%I #30 May 05 2024 09:49:24
%S 1,1,1,3,5,9,17,31,57,15,13,85,113,211,49,373,633,155,1161,1949,3265,
%T 6375,11589,21229,39193,7211,67633,11437,86281,165351,26369,2781,
%U 19451,4861,2793,2715,1369,6877,1961,127,8965,1153,1245,11363,13761,26369,51493,91623,169485,31261
%N Zeroless analog of tribonacci numbers.
%C At n = 208666297 this sequence enters a cycle that has a period of 300056874 and begins: 2847, 26331, 5851, ... (only the first 3 terms of the cycle are needed to reproduce the entire cycle).
%C This can be compared with the sequence A243063, which enters a cycle that has a (relatively) small period of 912.
%F a(n) = Zr(a(n-1) + a(n-2) + a(n-3)), where the function Zr(k) removes all zero digits from k.
%e a(9) = Zr(a(8) + a(7) + a(6)) = Zr(17 + 31 + 57) = Zr(105) = 15.
%t a[0]=a[1]=a[2]=1; a[n_]:=FromDigits[DeleteCases[IntegerDigits[a[n-1]+a[n-2]+a[n-3]],0]]; Array[a,50,0] (* _Stefano Spezia_, Apr 12 2024 *)
%o (Python)
%o def a(n):
%o a, b, c = 1, 1, 1
%o for _ in range(n):
%o a, b, c = b, c, int(str(a+b+c).replace('0', ''))
%o return a
%o (Python) # faster for initial segment of sequence
%o from itertools import islice
%o def agen(): # generator of terms
%o a, b, c = 1, 1, 1
%o while True: yield a; a, b, c = b, c, int(str(a+b+c).replace("0", ""))
%o print(list(islice(agen(), 50))) # _Michael S. Branicky_, Apr 13 2024
%Y Cf. A000045, A000213, A004719, A242350, A243657, A243658, A243063.
%K nonn,base
%O 0,4
%A _Bryle Morga_, Apr 11 2024