Kevin P. Thompson - cp's OEIS frontend

A363491 Numbers k such that 2^k - 5 is a semiprime.

7, 13, 14, 16, 19, 28, 30, 31, 40, 42, 51, 54, 55, 58, 62, 68, 85, 88, 96, 111, 112, 116, 128, 148, 160, 162, 188, 192, 198, 220, 222, 236, 242, 276, 300, 318, 319, 320, 332, 372, 373, 398, 420, 428, 432, 458, 460, 482, 505, 520, 532, 542, 546, 556, 650, 692, 714

Offset: 1

Kevin P. Thompson, Jun 05 2023

nonn

The numbers 806 and 811 are also terms with 770 being the only remaining unknown below them.

			13 is a term because 2^13 - 5 = 8187 = 3 * 2729 is a semiprime.

Cf. A001358, A085724, A363374.

Select[Range[715],PrimeOmega[2^#-5]==2&] (* