A093653 -id:A093653 - cp's OEIS frontend

A338452 Numbers k such that k and k+1 have the same total binary weight of their divisors (A093653).

3, 4, 7, 20, 31, 57, 94, 98, 118, 122, 127, 201, 213, 218, 230, 242, 243, 244, 334, 384, 393, 423, 429, 481, 565, 603, 633, 694, 704, 729, 766, 844, 921, 1138, 1141, 1221, 1262, 1401, 1533, 1654, 1726, 1761, 1837, 1838, 1862, 1882, 1942, 2162, 2245, 2361, 2362

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1).

The Mersenne primes (A000668) are terms since if 2^p - 1 is a prime then A093653(2^p-1) = A093653(2^p) = p+1.

			3 is a term since A093653(3) = A093653(4) = 3.

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

Cf. A000120, A093653.
A000668 is a subsequence.
Similar sequences: A002961, A005237, A054004, A064125, A238380, A293183, A306985.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; f1 = f[1]; Do[f2 = f[n]; If[f1 == f2, AppendTo[s, n - 1]]; f1 = f2, {n, 2, 240}]; s

A338453 Starts of runs of 3 consecutive numbers with the same total binary weight of their divisors (A093653).

Original entry on oeis.org

3, 242, 243, 1837, 2361, 3693, 3728, 6061, 6457, 9782, 11181, 11721, 13855, 15177, 20017, 22591, 28021, 31461, 31887, 33098, 33993, 38137, 52016, 52112, 60321, 76897, 78542, 78745, 80461, 108394, 116017, 119541, 124453, 125493, 127117, 127417, 145369, 151805, 154113

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1) = A093653(k+2).

			3 is a term since A093653(3) = A093653(4) = A093653(5) = 3.

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

Cf. A093653.
Subsequence of A338452.
Similar sequences: A005238, A006073, A045939.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; m = 3; fs = f /@ Range[m]; Do[If[Equal @@  fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 155000}]; s
SequencePosition[Table[Total[DigitCount[Divisors[n],2,1]],{n,160000}],{x_,x_,x_}][[All,1]] (* Harvey P. Dale, Feb 04 2023 *)

A338454 Starts of runs of 4 consecutive numbers with the same total binary weight of their divisors (A093653).

Original entry on oeis.org

242, 947767, 1041607, 2545015, 3275463, 8170983, 15720871, 21532430, 23752181, 25135885, 25595913, 27981703, 28226983, 30505142, 30962767, 33364805, 37264493, 49002661, 49766629, 52910454, 53408456, 57917191, 57952016, 58331576, 59230454, 60014053, 60723111, 63378005

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1) = A093653(k+2) = A093653(k+3).

			242 is a term since A093653(242) = A093653(243) = A093653(244) = A093653(245) = 18.

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

Cf. A093653.
Subsequence of A338452 and A338453.
Similar sequences: A006601, A045932, A045940.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; m = 4; fs = f /@ Range[m]; Do[If[Equal @@  fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 10^7}]; s

A338514 Numbers k such that k and k+1 are both divisible by the total binary weight of their divisors (A093653).

Original entry on oeis.org

1, 2, 54, 2119, 11100, 13727, 14382, 15799, 16399, 20159, 20950, 33421, 34617, 36328, 36396, 39400, 42198, 42438, 42650, 46253, 46873, 50370, 55368, 56600, 58793, 67013, 67320, 69023, 72325, 76057, 86393, 90781, 92906, 93216, 105909, 132088, 134028, 134823, 140466

Offset: 1

Amiram Eldar, Oct 31 2020

Numbers k such that k and k+1 are both in A093705, or, equivalently, k is divisible by A093653(k) and k+1 is divisible by A093653(k+1).

			1 is a term since 1 and 2 are both terms of A093705.

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

Cf. A000120, A093653, A093705.

Similar sequences: A330927, A330931, A334345, A338452.

divQ[n_] := Divisible[n, DivisorSum[n, DigitCount[#, 2, 1] &]]; q1 = divQ[1]; Reap[Do[q2 = divQ[n]; If[q1 && q2, Sow[n - 1]]; q1 = q2, {n, 2, 10^5}]][[2, 1]]
SequencePosition[Table[If[Divisible[n,Total[DigitCount[Divisors[n],2,1]]],1,0],{n,150000}],{1,1}][[All,1]] (* Harvey P. Dale, Jun 14 2022 *)

A373092 The number of iterations of the map x -> A093653(x) that are required to reach from n to one of the fixed points, 1, 2, 3 or 6.

