A355654 For any number n with runs in binary expansion (r_w, ..., r_0), let p(n) be the polynomial of a single indeterminate x where the coefficient of x^e is r_e for e = 0..w and otherwise 0, and let q be the inverse of p; a(n) = q(p(n)^2).

0, 1, 9, 15, 271, 313, 481, 511, 33279, 34785, 39993, 40719, 61455, 61689, 65409, 65535, 16842751, 17039233, 17809657, 17821711, 20455183, 20479033, 20842465, 20939263, 31457791, 31465441, 31584313, 31588111, 33488911, 33489913, 33553921, 33554431, 34393292799

Offset: 0

Author

Rémy Sigrist, Jul 12 2022

Keywords

• nonn
• base

Comments

In other words, a(n) encodes the square of the polynomial encoded by n.

Examples

			The first terms, alongside their binary expansions and corresponding polynomials, are:
  n  a(n)   bin(n)  bin(a(n))         p(n)           p(a(n))
  -  -----  ------  ----------------  -------------  -----------------------------
  0      0       0                 0              0                              0
  1      1       1                 1              1                              1
  2      9      10              1001          x + 1                  x^2 + 2*x + 1
  3     15      11              1111              2                              4
  4    271     100         100001111          x + 2                  x^2 + 4*x + 4
  5    313     101         100111001    x^2 + x + 1  x^4 + 2*x^3 + 3*x^2 + 2*x + 1
  6    481     110         111100001        2*x + 1                4*x^2 + 4*x + 1
  7    511     111         111111111              3                              9
  8  33279    1000  1000000111111111          x + 3                  x^2 + 6*x + 9
  9  34785    1001  1000011111100001  x^2 + 2*x + 1  x^4 + 4*x^3 + 6*x^2 + 4*x + 1
		

Links

• Rémy Sigrist, Binary plot of the sequence for n < 2^10
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A005811, A101211, A212739, A355653.

Programs

PARI

toruns(n) = { my (r=[]); while (n, my (v=valuation(n+n%2, 2)); n\=2^v; r=concat(v, r)); r }
fromruns(r) = { my (v=0); for (k=1, #r, v=(v+k%2)*2^r[k]-k%2); v }
a(n) = { fromruns(Vec(Pol(toruns(n))^2)) }

Formula

A005811(a(n)) = 2*A005811(n) - 1 for any n > 0.

a(2^n - 1) = 2^(n^2) - 1.