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 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 (list; graph; refs; listen; history; text; internal format)
 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 Andrica-Ionascu links for more comments. The integral formula (3.6) in Andrica-Vacaretu (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)). Andrica-Vacaretu used this generating function to prove the integral formula. Section 4 of Andrica-Vacaretu gives a table of the number of such representations of n for k=1,...,9. - Dorin Andrica, Nov 12 2013 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..20000 D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, Integers Conference 2013 Abstract. D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides. D. Andrica and D. Vacaretu, Representation theorems and almost unimodal sequences, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33. P. Erdős and J. Surányi, Egy additív számelméleti probléma (in Hungarian; Russian and German summaries), Mat. Lapok 10 (1959), pp. 284-290. 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, i-1) or b(abs(n-i^2), i-1)) end: a:= proc(n) local k; for k while not b(n, k) do od; k end: seq(a(n), n=0..100); # Alois P. Heinz, Nov 03 2013 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, i-1] || b[Abs[n-i^2], i-1])]; a[n_] := Module[{k}, For[k = 1, !b[n, k] , k++]; k]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 28 2014, after Alois P. Heinz *) CROSSREFS Cf. A000330, A158092, A231071, A231272. Sequence in context: A010504 A242132 A011450 * A183031 A021018 A010145 Adjacent sequences: A231012 A231013 A231014 * A231016 A231017 A231018 KEYWORD nonn,look AUTHOR Jonathan Sondow, Nov 02 2013 EXTENSIONS a(4) corrected and a(5)-a(85) from Donovan Johnson, Nov 03 2013 STATUS approved

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