Original entry on oeis.org

0, 0, 0, 1, 1, 0, 2, 2, 2, 1, 2, 3, 2, 3, 3, 2, 1, 2, 2, 3, 3, 3, 2, 4, 1, 3, 3, 4, 2, 3, 1, 1, 3, 1, 3, 4, 2, 3, 2, 4, 2, 3, 2, 4, 4, 2, 1, 4, 3, 4, 3, 4, 2, 3, 3, 3, 2, 2, 1, 4, 1, 4, 2, 3, 3, 3, 2, 3, 2, 3, 2, 4, 2, 3, 3, 4, 3, 4, 1, 4, 4, 3, 2, 4, 3, 2, 4

Offset: 1

Amiram Eldar, May 23 2024

			The iterations for the n = 1..7 are:
  n  a(n)  iterations
  -  ----  -----------
  1    0   1
  2    0   2
  3    0   3
  4    1   4 -> 3
  5    1   5 -> 3
  6    0   6
  7    2   7 -> 4 -> 3

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

Cf. A086793 (decimal analog), A093653, A373093, A373094.

d[n_] := DivisorSum[n, Plus @@ IntegerDigits[#, 2] &]; a[n_] := -2 + Length@ FixedPointList[d, n]; Array[a, 100]

a(n) = {my(c = 0); while(6 % n, n = sumdiv(n, d, hammingweight(d)); c++); c;}

A338515 Starts of runs of 3 consecutive numbers that are divisible by the total binary weight of their divisors (A093653).

Original entry on oeis.org

1, 348515, 8612344, 29638764, 30625110, 32039808, 32130600, 32481682, 43664313, 55318282, 55503719, 59671714, 69254000, 73152296, 93470904, 100366594, 103640097, 105026790, 109038462, 109212287, 122519464, 126667271, 147208982, 162007166, 169237545, 173392238

Offset: 1

Amiram Eldar, Oct 31 2020

			1 is a term since 1, 2 and 3 are terms of A093705.

Amiram Eldar, Table of n, a(n) for n = 1..5000

Subsequence of A338514.
Cf. A000120, A093653, A093705.
Similar sequences: A154701, A330932, A334346, A338453.

divQ[n_] := Divisible[n, DivisorSum[n, DigitCount[#, 2, 1] &]]; div = divQ /@ Range[3]; Reap[Do[If[And @@ div, Sow[k - 3]]; div = Join[Rest[div], {divQ[k]}], {k, 4, 10^7}]][[2, 1]]

A360641 Numbers k where A093653(k)/A000120(k) sets a new record.

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 24, 36, 66, 72, 132, 144, 264, 420, 528, 840, 1026, 1056, 1680, 2052, 4104, 8208, 16416, 32832, 65664, 73920, 84000, 110880, 118800, 131328, 133380, 237600, 263340, 266760, 526680, 533520, 1053360, 1067040, 2106720, 2134080, 3160080, 4213440

Offset: 1

Amiram Eldar, Feb 15 2023

Analogous to superabundant numbers (A004394) as A175526 is analogous to abundant numbers (A005101).
The corresponding record values are 1, 2, 3, 4, 9/2, 5, 6, 15/2, 8, 10, ... .
This sequence is infinite since A093653(k)/A000120(k) is unbounded: A093653(2^m)/A000120(2^m) = m+1 for all m >= 0.

			The values of A093653(k)/A000120(k) for k=1..10 are 1, 2, 3/2, 3, 3/2, 3, 4/3, 4, 5/2 and 3. The record values, 1, 2, 3 and 4, occur at 1, 2, 4 and 8, the first 4 terms of this sequence.

Amiram Eldar, Table of n, a(n) for n = 1..58 (terms below 10^10)

Cf. A000120, A004394, A005101, A093653, A093687, A175522, A175526, A360642.

seq[nmax_] := Module[{s = {}, rm = 0, r}, Do[If[(r = DivisorSum[n, DigitCount[#, 2, 1] &]/DigitCount[n, 2, 1]) > rm, rm = r; AppendTo[s, n]], {n, 1, nmax}]; s]; seq[10^4]

lista(kmax) = {my(rm = 0, r); for(k = 1, kmax, r = sumdiv(k, d, hammingweight(d))/hammingweight(k); if(r > rm, rm = r; print1(k, ", "))); }

# uses imports and definitions in A093653, A000120
from itertools import count, islice
def f(n): return A093653(n)/A000120(n)
def agen(r=0): yield from ((m, r:=fm)[0] for m in count(1) if (fm:=f(m))>r)
print(list(islice(agen(), 34))) # Michael S. Branicky, Feb 15 2023

A360642 a(n) is the least number k such that A093653(k)/A000120(k) = n.

Original entry on oeis.org

1, 2, 4, 8, 16, 24, 64, 66, 84, 72, 210, 132, 450, 792, 288, 264, 1044, 672, 5328, 528, 1344, 840, 1026, 1056, 4116, 1800, 4128, 2112, 5124, 3780, 6480, 2184, 3360, 8352, 11088, 8448, 4680, 50700, 4200, 4368, 20880, 8280, 13320, 13440, 12420, 4104, 46200, 8736

Offset: 1

Amiram Eldar, Feb 15 2023

a(n) exists for all n >= 1 since A093653(2^(k-1))/A000120(2^(k-1)) = k for all k >= 1.

