Another sequence idea

[Don Reble]

> k: solutions with a=...
> ------------------------------------
> 2: 3,  20,  119,  696,  4059

Some of those k-sequences are interesting. Let me expand them a bit.

   2: -1 0 3 20 119 696 4059 23660 137903 803760
  24: -12 -11 -8 -4 1 9 20 25 44 76 121 197 304 353 540 856 1301 2053
      3112 3597 5448 8576 12981 20425 30908 35709 54032 84996 128601
      202289 306060 353585 534964 841476
  33: -15 -4 7 27 60 181 227 612 1085 1985 3492 9047 11161 28860 50607
      91987 161276 416685 513883
  50: -21 -17 -12 -8 -1 7 28 44 67 87 124 168 287 379 512 628 843 1099
      1792 2328 3103 3779 5032 6524 10563 13687 18204 22144 29447 38143
      61684 79892 106219 129183 171748 222432 359639 465763 619208
      753052
  96: -39 -31 -28 -24 13 21 28 52 137 193 216 248 581 657 724 956 1789
      2341 2568 2884 6177 6929 7592 9888 18133 23597 25844 28972 61569
      69013 75576 98304 179921 234009 256252 287216 609893 683581 748548
      973532
The sequences have other negative values. If X is in a k-sequence, so is
(1-k-X).

They appear to have these recurrence relations:
   2: a(n) =  6a(n-1) - a(n- 2) +   2
  24: a(n) = 10a(n-6) - a(n-12) +  92
  33: a(n) = 46a(n-6) - a(n-12) + 704
  50: a(n) =  6a(n-6) - a(n-12) +  98
  96: a(n) = 10a(n-8) - a(n-16) + 380

The k=2 recurrence won't surprise many of us: it's just Pythagorean
stuff. But the others might. I didn't check whether the sparse sequences
have such relations.

Hmm... the recurrence constants are multiples of K-1.

    ---

The sequence of K consecutive squares starting at N^2, sums to
    k{N+[(k-1)/2]}^2 + (k^3-k)/12 = m^2
For odd values of K, that Diophantine equation looks rather Pellian.
Perhaps solutions can be found amongst continued-fraction convergents,
or the like, and maybe that explains the recurrences.

--
Don Reble       djr at nk.ca