1,1

From the 9th term onward the differences of this sequence appear to again be the squares.  Is there a simple explanation for this?

Similar examples are provided for the positive integers in A000124, the odd integers in A082562 and the primes in A082548.

For n > 9: a(n) - a(n-1) = n^2 up to at least n = 1785. - Zak Seidov and Jud McCranie, Feb 29 2012

To compute the terms in order, start with a list with the element 0.  Add 1^2 to each term of the list and add the sum to the list, if it isn't already on the list.  The cardinality of the list is a(1).  Then a(n+1) is computed by adding n^2 to each member of the list and adding the sum to the list, if it isn't already there.  The number of members of the list is a(2).  This is much faster than considering every subset. Jud McCranie, Mar 01 2012

Table of n, a(n) for n=1..40.

Toshitaka Suzuki, Proof of a conjectured formula for A208531, Jul 29 2019

Conjectures from Colin Barker, Feb 15 2016: (Start)

a(n) = (2*n^3+3*n^2+n-366)/6 for n>8.

a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>12.

G.f.: x*(2-4*x+4*x^2-2*x^4+8*x^5-7*x^6+7*x^7-10*x^8+6*x^9-6*x^10+4*x^11) / (1-x)^4. (End)

For a proof of these conjectures see the Suzuki (2019) link.

All subsets of {1,4,9,16} give distinct sums, so a(4)=16. Four pairs of subsets of {1,4,9,16,25} have the same sum, for example {9,16} and {25}, resulting in a(5)=28.

Table[Length[Union[Total /@ Subsets[Range[n]^2]]], {n, 17}] (* T. D. Noe, Feb 28 2012 *)

Cf. A000124, A082548, A082562.

Sequence in context: A209410 A228733 A318767 * A349052 A355969 A308542

Adjacent sequences:  A208528 A208529 A208530 * A208532 A208533 A208534

nonn

John W. Layman, Feb 27 2012

a(23)-a(26) from Zak Seidov, Feb 29 2012

a(27)-a(40) from Jud McCranie, Feb 29 2012

approved