Analogous to A007539 as A175522 is analogous to perfect numbers (A000396).

			a(1) = 1 since A093653(1)/A000120(1) = 1/1 = 1.
a(2) = 2 since A093653(2)/A000120(2) = 2/1 = 2, and 2 is the least number with this property.
a(3) = 4 since A093653(4)/A000120(4) = 3/1 = 3, and 4 is the least number with this property.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000120, A000396, A007539, A093653, A093687, A175522, A175526, A360641.

seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n <= nmax, i = DivisorSum[n, DigitCount[#, 2, 1] &]/DigitCount[n, 2, 1]; If[IntegerQ[i] && i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[50, 10^5]

lista(len, nmax) = {my(s = vector(len), c = 0, n = 1, i); while(c < len && n <= nmax, i = sumdiv(n, d, hammingweight(d))/hammingweight(n); if(denominator(i) == 1 && i <= len && s[i] == 0, c++; s[i] = n); n++); s }

# uses imports and definitions in A093653, A000120
from itertools import count, islice
def f(n): q, r = divmod(A093653(n), A000120(n)); return q if r == 0 else 0
def agen():
    n, adict = 1, dict()
    for k in count(1):
        v = f(k)
        if v not in adict: adict[v] = k
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 48))) # Michael S. Branicky, Feb 15 2023

a(n) <= 2^(n-1).

A373093 The fixed point of the iterations of the map x -> A093653(x) that start at n.

Original entry on oeis.org

1, 2, 3, 3, 3, 6, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 6, 6, 6, 3, 6, 3, 3, 3, 3, 6, 3, 3, 6, 3, 3, 3, 6, 6, 3, 3, 3, 3, 3, 3, 6, 3, 3, 6, 6, 6, 3, 6, 3, 3, 3, 3, 3, 3, 3, 6, 6, 3, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 6, 3

Offset: 1

Amiram Eldar, May 23 2024

Except for n = 1 and 2, all terms are either 3 or 6.

Do the asymptotic densities of the occurrences of 3 and 6 exist? The numbers of occurrences of 6 for n that do not exceed 10^k, for k = 1, 2, ..., are 2, 24, 234, 2735, 25321, 242398, 2605532, 27441386, 268518855, 2561508455, ... .

			The iterations for the n = 1..7 are:
  n  a(n)  iterations
  -  ----  -----------
  1    1   1
  2    2   2
  3    3   3
  4    3   4 -> 3
  5    3   5 -> 3
  6    6   6
  7    3   7 -> 4 -> 3

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

Cf. A093653, A373092.

d[n_] := DivisorSum[n, Plus @@ IntegerDigits[#, 2] &]; a[n_] := FixedPointList[d, n][[-1]]; Array[a, 100]

a(n) = {while(6 % n, n = sumdiv(n, d, hammingweight(d))); n;}

A338455 Starts of runs of 5 consecutive numbers with the same total binary weight of their divisors (A093653).

Original entry on oeis.org

1307029927, 2116078861, 2665774183, 2809370965, 4108623302, 4493733751, 5333670902, 5497285284, 5679049670, 8209799382, 9665369455, 9708528486, 10353426151, 10606564910, 12777118615, 12795699493, 13660293367, 13847206214, 14351020663, 15735895813, 17912257013

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1) = A093653(k+2) = A093653(k+3) = A093653(k+4).

Can 6 consecutive numbers have the same total binary weight of their divisors? If they exist, then they are larger than 10^11.

			1307029927 is a term since A093653(1307029927) = A093653(1307029928) = A093653(1307029929) = A093653(1307029930) = A093653(1307029931) = 72.

Cf. A093653.
Subsequence of A338452, A338453 and A338454.
Similar sequences: A045933, A045941, A049051.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; m = 5; fs = f /@ Range[m]; Do[If[Equal @@  fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 10^7}]; s

cp's OEIS Frontend

A338452 Numbers k such that k and k+1 have the same total binary weight of their divisors (A093653).

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A338453 Starts of runs of 3 consecutive numbers with the same total binary weight of their divisors (A093653).

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A338454 Starts of runs of 4 consecutive numbers with the same total binary weight of their divisors (A093653).

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A338514 Numbers k such that k and k+1 are both divisible by the total binary weight of their divisors (A093653).

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A373092 The number of iterations of the map x -> A093653(x) that are required to reach from n to one of the fixed points, 1, 2, 3 or 6.

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A338515 Starts of runs of 3 consecutive numbers that are divisible by the total binary weight of their divisors (A093653).

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A360641 Numbers k where A093653(k)/A000120(k) sets a new record.

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A360642 a(n) is the least number k such that A093653(k)/A000120(k) = n.

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