|
Search: id:A077221
|
|
|
| A077221 |
|
a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square. |
|
+0
11
|
|
| 0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232
(list; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
The sequence 0,1,8,17,32,49,72,97,128,... arises from reading the line from 0, in the direction 0, 1,... and the same line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol (info(AT)polprimos.com), May 03 2008
Contribution from Peter Luschny (peter(AT)luschny.de), Jul 12 2009: (Start)
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus
a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|.
|
|
FORMULA
|
a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) +1.
2n^2+4n+1+[n odd]. G.f.: (x^2+6x+1)/(1-x)^3/(1+x). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Mar 31 2003
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 29 2007
|
|
MAPLE
|
a := n -> 2*n^2 - (n mod 2); [From Peter Luschny (peter(AT)luschny.de), Jul 12 2009]
|
|
MATHEMATICA
|
a=1; lst={a}; Do[b=n^2-a; AppendTo[lst, b]; a=b, {n, 3, 6!, 2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), May 18 2009]
|
|
CROSSREFS
|
Cf. A077222.
Cf. A131925.
Sequence in context: A028884 A099358 A077222 this_sequence A106648 A076980 A159696
Adjacent sequences: A077218 A077219 A077220 this_sequence A077222 A077223 A077224
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Nov 03 2002
|
|
EXTENSIONS
|
Extended by Ralf Stephan (ralf(AT)ark.in-berlin.de), Mar 31 2003
|
|
|
Search completed in 0.002 seconds
|
