A360643 a(n) is the least A000120-perfect number (A175522) whose binary weight (A000120) is n, or 0 if no such number exists.

2, 0, 25, 169, 841, 95, 247, 943, 767, 5999, 6139, 16123, 30655, 90109, 122847, 245695, 522237, 1572591, 1966015, 3932095, 12582651, 28311519, 33423343, 100663023, 133693435, 402128831, 931135479, 1069547515, 1610612607, 11802771447, 12884901567, 25736249279

Offset: 1

Author

Amiram Eldar, Feb 15 2023

Keywords

• nonn
• base

Comments

Apparently, the sequence is increasing after n = 6.

Examples

			a(1) = 2 since A000120(2) = 1 and A093653(2)/A000120(2) = 4/2 = 2.
a(2) = 0 since there is no number m with binary weight 2 and with A093653(m) = 4.
a(3) = 25 since A000120(25) = 3 and A093653(25)/A000120(25) = 6/3 = 2, and 25 is the least number with this property.
		

Crossrefs

Cf. A000120, A093653, A175522.

Programs

Mathematica

seq[len_, nmax_] := Module[{s = Table[-1, {len}], n = 3, c = 2, bw, dbw}, s[[1]] = 2; While[c < len && n <= nmax, bw = DigitCount[n, 2, 1]; If[bw <= len && s[[bw]] < 0, dbw = DivisorSum[n, DigitCount[#, 2, 1] &]; If[dbw == 2*bw, c++; s[[bw]] = n]]; n += 2]; s]; seq[16, 10^6]

PARI

lista(len, nmax) = {my(s = vector(len,i,-1), n = 3, c = 2, bw, dbw); s[1] = 2; while(c < len && n <= nmax, bw = hammingweight(n); if(bw <= len && s[bw] < 0, dbw = sumdiv(n, d, hammingweight(d)); if(dbw == 2*bw, c++; s[bw] = n)); n += 2); s};