A355489 Numbers k such that A000120(k) = A007814(k) + 2.

3, 5, 9, 14, 17, 22, 26, 33, 38, 42, 50, 60, 65, 70, 74, 82, 92, 98, 108, 116, 129, 134, 138, 146, 156, 162, 172, 180, 194, 204, 212, 228, 248, 257, 262, 266, 274, 284, 290, 300, 308, 322, 332, 340, 356, 376, 386, 396, 404, 420, 440, 452, 472, 488, 513, 518

Offset: 1

Author

Mikhail Kurkov, Jul 04 2022 [verification needed]

Keywords

• nonn
• base

Comments

Each term k, doubled, can be put into a one-to-one correspondence with a maximal Schreier set (a subset of the positive integers with cardinality equal to the minimum element in the set) by interpreting the 1-based position of the ones in the binary expansion of 2*k (where position 1 corresponds to the least significant bit) as the elements of the corresponding maximal Schreier set. See A373556 for more information. Cf. also A371176. - Paolo Xausa, Jun 13 2024

Crossrefs

Cf. A000045, A000120, A007814, A010056, A025480, A048679, A072649, A371176, A373556.

Programs

Mathematica

Select[Range[500], DigitCount[#, 2, 1] == IntegerExponent[#, 2] + 2 &] (* Amiram Eldar, Jul 04 2022 *)

PARI

r=quadgen(5);
A355489_upto(nMax)={my(v1,v2,v3,v4); v1=vector(nMax,i,0); v1[1]=1; for(i=1,nMax-1,v1[i+1]=v1[i\r+1]+1); v2=vector(nMax,i,0); v2[1]=1; for(i=2,nMax,v2[i]=v1[i]-v1[i-1]); v3=vector(nMax,i,0); for(i=1,3,v3[i]=2^(i-1)); for(i=4,nMax,v3[i]=if(v2[i-1]==1,5,2*v3[i-fibonacci(v1[i-1]+1)]-if(v2[i]==1,1,0))); v4=vector(nMax,i,0); v4[1]=3; for(i=2,nMax,v4[i]=v4[i-1]+v3[i]); v4}

PARI

isok(k) = hammingweight(k) == valuation(k, 2) + 2; \\ Michel Marcus, Jul 06 2022
(Python 3.10+)
from itertools import count, islice
def A355489_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:n.bit_count()==(n&-n).bit_length()+1,count(max(startvalue,1)))
A355480_list = list(islice(A355489_gen(),30)) # Chai Wah Wu, Jul 15 2022

Formula

a(n) = a(n-1) + b(n) for n > 1 with a(1) = 3 where b(n) = {2^(n-1) if n < 4; 5 if c(n-1) = 1; otherwise 2*b(n - A000045(A072649(n-1) + 1)) - [c(n) = 1]} and where c(n) = A010056(n).

A025480(a(n)-1) = A048679(n) for n > 0.

a(A000045(n)) = 2^(n-1) + 1 for n > 1.