A380844 -id:A380844 - cp's OEIS frontend

A380845 The sum of divisors of n that have the same binary weight as n.

1, 3, 3, 7, 5, 9, 7, 15, 12, 15, 11, 21, 13, 21, 15, 31, 17, 36, 19, 35, 28, 33, 23, 45, 25, 39, 27, 49, 29, 45, 31, 63, 36, 51, 42, 84, 37, 57, 39, 75, 41, 84, 43, 77, 60, 69, 47, 93, 56, 75, 51, 91, 53, 81, 55, 105, 57, 87, 59, 105, 61, 93, 63, 127, 70, 108

Offset: 1

Amiram Eldar, Feb 05 2025

The number of these divisors is A380844(n).

			a(6) = 9 because 6 = 110_2 has binary weight 2, 2 of its divisors, 3 = 11_2 and 6, have the same binary weight, and 3 + 6 = 9.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A000203, A000265, A000396, A000668, A007814, A038712, A380844.

a[n_] := Module[{h = DigitCount[n, 2, 1]}, DivisorSum[n, # &, DigitCount[#, 2, 1] == h &]]; Array[a, 100]

a(n) = {my(h = hammingweight(n)); sumdiv(n, d, d * (hammingweight(d) == h));}

a(n) = Sum_{d|n} d * [A000120(d) = A000120(n)], where [ ] is the Iverson bracket.
a(2^n) = 2^(n+1) - 1.
a(n) <= A000203(n) with equality if and only if n is a power of 2.
a(n) = a(A000265(n)) * (2^(A007814(n)+1)-1) = a(A000265(n)) * A038712(n), or equivalently, a(k*2^n) = a(k)*(2^(n+1)-1) for k odd and n >= 0.
In particular, since a(p) = p for an odd prime p, a(p*2^n) = p*(2^(n+1)-1) for an odd prime p and n >= 0.
a(A000396(n)) = A000668(n)^2, assuming that odd perfect numbers do no exist.

A383363 Composite numbers k all of whose proper divisors have binary weights that are not equal to the binary weight of k.

Original entry on oeis.org

15, 25, 27, 39, 51, 55, 57, 63, 69, 77, 81, 85, 87, 91, 95, 99, 111, 115, 117, 119, 121, 123, 125, 141, 143, 145, 147, 159, 169, 171, 175, 177, 183, 185, 187, 201, 203, 205, 207, 209, 213, 215, 219, 221, 231, 235, 237, 243, 245, 247, 249, 253, 255, 261, 265, 275

Offset: 1

Amiram Eldar, Apr 24 2025

First differs from A325571 at n = 56: A325571(56) = 267 is not a term of this sequence. Differs from A325571 by having the terms 16849, 35235, 101265, 268357, 295717, ..., and not having the terms 267, 295, 327, 387, 395, ... .
Composite numbers k such that A380844(k) = 1.
All the odd primes p have A380844(p) = 1.
All the terms are odd numbers since for an even number k, A000120(k) = A000120(k/2).

			15 = 3 * 5 is a term since it is composite, and its binary weight, A000120(15) = 4 is different from the binary weights of its proper divisors: A000120(1) = 1, A000120(3) = 2, and A000120(5) = 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A325571, A380844, A383364, A383365.

q[k_] := CompositeQ[k] && DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; Select[Range[1, 300, 2], q]

isok(k) = if(k == 1 || isprime(k), 0, my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) == 1);

A383364 a(n) is the least number k with exactly n proper divisors, where all of them have binary weights that are different from the binary weight of k.

Original entry on oeis.org

1, 3, 25, 15, 81, 63, 15625, 231, 1225, 405, 59049, 495, 531441, 5103, 2025, 1485, 33232930569601, 2475, 3814697265625, 6237, 18225, 295245, 31381059609, 4095, 1500625, 2657205, 81225, 25515, 22876792454961, 14175, 931322574615478515625, 21735, 31236921, 301327047

Offset: 0

Amiram Eldar, Apr 24 2025

			a(0) = 1 since 1 has no proper divisors.
a(1) = 3 since 3 has one proper divisor, 1, and A000120(1) = 1 != A000120(3) = 2, while 2 also has one proper divisor, 1, but A000120(2) = A000120(1) = 1.

