(n-1) which is a unique sum of 4
> distinct earlier terms.
>
> 1, 2, 3, 4, 10, 16, 17, 18, 19, 22, 63?
>
> n...a(n)
> 1...1, by definition
> 2...2, by definition
> 3...3, by definition
> 4...4, by definition
> 5...10, because 1+2+3+4=10 uniquely, and none of {5,6,7,8,9} are in
> the set of sums of 4 unique terms of {1,2,3,4} as that set is a
> singlet.
> 6...16, because 1+2+3+10 = 16, and none of {11,12,13,14,15} seem to be
> in the set of sums of 4 unique terms of {1,2,3,4,10}
> 7...17, based on back of the envelope, not confirmed
> 8...18, ditto
> 9...19, ditto
> 10..22, I think
> 11..63 maybe? 4+8+19+22 = 63; I've found witnesses to nonuniqueness of
> 23 through 62
>
> This is not hard to code, as Ulam and Knuth did…
>