A371176 - OEIS

%I #59 Jun 13 2024 03:35:12

%S 1,2,4,6,8,10,12,16,18,20,24,28,32,34,36,40,44,48,52,56,64,66,68,72,

%T 76,80,84,88,96,100,104,112,120,128,130,132,136,140,144,148,152,160,

%U 164,168,176,184,192,196,200,208,216,224,232,240,256,258,260,264,268

%N Numbers k such that A000120(k) <= A001511(k).

%C It appears that this sequence is obtained when ordering Schreier sets as explained in the Bird link. See decM(n) PARI code. - _Michel Marcus_, May 31 2024

%C That is correct since the binary representation of these numbers can be put into 1-to-1 correspondence with Schreier sets, which satisfy |X| <= min X, using the indicator function of X as the bits (starting from the right, LSB). The reason is that A000120 then computes |X| and A001511 computes min X. For example, the Schreier set X = {2, 5} can be mapped to 10010_2 = 18. - _Michael S. Branicky_, May 31 2024

%C From _David A. Corneth_, May 31 2024: (Start)

%C If k is in the sequence then so is 2*k.

%C a(A000045(k)) = 2^(k-2) for k >= 2. (End)

%C Apart from a(1), all terms are even. - _Paolo Xausa_, May 31 2024

%H Michel Marcus, <a href="/A371176/b371176.txt">Table of n, a(n) for n = 1..10945</a>

%H Alistair Bird, <a href="https://outofthenormmaths.wordpress.com/2012/05/13/jozef-schreier-schreier-sets-and-the-fibonacci-sequence/">Jozef Schreier, Schreier sets and the Fibonacci sequence</a>, Out Of The Norm blog, May 13 2012.

%F a(n) = b(n)*A001316(b(n))/2 where b(n) = A048679(n).

%F a(n) = Sum_{i=0..n-1} 2^A213911(i).

%F a(n) = 2^(A072649(n) - 1) + [c(n) > 0]*2*a(c(n)) where c(n) = A066628(n).

%F a(n) = 2*a(A005206(n)) + [A003849(n) = 1]*2^A007895(n-1) for n > 1 with a(1) = 1.

%t Join[{1}, Select[Range[2, 1000, 2], DigitCount[#, 2, 1] <= IntegerExponent[#, 2] + 1 &]] (* _Paolo Xausa_, May 31 2024 *)

%o (PARI) isok(n) = hammingweight(n) <= (valuation(n, 2) + 1)

%o (PARI) b1(n) = if(n == 0, 0, my(A = 0); forstep(i = logint(n, 2), 0, -1, if(A >= i, break); if(bittest(n, i), A++)); (n >> A)*(1 << A))

%o upto(n, m) = my(A = b1(m), B, v1); v1 = vector(n, i, 0); for(i = 1, n, B = hammingweight(A) - 1; A += 2^(B - (A/2^B)%2 + 1); v1[i] = A); v1 \\ first n terms greater than m \\  [verification needed]

%o (PARI) Zeckendorf(n) = my(A = n, B1, B2, C, m = 1, v1); v1 = [n > 0]; while(fibonacci(m+1) <= n, m++); m--; B1 = fibonacci(m+1); B2 = fibonacci(m); A -= B1; while(m > 1, m--; C = 0; if(A >= B2, A -= B2; C = 1); [B1, B2] = [B2, B1 - B2]; v1 = concat(v1, C)); v1

%o a(n) = my(v1); v1 = Zeckendorf(n); for(i = 2, #v1, if(v1[i], v1 = concat(vector(#v1-1, j, v1[j + (j >= (i-1))]), 0))); fromdigits(v1, 2) \\  [verification needed]

%o (PARI) M(n) = my(list=List()); for (i=1, n, forsubset(i, s, my(bOk = if (#s && (vecmax(s) == n), #s <= vecmin(s), 0)); if (bOk, listput(list, vecsort(Vec(s),,4))););); Vec(list);

%o decM(nn) = my(v = vector(nn, k, M(k)), list=List()); for (i=1, #v, my(vi = v[i]); for (j=1, #vi, my(s = vecsort(vi[j]), slist=List(), m = vecmax(s)); forstep(k=m, 1, -1, listput(slist, sign(vecsearch(s, k)))); listput(list, fromdigits(Vec(slist), 2)););); vecsort(Vec(list)); \\ _Michel Marcus_, May 31 2024

%o (Python)

%o def ok(n): return n.bit_count() <= (-n&n).bit_length()

%o print([k for k in range(1, 300) if ok(k)]) # _Michael S. Branicky_, May 31 2024

%o (Python)  # Assuming the list starts with 0.

%o def a():

%o     n = na = nb = 1

%o     while True:

%o         yield not(nb < (na - 1) << 1)

%o         nb, na = na, n.bit_count()

%o         n += 1

%o aList = a(); print([n for n in range(77) if next(aList)]) # _Peter Luschny_, Jun 07 2024

%Y Cf. A000045, A000120, A001316, A001511, A003849, A005206, A007895, A048679, A066628, A072649, A213911.

%Y Cf. A355489, A373345, A373347 (complement), A373557.

%K nonn,base

%O 1,2

%A _Mikhail Kurkov_, Mar 14 2024