A362531 The smallest integer m such that m mod 2k >= k for k = 1, 2, 3, ..., n.

1, 3, 3, 15, 15, 47, 95, 95, 287, 335, 1199, 1199, 1295, 2015, 2879, 2879, 2879, 2879, 2879, 2879, 2879, 43199, 211679, 211679, 211679, 211679, 211679, 211679, 211679, 211679, 3084479, 3084479, 3084479, 3084479, 3084479, 3084479, 302702399, 469909439

Offset: 1

Author

Tomohiro Yamada, Apr 24 2023

Keywords

• nonn

Comments

Take the square array A(k, l) with k= 1, 2, ... and l = 0, 1, ... such that for each k, A(k, l) takes k zeros and then k ones alternately:

0, 1, 0, 1, 0, 1, 0, 1, ...

0, 0, 1, 1, 0, 0, 1, 1, ...

0, 0, 0, 1, 1, 1, 0, 0, ...

...

Then the a(n)-th column is the first column which begins with n ones.

Examples

			a(3) = 3 since 3 mod 2 = 1, 3 mod 4 = 3 >= 2, 3 mod 6 = 3 (but 3 mod 8 = 3 < 4) while 1 mod 4 = 1 < 2, 2 mod 2 = 0 < 1.
		

Links

• Tomohiro Yamada, Fast Python program, the original program written by nazratt2.

Crossrefs

Cf. A053664.

Programs

PARI

a(n)={my(m);m=1;while(vecmin(vector(n,j,(m%(2*j))/j))<1,m=m+1);m}

PARI

n=1;for(m=1,10^9,while(vecmin(vector(n,j,(m%(2*j))/j))>=1,print(n," ",m);n=n+1))

Python

def A362531(n):
    m = 1
    while True:
        for k in range(n,0,-1):
            if (l:=k-m%(k<<1))>0:
                break
        else:
            return m
        m += l # Chai Wah Wu, Jun 21 2023

Extensions

a(37)-a(38) from Yifan Xie, Apr 24 2023