Karl-Heinz Hofmann - cp's OEIS frontend

A385114 a(n) is the least k with k > a(n-1) such that A000120(k) = A002487(n), with a(0) = 0.

0, 1, 2, 3, 4, 7, 9, 11, 16, 23, 25, 31, 33, 47, 49, 51, 64, 79, 83, 127, 131, 255, 271, 319, 320, 351, 355, 383, 385, 415, 417, 419, 512, 543, 551, 767, 771, 2047, 2111, 2559, 2561, 3071, 3135, 8191, 8207, 10239, 10271, 10367, 12288, 12415, 12431, 13311, 13315, 14335

Offset: 0

Karl-Heinz Hofmann, Jun 18 2025

			  n |  a(n)_2   a(n)   A002487(n)
----+-----------------------------
  0 |       0     0       0
  1 |       1     1       1
  2 |      10     2       1
  3 |      11     3       2
  4 |     100     4       1
  5 |     111     7       3
  6 |    1001     9       2
  7 |    1011    11       3
  8 |   10000    16       1
  9 |   10111    23       4
 10 |   11001    25       3
 11 |   11111    31       5
 12 |  100001    33       2
 13 |  101111    47       5
 14 |  110001    49       3

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000. (Terms 0..293 from Hugo Pfoertner.)
Karl-Heinz Hofmann, Python code.
Michael De Vlieger, Plot the i-th bit of a(n) at (x,y) = (n,i), n = 1..2^11, reading bits from the least significant to the most.
Karl-Heinz Hofmann, Vertical visualization of terms in binary, reading bits from the most significant to the least.

Cf. A000120, A002487, A101624, A168081, A277020.

Cf. A212288, A212289.

k = s[0] = 0; s[1] = 1; s[n_] := s[n] = If[EvenQ[n], s[n/2], s[(n - 1)/2] + s[(n + 1)/2]]; Array[s, 2^10]; {k}~Join~Monitor[Table[Set[k, Max[k + 1, 2^s[n] - 1]]; While[DigitCount[k, 2, 1] != s[n], k++]; k, {n, 100}], n] (* Michael De Vlieger, Jul 14 2025 *)

lista(nn) = my(list = List([0]), last = 0); for (n=1, nn, my(k=last+1, d=dia(n)); while (hammingweight(k) != d, k++); last = k; listput(list, k);); Vec(list); \\ Michel Marcus, Jul 12 2025

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a(A212288(n)) >= 2^A212289(n+1) - 1. - Hugo Pfoertner, Jul 15 2025

A384441 Binary XOR of n and the prime factors of n.

Original entry on oeis.org

1, 0, 0, 6, 0, 7, 0, 10, 10, 13, 0, 13, 0, 11, 9, 18, 0, 19, 0, 19, 17, 31, 0, 25, 28, 21, 24, 25, 0, 26, 0, 34, 41, 49, 33, 37, 0, 55, 41, 47, 0, 44, 0, 37, 43, 59, 0, 49, 54, 53, 33, 59, 0, 55, 57, 61, 41, 37, 0, 56, 0, 35, 59, 66, 73, 72, 0, 87, 81, 70, 0, 73, 0, 109

Offset: 1

Karl-Heinz Hofmann, May 30 2025

			For n = 12 the prime factors are {2,3} -> a(12) = 12 XOR 2 XOR 3 = 13.
a(13) = 13 XOR 13 = 0.

Alois P. Heinz, Table of n, a(n) for n = 1..16384
Karl-Heinz Hofmann, Graph up to n = 2^17
Karl-Heinz Hofmann, Zoom trip of the graph with number of prime factors of n colored.
Wikipedia, Bitwise operation: XOR

Cf. A000040, A052548, A178910, A293212.

f:= l-> `if`(l=[], 0, Bits[Xor](l[1], f(l[2..-1]))):
a:= n-> f([n, map(i-> i[1], ifactors(n)[2])[]]):
seq(a(n), n=1..74);  # Alois P. Heinz, May 30 2025

a[n_] := BitXor @@ Join[{n}, FactorInteger[n][[;; , 1]]]; a[1] = 1; Array[a, 100] (* Amiram Eldar, May 30 2025 *)

a(n) = my(f=factor(n)[,1]); my(b=n); for (k=1, #f, b=bitxor(b, f[k])); b; \\ Michel Marcus, May 30 2025

from sympy import primefactors
def A384441(n):
    result = n
    for pf in primefactors(n): result ^= pf
    return result

a(n) = XOR(n,A293212(n)).
a(n) = 0 <=> n is prime.
a(2^n) = A052548(n) for n>=2.

