A055042 Numbers of the form 2^(2i+1)*(8*j+5).
10, 26, 40, 42, 58, 74, 90, 104, 106, 122, 138, 154, 160, 168, 170, 186, 202, 218, 232, 234, 250, 266, 282, 296, 298, 314, 330, 346, 360, 362, 378, 394, 410, 416, 424, 426, 442, 458, 474, 488, 490, 506, 522, 538, 552, 554, 570, 586, 602, 616
Offset: 1
Examples
42 = 21*2 = 2^(2*0 + 1)*(8*2 + 5) is in the sequence. - _David A. Corneth_, Apr 18 2021
References
- Burton W. Jones, The Arithmetic of Quadratic Forms, Carus Monograph 10, Math. Assoc. America, 1967; Problem 60, p. 204.
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000
- L. J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).
Crossrefs
Cf. A055045.
Programs
-
Mathematica
With[{max = 700}, Flatten[Table[2^(2*i + 1)*(8*j + 5), {i, 0, (Log2[max] - 1)/2}, {j, 0, Floor[(max/2^(2*i + 1) - 5)/8]}]] // Sort] (* Amiram Eldar, Mar 29 2025 *)
-
PARI
upto(n) = { my(res = List()); for(i = 0, logint(n\2, 2), forstep(j = 5, n>>(2*i+1), 8, listput(res, 4^i*2*j) ) ); Set(res) } \\ David A. Corneth, Apr 18 2021
-
Python
def A055042(n): def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 kmin = kmax >> 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def f(x): return n+x-sum(((x>>(i<<1)+1)-5>>3)+1 for i in range(x.bit_length()-1>>1)) return bisection(f,n,n) # Chai Wah Wu, Mar 19 2025
Formula
a(n) = 2*A055045(n). - Chai Wah Wu, Mar 19 2025
Comments