Amiram Eldar, Table of n, a(n) for n = 0..99

Cf. A000120, A032741, A380844, A383363, A383365.

q[k_] := DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; seq[len_] := Module[{s = Table[0, {len}], c = 0, k = 1, i}, While[c < len, i = DivisorSigma[0, k]; If[i <= len && s[[i]] == 0 && q[k], c++; s[[i]] = k]; k++]; s]; seq[16]

is1(k) = {my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) ==  1};
list(len) = {my(s = vector(len), c = 0, k = 1, i); while(c < len, i = numdiv(k); if(i <= len && s[i] == 0 && is1(k), c++; s[i] = k); k++); s;}

A383365 Numbers k with a record number of proper divisors, where all of them have binary weights that are different from the binary weight of k.

Original entry on oeis.org

1, 3, 15, 63, 231, 405, 495, 1485, 2475, 4095, 14175, 21735, 24255, 31185, 79695, 190575, 218295, 239085, 294525, 904365, 1276275, 2789325, 3586275, 4937625, 6912675, 10072755, 17342325, 17972955, 26801775, 46621575, 80405325, 192567375, 326351025, 333107775, 654729075

Offset: 1

Amiram Eldar, Apr 24 2025

The corresponding record values are 0, 1, 3, 5, 7, 9, 11, 15, 17, ... (see the link for more values).

All terms are odd as an even number k has proper divisor k/2 with the same binary weight. - David A. Corneth, Apr 24 2025

			a(1) = 1 since 1 has no proper divisors.
a(2) = 3 since 3 has one proper divisor, 1, and A000120(1) = 1 != A000120(3) = 2, while 2 also has one proper divisor, 1, but A000120(2) = A000120(1) = 1.
a(3) = 15 since 15 has 3 proper divisors, 1, 3 and 5, and A000120(1) = 1 and A000120(3) = A000120(5) = 2 are different from A000120(15) = 4. All the numbers below 15 have fewer proper divisors with this property.

David A. Corneth, Table of n, a(n) for n = 1..91 (first 46 terms from Amiram Eldar)
Amiram Eldar, Table of n, a(n), A032741(a(n)) for n = 1..46.

Cf. A000120, A032741, A380844, A383363, A383364.

q[k_] := DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; seq[kmax_] := Module[{s = {}, d, dm = 0}, Do[d = DivisorSigma[0, k]; If[d > dm && q[k], dm = d; AppendTo[s, k]], {k, 1, kmax}]; s]; seq[10^5]

is1(k) = {my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) ==  1};
list(kmax) = {my(d, dm = 0); for(k = 1, kmax, d = numdiv(k); if(d > dm && is1(k), dm = d; print1(k, ", ")));}

A381069 Numbers k that have a record number of divisors that have the same binary weight as k.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 72, 144, 288, 576, 1080, 2160, 4320, 8640, 17280, 34560, 69120, 99360, 136080, 198720, 272160, 397440, 529200, 544320, 1058400, 2116800, 3160080, 4233600, 6320160, 8467200, 12640320, 16934400, 25280640, 50561280, 76744800, 101122560, 102816000

Offset: 1

Amiram Eldar, Feb 12 2025

Indices of records of A380844, i.e., numbers k such that A380844(k) > A380844(m) for all m < k.
This sequence is infinite since A380844 is unbounded (e.g., A380844(2^n) = n+1).
Analogous to highly composite numbers (A002182) with the number of divisors with the same binary weight (A380844) instead of the number of divisors (A000005).
The corresponding record values are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 20, 24, 28, ... (see the link for more values).

Amiram Eldar, Table of n, a(n) for n = 1..60
Amiram Eldar, Table of n, a(n), A380844(a(n)) for n = 1..60

Cf. A000005, A002182, A380844, A381070.

seq[lim_] := Module[{h, d, dmax = 0, s = {}}, Do[h = DigitCount[k, 2, 1]; d = DivisorSum[k, 1 &, DigitCount[#, 2, 1] == h &]; If[d > dmax, dmax = d; AppendTo[s, k]], {k, 1, lim}]; s]; seq[10^5]

list(lim) = {my(h, d, dmax = 0); for(k = 1, lim, h = hammingweight(k); d = sumdiv(k, d, hammingweight(d) == h); if(d > dmax, dmax = d; print1(k, ", ")));}

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A380845 The sum of divisors of n that have the same binary weight as n.

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A383363 Composite numbers k all of whose proper divisors have binary weights that are not equal to the binary weight of k.

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A383364 a(n) is the least number k with exactly n proper divisors, where all of them have binary weights that are different from the binary weight of k.

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A383365 Numbers k with a record number of proper divisors, where all of them have binary weights that are different from the binary weight of k.

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A381069 Numbers k that have a record number of divisors that have the same binary weight as k.

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