A383413 Area A of triangles such that the sides are distinct integers and A is an integer.

Original entry on oeis.org

6, 24, 30, 36, 42, 54, 60, 66, 72, 84, 90, 96, 114, 120, 126, 132, 144, 150, 156, 168, 180, 198, 204, 210, 216, 234, 240, 252, 264, 270, 288, 294, 300, 306, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 456, 462, 468, 480, 486, 504, 510, 522, 528, 540, 546, 576, 594

Offset: 1

Karl-Heinz Hofmann, Apr 26 2025

nonn

All terms are multiples of 6.

Subsequence of A188158.

			72 is in the sequence because the triangle with sides {a=5, b=29, c=30} has an area of exactly 72 and all sides are distinct.
12 is not in the sequence because this area is only possible with the isosceles triangles {a=5, b=5, c=6} and {a=5, b=5, c=8} with a and b not distinct.

Wikipedia, Heronian triangle

Cf. A188158, A316853.

nn = 450; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[a!=b!=c&&0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a-1}, {c, b-1}]; Union[lst]; lst (* James C. McMahon, May 10 2025 *)

A382509 Integers s = (p1+p2)/4 such that p1 and p2 are consecutive primes and s can be written in the form p*2^k with k>=0 and p>2 prime.

Original entry on oeis.org

3, 6, 13, 17, 28, 38, 43, 67, 80, 88, 96, 118, 127, 137, 167, 178, 188, 193, 218, 223, 272, 283, 298, 302, 328, 368, 472, 487, 508, 563, 592, 613, 617, 634, 643, 647, 662, 718, 773, 778, 802, 808, 872, 878, 932, 1033, 1142, 1168, 1172, 1187, 1193, 1198, 1256, 1277

Offset: 1

Karl-Heinz Hofmann and Hugo Pfoertner, Apr 18 2025

			For n = 2: a(n) = 6 because 4 * 6 = 24 and 24 is the sum of the two consecutive primes 11 and 13 and the factorization of 6 is 3 * 2^1.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Cf. A001043, A118134.

Select[Plus @@@ Partition[Prime[Range[400]], 2, 1]/4, IntegerQ[#] && PrimeQ[#/2^IntegerExponent[#, 2]] &] (* Amiram Eldar, Apr 21 2025 *)

is(n) = my(v=valuation(n, 2), n2);if(!isprime(n>>v), return(0)); n2 = 2*n; n2 - precprime(n2) == nextprime(n2) - n2 \\ David A. Corneth, Apr 21 2025

from sympy import isprime, sieve as prime
A382509 = []
for x in range(2,1000):
    if (totest := (prime[x] + prime[x+1])) % 4 == 0:
        s = totest // 4
        while totest % 2 == 0: totest //= 2
        if isprime(totest): A382509.append(s)
print(A382509)

A378719 Numbers k for which k/w = 3/4, where w is the number of numbers in the range {1..k} containing at least one decimal digit 1.

Original entry on oeis.org

19131872, 20809344, 20811488, 21257636, 22732948, 172186880, 1549681952, 2870345516, 2870345520, 2871907160, 2872143356, 2872251248, 13947137600, 125524238432, 496384984980, 516044091344, 530010360824, 530030555544, 530031618464, 530032461188, 530163415832, 530260860088

Offset: 1

Karl-Heinz Hofmann and Hugo Pfoertner, Dec 05 2024

Karl-Heinz Hofmann and Hugo Pfoertner, Table of n, a(n) for n = 1..37
IBM Research, Counting numbers with specific digits, Ponder This Challenge December 2024, with a bonus question asking for the last term a(37) of this sequence.
Hugo Pfoertner, PARI program (2025).
Karl-Heinz Hofmann, Python program.
Karl-Heinz Hofmann, Graph zoom to a(8) and a(9) (animated gif).
Karl-Heinz Hofmann, Whole Graph up to a(37) (then diverging) .

Cf. A004835, A346685.

PARI
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A371000 a(n) are the records of the number of addends of the form j*(j+1)/2 (A000217) producing a sequence of consecutive composites starting at A370999(n).

