

A231015


Least k such that n = + 1^2 + 2^2 + 3^2 + 4^2 + ... + k^2 for some choice of + signs.


6



7, 1, 4, 2, 3, 2, 3, 6, 7, 6, 4, 6, 3, 5, 3, 5, 7, 6, 7, 6, 4, 5, 4, 5, 7, 9, 7, 5, 4, 5, 4, 6, 7, 6, 7, 5, 7, 5, 7, 6, 7, 6, 7, 9, 8, 5, 8, 5, 7, 6, 7, 6, 11, 5, 8, 5, 7, 6, 7, 6, 7, 10, 7, 6, 7, 6, 7, 9, 7, 10, 7, 6, 7, 6, 8, 9, 8, 9, 8, 9, 7, 6, 7, 6, 11, 9
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OFFSET

0,1


COMMENTS

Erdős and Surányi proved that for each n there are infinitely many k satisfying the equation.
A158092(k) is the number of solutions to 0 = +1^2 + 2^2 + ... + k^2. The first nonzero value is A158092(7) = 2, so a(0) = 7.
a(n) is also defined for n < 0, and clearly a(n) = a(n).
See A158092 and the AndricaIonascu links for more comments.
The integral formula (3.6) in AndricaVacaretu (see Theorem 3 of the INTEGERS 2013 slides which has a typo) gives in this case the number of representations of n as + 1^2 + 2^2 + ... + k^2 for some choice of + signs. This integral formula is (2^n/2*Pi)*Integral_{t=0..2*Pi} cos(n*t) * Product_{j=1..k} cos(j^2*t) dt. Clearly the number of such representations of n is the coefficient of z^n in the expansion (z^(1^2) + z^(1^2))*(z^(2^2) + z^(2^2))*...*(z^(k^2) + z^(k^2)). AndricaVacaretu used this generating function to prove the integral formula. Section 4 of AndricaVacaretu gives a table of the number of such representations of n for k=1,...,9.  Dorin Andrica, Nov 12 2013


LINKS



FORMULA

a(n(n+1)(2n+1)/6) = a(A000330(n)) = n for n > 0.
a((n(n+1)(2n+1)/6)2) = a(A000330(n)2) = n for n > 0.


EXAMPLE

0 = 1^2 + 2^2  3^2 + 4^2  5^2  6^2 + 7^2.
1 = 1^2.
2 =  1^2  2^2  3^2 + 4^2.
3 =  1^2 + 2^2.
4 =  1^2  2^2 + 3^2.


MAPLE

b:= proc(n, i) option remember; local m; m:=i*(i+1)*(2*i+1)/6;
n<=m and (n=m or b(n+i^2, i1) or b(abs(ni^2), i1))
end:
a:= proc(n) local k; for k while not b(n, k) do od; k end:


MATHEMATICA

b[n_, i_] := b[n, i] = Module[{m}, m = i*(i+1)*(2*i+1)/6; n <= m && (n == m  b[n+i^2, i1]  b[Abs[ni^2], i1])]; a[n_] := Module[{k}, For[k = 1, !b[n, k] , k++]; k]; Table[a[n], {n, 0, 100}] (* JeanFrançois Alcover, Jan 28 2014, after Alois P. Heinz *)


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STATUS

approved



