A342728 a(n) is the least number k such that A066323(k) = n.

0, 1, 2, 3, 4, 5, 6, 7, 23, 39, 55, 71, 87, 103, 359, 615, 871, 1127, 1383, 1639, 5735, 9831, 13927, 18023, 22119, 26215, 91751, 157287, 222823, 288359, 353895, 419431, 1468007, 2516583, 3565159, 4613735, 5662311, 6710887, 23488103, 40265319, 57042535, 73819751

Offset: 0

Author

Amiram Eldar, Mar 19 2021

Keywords

• nonn
• base
• easy

Comments

a(n) is the least number k whose sum of digits in base i-1 (or in base -4) is n.

Links

• Amiram Eldar, Table of n, a(n) for n = 0..1000
• Walter Penney, A "binary" system for complex numbers, Journal of the ACM, Vol. 12, No. 2 (1965), pp. 247-248.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,16,-16).

Crossrefs

Cf. A007608, A066321, A066323, A271472, A342725, A342726, A342727, A342729.

Programs

Mathematica

Join[{0}, LinearRecurrence[{1,0,0,0,0,16,-16}, Range[7], 50]]

Formula

a(n) = n for n <= 7, and a(n) = a(n-1) + 16*a(n-6) - 16*a(n-7) for n > 7.

G.f.: x*(1 + x + x^2 + x^3 + x^4 + x^5 - 15*x^6)/(1 - x - 16*x^6 + 16*x^7). - Stefano Spezia, Mar 20 2021

From Greg Dresden, Jun 21 2021: (Start)

a(3*n+1) = (24 + (4^n)*(25 - 9*(-1)^n))/40.

a(3*n+2) = (24 + (4^n)*(50 + 6*(-1)^n))/40.

a(3*n+3) = (24 + (4^n)*(75 + 21*(-1)^n))/40. (End)