A332581 a(0)=0, a(1)=1; for n>1, a(n) = max(sum0,sum1) mod n, where sum0 is the sum of all previous even terms, sum1 is the sum of all previous odd terms.

0, 1, 1, 2, 2, 4, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 5, 9, 17, 9, 16, 14, 12, 10, 16, 0, 28, 22, 8, 12, 20, 0, 32, 22, 0, 34, 22, 38, 27, 20, 33, 26, 45, 38, 20, 32, 5, 48, 35, 26, 43, 34, 2, 52, 36, 2, 54, 36, 61, 50, 24, 36, 60, 40, 67, 54, 24, 34, 54, 20, 25, 10, 5

Offset

0,4

Comments

For n > 2842, sum0 > sum1.

For all n > 54388, a(n) = 9514.

Indices of zeros: 0, 30, 36, 39, 3552, 8526, 9156, 11143, 12775, 36077, 38141.

Links

Robert Israel, Table of n, a(n) for n = 0..54388

Example

a(10) = max(2+2+4+2+2, 1+1+3+3) mod 10 = 12 mod 10 = 2.

Maple

R:= 0, 1: s0:= 0: s1:= 1:
for n from 2 to 100 do
  v:= max(s0, s1) mod n;
  R:= R, v;
  if v::odd then s1:= s1+v else s0:= s0+v fi
od:
R; # Robert Israel, Feb 16 2024

Prog

(Python)
a = [0, 1]
s0, s1 = 0, 1
for n in range(2, 1000):
    v = max(s0, s1) % n
    a.append(v)
    if (v & 1): s1 += v
    else:       s0 += v
print(a)

Crossrefs

Cf. A300392.

Sequence in context: A127796 A131287 A102640 * A328059 A123674 A363219

Adjacent sequences: A332578 A332579 A332580 * A332582 A332583 A332584

Keyword

nonn

Author

Alex Ratushnyak, Feb 16 2020

Status

approved