A334535 a(1) = 1, a(2) = 2, and for n > 2, a(n) = 2*a(a(n-i))+a(n-1)-a(n-2) where i >= 2 is the smallest number that satisfies a(n-i) < n.

1, 2, 3, 5, 8, 19, 27, 24, 45, 69, 72, 51, 27, 24, 45, 69, 72, 51, 27, 30, 57, 81, 78, 51, 75, 126, 153, 333, 486, 459, 891, 1350, 1377, 945, 486, 459, 891, 1350, 1377, 945, 486, 459, 891, 1350, 1377, 945, 486, 459, 891, 1350, 1377, 2781, 4158, 4131, 2727, 1350

Offset

1,2

Comments

The sequence is piecewise-periodic.

Links

Smirnov Vladimir, Table of n, a(n) for n = 1..3000

Formula

a(n) = 2*a(a(n-i))+a(n-1)-a(n-2) where i = 2, unless a(n-i) >= n, in which case i = 3,4,5,6...

Example

For n = 6, a(n-i) = a(6-2) = a(4) = 5; a(6) = 2*a(5)+a(5)-a(4) = 19.
For n = 7, a(n-i) = a(7-2) = a(5) = 8; but a(n-i)>n, then a(n-i) = a(7-3) = a(4) = 5; a(7) = 2*a(5)+a(6)-a(5) = 27.
For n = 9, a(n-i) = a(9-2) = a(7) = 27; but a(n-i)>n, then a(n-i) = a(9-3) = a(6) = 19; but a(n-i)>n, then a(n-i) = a(9-4) = a(5) = 8; a(9) = 2*a(8)+a(8)-a(7) = 45.
Simplified calculation option for n = 31, a(n-i) = a(31-2) = a(29) = 486; but a(n-i)> n, visually find in the sequence such a(n) that is located closest to the end of the sequence and less than n: this is a(20) = 30, then a(n-i) = 30; a(31) = 2 * a(30) + a(30)- a(29) = 891.

Maple

a[1] := 1:
a[2] := 2:
for n from 3 to 100 do
  i := 2;
  while a[n-i] >= n do i := i+1;
  end do:
  a[n] := 2*a[a[n-i]]+a[n-1]-a[n-2]
end do:
seq(a[n], n=1..100);

Mathematica

a[1]=1; a[2]=2; For[n=3, n<=100, n++, i=2; While[a[n-i]>=n, i++]; a[n]= 2*a[a[n-i]]+a[n-1]-a[n-2]]; Table[a[n], {n, 1, 100}]

Prog

(C) int main() {
int a[100];
a[1]=1;
a[2]=2;
printf("%d\n", 1);
printf("%d\n", 2);
for (int n=3; n<=99; n++)
{int i=2;
while (a[n-i]>=n) {i++; }
a[n]=2*a[a[n-i]]+a[n-1]-a[n-2];
printf("%d\n", a[n]); }
return 0; }
(PARI) lista(nn) = {my(va = vector(nn)); va[1] = 1; va[2] = 2; for (n=3, nn, my(i = 2); while(va[n-i] >= n, i++); va[n] = 2*va[va[n-i]]+va[n-1]-va[n-2]; ); va; } \\ Michel Marcus, May 09 2020
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
def A334535(n):
    if n <= 2:
        return n
    i, a, b = 2, A334535(n-1), A334535(n-2)
    q = b
    while q >= n:
        i += 1
        q = A334535(n-i)
    return 2*A334535(q)+a-b # Chai Wah Wu, Jun 30 2020

Crossrefs

Cf. A030118.

Sequence in context: A025071 A319644 A049908 * A135568 A103004 A102016

Adjacent sequences: A334532 A334533 A334534 * A334536 A334537 A334538

Keyword

nonn

Author

Smirnov Vladimir, May 05 2020

Status

approved