A141435 a(1) = 1, a(2) = 2; a(n) = a(n-a(1)) + a(n-a(2)) + a(n-a(3)) + a(n-a(4)) + ...

1, 2, 3, 6, 11, 20, 38, 71, 132, 247, 461, 861, 1609, 3005, 5613, 10485, 19584, 36581, 68330, 127632, 238404, 445314, 831798, 1553712, 2902170, 5420945, 10125754, 18913838, 35329048, 65990929, 123264078, 230244265, 430071949, 803328933

Offset: 1

Author

Raes Tom (tommy1729(AT)hotmail.com), Aug 06 2008

Keywords

• easy
• nonn

Comments

Thus we get a self-reference sequence that grows exponentially. a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-6) + a(n-11) + a(n-20) + ...

A Fibonacci-like sequence, even closer to the tribonacci numbers.

Lim n-> oo log (a(n))/n converges.

Examples

			a(6) = 20 because 20 = a(5) + a(4) + a(3) = 11 + 6 + 3
a(8) = 71 because 71 = a(7) + a(6) + a(5) + a(2) = 38 + 20 + 11 + 2
		

Crossrefs

Cf. A058265, A008937, A001590, A000213, A000073, A096436, A102575, A111129.

Programs

Maple

A141435 := proc(n) option remember; local a,i; if n <= 3 then RETURN(n); else a :=0 ; for i from 1 to n-1 do if n-procname(i) < 1 then RETURN(a); else a := a+procname(n-procname(i)) ; fi; od; RETURN(a); fi; end: for n from 1 to 80 do printf("%d,",A141435(n)) ; od: # R. J. Mathar, Nov 03 2008

Python

def A141435(terms):
    seq = [1, 2]
    for n in range(3, terms):
        s = 0
        for m in seq:
            if (n - m) > 0:
                s += seq[n - m - 1] #fix for python indexing
        seq.append(s)
    return seq
print(A141435(40)) # Andres Cruz y Corro A, Jun 19 2019

Extensions

More terms from R. J. Mathar, Nov 03 2008