A327759 a(1) = 1; a(2) = 2; a(n) = n - max{k

1, 2, 2, 3, 1, 2, 1, 4, 5, 1, 2, 4, 1, 5, 1, 2, 5, 1, 2, 8, 4, 5, 3, 4, 3, 6, 1, 8, 3, 2, 5, 4, 7, 6, 2, 3, 9, 1, 2, 12, 4, 5, 3, 4, 8, 9, 7, 8, 3, 10, 1, 12, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 2, 3, 13, 1, 2, 16, 4, 5, 3, 4, 8, 9, 7, 8, 12, 13, 11, 12, 3

Offset: 1

Author

Ali Sada, Sep 24 2019

Keywords

• nonn

Comments

From M. F. Hasler, Sep 29 2019: (Start)

The sequence may be seen as a table with rows of length |2n-1|, n = 0, 1, ...

Then from n = 5, a(18) = 1 on, the rows are of the form

row(n) = (1, 2, 2n-2, ((4k, 4k+1, 4k-1, 4k), k=1..(n-3)/2), 3, 2n-4) for odd n,

row(n) = (1, 2n-4, ((2k+1, 2k), k=1..n-3), 2, 3, 2n-3) for even n.

All rows n >= 5 start with a((n-1)^2 + 2) = 1, and there are no other '1's beyond a(15). (End)

Examples

			a(9) is odd. The largest term up to that point is 5. The largest index of 5 is 9. a(10) = 10 - 9 = 1.
a(16) is even. The second largest term up to that point is 4. The largest index of 4 is 12. a(17) = 17 - 12 = 5.
From _M. F. Hasler_, Sep 29 2019: (Start)
Written as a table with rows of length |2n-1|, n = 0, 1, ...:
   1,  /* row n=0 */
   2,  /* row n=1; from here on, length = 2n-1 */
   2, 3, 1,  /* row n=2 */
   2, 1, 4, 5, 1,  /* row n=3 */
   2, 4, 1, 5, 1, 2, 5,  /* n=4 */
   1, 2, 8, 4, 5, 3, 4, 3, 6,  /* n=5. Here starts the regular pattern. */
   1, 8, 3, 2, 5, 4, 7, 6, 2, 3, 9,  /* n=6 */
   1, 2, 12, 4, 5, 3, 4, 8, 9, 7, 8, 3, 10,  /* n=7 */
   1, 12, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 2, 3, 13,  /* n=8 */
   1, 2, 16, 4, 5, 3, 4, 8, 9, 7, 8, 12, 13, 11, 12, 3, 14,  /* n=9 */
   1, 16, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14, 2, 3, 17,  /* n=10 */
   ...
(End)
		

Links

• John Tyler Rascoe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A059100 (indices of '1's, starting with 18), A141044 (col.1, starting at row 3).

Programs

Mathematica

s={1, 2}; sm = 2; sm2 = 1; Do[a = Length[s] + 1 - If[OddQ[s[[-1]]], Position[s, ?(# == sm &)], Position[s, ?(# == sm2 &)]][[-1, 1]]; AppendTo[s, a]; If[a > sm, sm2 = sm; sm = a,If[a < sm && a > sm2, sm2 = a]], {100}]; s (* Amiram Eldar, Sep 28 2019 *)

PARI

A327759_upto(N=99, idx=[0,0], L, S, a)=vector(N,n,a=n-if(n>2,idx[2-a%2]); LM. F. Hasler, Sep 29 2019

PARI

A327759(n)={my(r=sqrtint(abs(n-2))+1,c=n-(r-1)^2-1); if(n<17, digits(1223121451241512)[n], c==1, 1, c==2*r-2, 3,c==2*r-1, 2*r-3-r%2, r%2, if(c==3, 2*r-2, c>2, c\4*4+[0,1,-1,0][c-c\4*4+1], 2), c==2, 2*r-4, c<2*r-3, c\/2*2+(c%2)-2,2)} \\ M. F. Hasler, Sep 30 2019

Python

def A327759list(nmax):
    A = [1,2]
    for n in range(3,nmax+1):
        if A[-1]%2 == 0:
            A2 = list(set(A))
            A2.sort()
            m = A2[-2]
        else: m = max(A)
        i = len(A) - 1
        while A[i] != m: i -= 1
        A.append(n-i-1)
    print(A) # John Tyler Rascoe, Jan 13 2023

Formula

a(n) = 1 iff n is in {1, 5, 7, 10, 13, 15} union A059100 \ { 2, 3, 6, 11 }.