Original entry on oeis.org

0, 1, 3, 6, 9, 12, 15, 21, 27, 33, 45, 87, 99, 102, 123, 156, 246, 273, 282, 330, 429, 465, 477, 561, 681, 891, 1050, 1206, 1338, 1443, 1449, 1479, 1656, 1701, 1836, 1899, 1941, 2151, 2199, 2280, 2301, 2517, 2592, 2595, 2709, 2724, 2751, 2760, 2934, 3045, 3240, 3333

Offset: 1

Karl-Heinz Hofmann and Hugo Pfoertner, Mar 09 2024

nonn

The "Ponder This Challenge" of March 2024 asked for the smallest start value of sequences of at least 1000 (or 2024 to gain the bonus) composite numbers in a progression defined by A000217.

Karl-Heinz Hofmann and Hugo Pfoertner, Table of n, a(n) for n = 1..59
IBM Research, Composite Sequences, Ponder This Challenge March 2024.

Cf. A000217, A370999, A371001, A371002.

a371000(upto) = {my(m=0); forcomposite (k=4, upto, for (j=1, oo, if (isprime(k+(j*(j+1))/2), if (j>m, print1(j-1, ", "); m=j); break)))};
a371000(10^7)

A370999 Least composite number k such that the number m of consecutive composite sums k + j*(j+1)/2, j = 1, ..., m is a new maximum.

Original entry on oeis.org

4, 8, 9, 15, 24, 90, 105, 114, 225, 264, 300, 945, 5349, 7035, 11739, 17280, 35475, 46914, 190365, 351645, 603054, 1209900, 3146220, 3279864, 6407664, 26447649, 115192665, 408291345, 1080430119, 1298351109, 6459163344, 7731193299, 8096124894, 8884256514, 18927105834

Offset: 1

Karl-Heinz Hofmann and Hugo Pfoertner, Mar 09 2024

nonn

a(60) > 10^15.

Karl-Heinz Hofmann and Hugo Pfoertner, Table of n, a(n) for n = 1..59
IBM Research, Composite Sequences, Ponder This Challenge March 2024.

A371000 gives the corresponding counts m.

Cf. A000217, A371001, A371002.

a(43) in b-file corrected by Kebbaj Mohamed Reda, May 29 2024

A365933 a(n) is the period of the remainders when repdigits are divided by n.

Original entry on oeis.org

1, 9, 27, 9, 9, 27, 54, 9, 81, 9, 18, 27, 54, 54, 27, 9, 144, 81, 162, 9, 54, 18, 198, 27, 9, 54, 243, 54, 252, 27, 135, 9, 54, 144, 54, 81, 27, 162, 54, 9, 45, 54, 189, 18, 81, 198, 414, 27, 378, 9, 432, 54, 117, 243, 18, 54, 162, 252, 522, 27, 540, 135, 162, 9, 54

Offset: 1

Karl-Heinz Hofmann, Nov 07 2023

For n>1: Periods are divisible by 9 (= a full cycle in the sequence of repdigits). a(n)/9 is the period of the remainders when repunits are divided by n. So the digit part of the repdigits has no effect on periods generally. For most n the beginning of the periodic part is always A010785(1). If n is a term of A083118 the periodic part starts later after some initial remainders that do not repeat.

			For n = 6:                Remainders of A010785(1..54) mod n.
A010785( 1...9) mod n:      [1, 2, 3, 4, 5, 0, 1, 2, 3]
A010785(10..18) mod n:      [5, 4, 3, 2, 1, 0, 5, 4, 3]
A010785(19..27) mod n:      [3, 0, 3, 0, 3, 0, 3, 0, 3]
So the period is 3*9 = 27. Thus a(n) = 27. And the pattern seen above starts again:
A010785(28..36) mod n:      [1, 2, 3, 4, 5, 0, 1, 2, 3]
A010785(37..45) mod n:      [5, 4, 3, 2, 1, 0, 5, 4, 3]
A010785(46..54) mod n:      [3, 0, 3, 0, 3, 0, 3, 0, 3]

Karl-Heinz Hofmann, Table with additional information.

Cf. A010785, A002275.
Cf. A305322 (divisor 3), A002279 (divisor 5), A366596 (divisor 7).
Cf. A083118 (the impossible divisors).

def A365933(n):
    if n == 1: return 1
    remainders, exponent = [], 1
    while (rem:=(10**exponent // 9 % n)) not in remainders:
        remainders.append(rem); exponent += 1
    return (exponent - remainders.index(rem) - 1) * 9

def A365933(n):
    if n==1: return 1
    a,b,x,y=1,1,1%n,11%n
    while x!=y:
        if a==b:
            a<<=1
            x,b=y,0
        y = (10*y+1)%n
        b+=1
    return 9*b # Chai Wah Wu, Jan 23 2024

A365932 a(n) = the number of cubes (of integers > 0) that have bit length n.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 3, 5, 5, 6, 9, 10, 13, 17, 21, 26, 34, 42, 52, 67, 84, 105, 134, 167, 211, 267, 335, 422, 533, 670, 845, 1065, 1341, 1690, 2130, 2682, 3380, 4259, 5365, 6760, 8518, 10730, 13520, 17035, 21461, 27040, 34069, 42923, 54080, 68137, 85847

Offset: 1

Karl-Heinz Hofmann, Oct 05 2023

Number of cubes in the range: 2^(n-1) <= k^3 < 2^n-1.

