%I #30 Sep 29 2025 23:26:57

%S 3,5,9,14,17,22,26,33,38,42,50,60,65,70,74,82,92,98,108,116,129,134,

%T 138,146,156,162,172,180,194,204,212,228,248,257,262,266,274,284,290,

%U 300,308,322,332,340,356,376,386,396,404,420,440,452,472,488,513,518

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

%C 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

%F 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).

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

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

%t Select[Range[500], DigitCount[#, 2, 1] == IntegerExponent[#, 2] + 2 &] (* _Amiram Eldar_, Jul 04 2022 *)

%o (PARI) r=quadgen(5);

%o 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}

%o (PARI) isok(k) = hammingweight(k) == valuation(k, 2) + 2; \\ _Michel Marcus_, Jul 06 2022

%o (Python)

%o from itertools import count, islice

%o def A355489_gen(startvalue=1): # generator of terms >= startvalue

%o return filter(lambda n:n.bit_count()==(n&-n).bit_length()+1,count(max(startvalue,1)))

%o A355480_list = list(islice(A355489_gen(),30)) # _Chai Wah Wu_, Jul 15 2022

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

%K nonn,base

%O 1,1

%A _Mikhail Kurkov_, Jul 04 2022 [verification needed]