A381580 Numbers whose Chung-Graham representation (A381579) is palindromic.

0, 1, 2, 4, 9, 12, 15, 18, 22, 33, 44, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 145, 174, 203, 232, 261, 290, 319, 348, 378, 399, 420, 441, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 988

Offset: 1

Author

Amiram Eldar, Feb 28 2025

Keywords

• nonn
• easy
• base

Comments

The numbers of the form Fibonacci(2*k) + 1 (A055588) are all terms since A381579(A055588(0)) = 1, A381579(A055588(1)) = 2, and A381579(A055588(k)) = 10^(k-1)+1 (i.e., two 1's with k-2 0's between them) for k >= 2.

Examples

			The first 10 terms are:
   n  a(n) A381579(a(n))
   ---------------------
   1   0               0
   2   1               1
   3   2               2
   4   4              11
   5   9             101
   6  12             111
   7  15             121
   8  18             202
   9  22            1001
  10  33            1111
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A381579.

Subsequence: A055588.

Similar sequences: A002113, A006995, A094202, A331191.

Programs

Mathematica

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2*10^(k-1); m -= 2*f[k], s += 10^(k-1); m -= f[k]]]; PalindromeQ[s]]; Select[Range[0, 1000], q]

PARI

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
isok(n) = {my(s = 0, m = n, k, d); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2*10^(k-1); m -= 2*f(k), s += 10^(k-1); m -= f(k))); d = digits(s); Vecrev(d) == d;}