There is no need to include 2^n-1 because it is a Mersenne number and it cannot be a power anyway.

			For n = 13; a(n) = 5; following 5 cubes have a bit length of 13: 16^3, 17^3, 18^3, 19^3 and 20^3.

Cf. A004632.
Cf. A017981 (partial sums).
Cf. A002580, A126726, A365930.

a[n_] := Floor[Surd[2^n-1, 3]] - Floor[Surd[2^(n-1)-1, 3]]; Array[a, 56] (* Amiram Eldar, Oct 30 2023 *)

from sympy import integer_nthroot
def A365932(n):
    return integer_nthroot(2**n-1, 3)[0] - integer_nthroot(2**(n-1)-1, 3)[0]
print([A365932(n) for n in range(1,57)])

a(n) = floor((2^n-1)^(1/3)) - floor((2^(n-1)-1)^(1/3)) for n > 0.

Limit_{n->oo} a(n)/a(n-1) = 2^(1/3) = A002580.

A365931 a(n) = number of pairs {x,y} with (x,y > 1) such that x^y (= terms of A072103) has bit length <= n.

Original entry on oeis.org

0, 0, 1, 3, 7, 10, 18, 25, 35, 50, 69, 94, 132, 178, 244, 334, 460, 629, 869, 1201, 1668, 2314, 3223, 4493, 6280, 8793, 12322, 17288, 24286, 34139, 48036, 67630, 95274, 134285, 189349, 267090, 376880, 531942, 750991, 1060463, 1497741, 2115669, 2988957, 4223225, 5967822, 8433889

Offset: 1

Karl-Heinz Hofmann, Oct 07 2023

Number of pairs {x,y} with (x,y > 1) for which x^y < 2^n-1.
In some special cases different pairs have the same result (see A072103 and the example here) and those multiple representations are counted separately.
There is no need to include 2^n-1 because it is a Mersenne number and it cannot be a power anyway.
Limit_{n->oo} a(n)/a(n-1) = sqrt(2) = A002193.
Partial sums of A365930.

			For n = 6: the Mersenne number 2^6-1 = 63 is the largest number with bit length 6 and the upper bound for the following a(6) = 10 powers: 2^2, 2^3, 2^4, 2^5, 3^2, 3^3, 4^2, 5^2, 6^2, 7^2.

Cf. A072103, A002193, A365930 (first differences).

Cf. A017912 (squares), A017981 (cubes).

a[n_] := Sum[Ceiling[2^(n/k)] - 2, {k, 2, n}]; Array[a, 47]

from sympy import integer_nthroot, integer_log
def A365931(n):
    result, nMersenne, new = 0, (1<

a(n) = Sum_{y = 2..n} (ceiling(2^(n/y)) - 2)
a(n) = Sum_{y = 2..n} (floor((2^n-1)^(1/y)) - 1)
a(n) = Sum_{k = 1..n} A365930(k).

cp's OEIS Frontend

User: Karl-Heinz Hofmann

A385114 a(n) is the least k with k > a(n-1) such that A000120(k) = A002487(n), with a(0) = 0.

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A384441 Binary XOR of n and the prime factors of n.

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Maple

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A383413 Area A of triangles such that the sides are distinct integers and A is an integer.

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Mathematica

A382509 Integers s = (p1+p2)/4 such that p1 and p2 are consecutive primes and s can be written in the form p*2^k with k>=0 and p>2 prime.

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Mathematica

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A378719 Numbers k for which k/w = 3/4, where w is the number of numbers in the range {1..k} containing at least one decimal digit 1.

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PARI

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A371000 a(n) are the records of the number of addends of the form j*(j+1)/2 (A000217) producing a sequence of consecutive composites starting at A370999(n).

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PARI

A370999 Least composite number k such that the number m of consecutive composite sums k + j*(j+1)/2, j = 1, ..., m is a new maximum.

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A365933 a(n) is the period of the remainders when repdigits are divided by n.

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