A381580 - OEIS

%I #7 Feb 28 2025 12:08:12

%S 0,1,2,4,9,12,15,18,22,33,44,56,64,72,80,88,96,104,112,120,128,136,

%T 145,174,203,232,261,290,319,348,378,399,420,441,462,483,504,525,546,

%U 567,588,609,630,651,672,693,714,735,756,777,798,819,840,861,882,903,924,945,966,988

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

%C 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.

%H Amiram Eldar, <a href="/A381580/b381580.txt">Table of n, a(n) for n = 1..10000</a>

%e The first 10 terms are:

%e    n  a(n) A381579(a(n))

%e    ---------------------

%e    1   0               0

%e    2   1               1

%e    3   2               2

%e    4   4              11

%e    5   9             101

%e    6  12             111

%e    7  15             121

%e    8  18             202

%e    9  22            1001

%e   10  33            1111

%t 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]

%o (PARI) mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));

%o 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;}

%Y Cf. A000045, A381579.

%Y Subsequence: A055588.

%Y Similar sequences: A002113, A006995, A094202, A331191.

%K nonn,easy,base,new

%O 1,3

%A _Amiram Eldar_, Feb 28 2025