A001033 Numbers n such that the sum of the squares of n consecutive positive odd numbers x^2 + (x+2)^2 + ... + (x+2n-2)^2 = k^2 for some integer k. The least values of x and k for each n are in A056131 and A056132, respectively.
1, 16, 25, 33, 49, 52, 64, 73, 97, 100, 121, 148, 169, 177, 193, 196, 241, 244, 249, 256, 276, 289, 292, 297, 313, 337, 361, 388, 393, 400, 409, 457, 481, 484, 528, 529, 537, 577, 592, 625, 628, 649, 673, 676, 708, 724, 753, 772, 784, 793, 832, 841, 852, 897, 913, 961, 964, 976, 996
Offset: 1
Examples
a(1) = 1 from 1^2. a(2) = 16 from 27^2 + 29^2 + ... + 55^2 + 57^2 = 172^2. a(4) = 33 from 91^2 + 93^2 + ... + 153^2 + 155^2 = 715^2.
References
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Christopher E. Thompson, Table of n, a(n) for n = 1..7103 (up to 250000, extending first 100 terms computed by T. D. Noe).
- U. Alfred, Sums of squares of consecutive odd integers, Math. Mag., 40 (1967), 194-199.
- J. A. H. Hunter, A note on sums of squares of consecutive odd numbers, Math. Mag. 42 (1969), 145.
- Andrzej Makowski, Remark on the paper "Sums of squares of consecutive odd numbers", Math. Mag. 43 (1970), 212-213.
- William Sollfrey, Note on sums of squares of consecutive odd integers, Math. Mag. 41 (1968), 255-258.
Programs
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Mathematica
r[1] = {True, {1, 1}}; r[n_] := (rn = Reduce[x > 0 && k > 0 && Sum[(x + 2*j)^2, {j, 0, n - 1}] == k^2, {x, k}, Integers]; srn = Simplify[(rn /. C[1] -> 0) || (rn /. C[1] -> 1) || (rn /. C[1] -> 2)]; rnOdd = Which[rn === False, False, srn[[0]] === And, srn, True, Select[srn, OddQ[x /. ToRules[#1]] & ]]; If[ rnOdd === False, {False, {0, 0}}, {True, {x, k} /. Flatten[{ToRules[rnOdd]}]}]); A001033 = Reap[Do[rn = r[n]; {x0, k0} = rn[[2]]; If[rn[[1]] && OddQ[x0], Print[{n, x0, k0}]; Sow[n]], {n, 1, 1000}]][[2, 1]] (* Jean-François Alcover, Mar 14 2012 *)
Formula
We must solve m*(3*x^2 + 6*m*x - 6*x + 4*m^2 - 6*m + 2)/3 = k^2 in integers (x, m, k). - N. J. A. Sloane
For a given n, we must determine whether the generalized Pell equation 4n*y^2 + 4y*n^2 + n(4n^2-1)/3 = k^2 has any integer solutions with y >= 0. Note that x = 2y+1 will be the first odd number being squared. If there are solutions then n is in this sequence. - T. D. Noe, Oct 24 2007
Extensions
More terms from Robert G. Wilson v
Corrected and extended by T. D. Noe, Oct 24 2007
1024 was missing from b-file. - Christopher E. Thompson, Feb 05 2016
Comments