A231272 -id:A231272 - cp's OEIS frontend

A231015 Least k such that n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for some choice of +- signs.

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

Offset: 0

Jonathan Sondow, Nov 02 2013

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

			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.

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.

Cf. A000330, A158092, A231071, A231272.

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

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 *)

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.

a(4) corrected and a(5)-a(85) from Donovan Johnson, Nov 03 2013

A231071 Number of solutions to n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for minimal k giving at least one solution.

Original entry on oeis.org

2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 9, 1, 3, 1, 1, 1, 2, 1, 1, 6, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1, 1, 1, 4, 3, 1, 2, 1, 2, 2, 1, 2, 1, 14, 2, 1, 3, 2, 1, 2, 1, 1, 7, 1, 3, 2, 5, 1, 2, 1

Offset: 0

Alois P. Heinz, Nov 03 2013

This type of sequence was first studied by Andrica and Vacaretu. - Jonathan Sondow, Nov 06 2013

			a(8) = 3: 8 = -1-4-9-16+25-36+49 = -1-4+9+16-25-36+49 = -1+4+9-16+25+36-49.
a(9) = 2: 9 = -1-4+9+16+25-36 = 1+4+9-16-25+36.
a(10) = 1: 10 = -1+4-9+16.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Dorin Andrica and Daniel Vacaretu, Representation theorems and almost unimodal sequences, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33.

Cf. A000330, A231015, A231272.

Cf. A083527, A158092 (extremal sums).

b:= proc(n, i) option remember; (m->`if`(n>m, 0, `if`(n=m, 1,
      b(n+i^2, i-1) +b(abs(n-i^2), i-1))))((1+(3+2*i)*i)*i/6)
    end:
a:= proc(n) local k; for k while b(n, k)=0 do od; b(n, k) end:
seq(a(n), n=0..100);

b[n_, i_] := b[n, i] = Function[m, If[n > m, 0, If[n == m, 1,
   b[n+i^2, i-1] + b[Abs[n-i^2], i-1]]]][(1+(3+2*i)*i)*i/6];
a[n_] := Module[{k}, For[k = 1, b[n, k] == 0, k++]; b[n, k]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Sep 01 2022, after Alois P. Heinz *)

From Jonathan Sondow, Nov 03 2013: (Start)
a(n(n+1)(2n+1)/6) = 1 for n > 0: n(n+1)(2n+1)/6 = 1+4+9+...+n^2. See A000330.
a(n(n+1)(2n+1)/6 - 2) = 1 for n > 1: n(n+1)(2n+1)/6 - 2 = -1+4+9+...+n^2. (End)

A231016 Numbers m with non-unique solution to m = +- 1^2 +- 2^2 +- ... +- k^2 with minimal k giving at least one solution.

Original entry on oeis.org

0, 8, 9, 16, 18, 25, 31, 32, 33, 34, 39, 40, 41, 42, 43, 46, 48, 50, 52, 54, 58, 61, 67, 69, 74, 75, 77, 79, 80, 82, 84, 85, 87, 88, 90, 93, 95, 96, 97, 99, 101, 103, 104, 105, 107, 110, 111, 113, 115, 116, 117, 118, 121, 123, 127, 129, 131, 133, 135, 137, 141

Offset: 1

Jonathan Sondow, Nov 06 2013

nonn

The minimal k = A231015(m).

Complement of A231272.

			0 = 1 + 4 - 9 + 16 - 25 - 36 + 49 = sum with signs reversed, so 0 is a member.
9 = - 1 - 4 + 9 + 16 + 25 - 36 = 1 + 4 + 9 - 16 - 25 + 36, so 9 is a member.
A000330(k) = k(k+1)(2k+1)/6 = 1^2 + 2^2 + ... + k^2 is not a member, for k > 0.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Andrica, D., Vacaretu, D., Representation theorems and almost unimodal sequences, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33.

Cf. A000330, A231015, A231071, A231272.

b:= proc(n, i) option remember; local m, t; m:= (1+(3+2*i)*i)*i/6;
      if n>m then 0 elif n=m then 1 else
         t:= b(abs(n-i^2), i-1);
         if t>1 then return 2 fi;
         t:= t+b(n+i^2, i-1); `if`(t>1, 2, t)
      fi
    end:
a:= proc(n) option remember; local m, k;
      for m from 1+ `if`(n=1, -1, a(n-1)) do
        for k while b(m, k)=0 do od;
        if b(m, k)>1 then return m fi
      od
    end:
seq(a(n), n=1..80);  # Alois P. Heinz, Nov 06 2013

b[n_, i_] := b[n, i] = Module[{m, t}, m = (1+(3+2*i)*i)*i/6; Which[n>m, 0, n == m, 1, True, t = b[Abs[n-i^2], i-1]; If[t>1, Return[2]]; t = t + b[n+i^2, i-1]; If[t>1, 2, t]]]; a[n_] := a[n] = Module[{m, k}, For[m = 1 + If[n == 1, -1, a[n-1]], True, m++, For[k = 1, b[m, k] == 0, k++]; If[b[m, k]>1, Return[m]]]]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jan 28 2014, after Alois P. Heinz *)

{ n : A231071(n) > 1 }.

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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.

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A231071 Number of solutions to n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for minimal k giving at least one solution.

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A231016 Numbers m with non-unique solution to m = +- 1^2 +- 2^2 +- ... +- k^2 with minimal k giving at least one solution.

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