A100476 a(n) = A000720(Sum_{j=1..4} a(n-j)) with a(1)=a(2)=a(3)=a(4)=1.

1, 1, 1, 1, 2, 3, 4, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 16, 18, 18, 19, 20, 21, 21, 22, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Author

Jonathan Vos Post, Nov 22 2004

Keywords

• easy
• nonn

Comments

For n > 29 we have a(n) = 24. Starting with other values of a(1), a(2), a(3), a(4) what behaviors are possible? The sequence is in any case bounded. If for some k a(k) + a(k+1) + a(k+2) + a(k+3) > 400, then a(k+4) is smaller than the average of a(k), a(k+1), a(k+2) and a(k+3), which means that the sequence will always stick at a single integer after some point or go into a loop. Are there values a(1), a(2), a(3), a(4) such that the sequence would indeed exhibit cyclic behavior?

a(n) = 24 for 30 <= n <= 10^7. - G. C. Greubel, Apr 06 2023

Examples

			a(6) = A000720(a(2)+a(3)+a(4)+a(5)) = A000720(5) = 3.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1).

Crossrefs

Cf. A000720, A100478.

Programs

Mathematica

a={1,1,1,1}; Do[ AppendTo[a,PrimePi[a[[-1]]+a[[-2]]+a[[-3]]+a[[-4]]]], {70}]; a
RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==PrimePi[a[n-1]+ a[n-2]+ a[n-3]+a[n-4]]},a[n],{n,80}] (* Harvey P. Dale, Sep 19 2011 *)

SageMath

@CachedFunction
def a(n): # a = A100476
    if (n<5): return 1
    else: return prime_pi( sum(a(n-j) for j in range(1,5)) )
[a(n) for n in range(1,81)] # G. C. Greubel, Apr 06 2023

Extensions

Edited by Stefan Steinerberger, Aug 08 2007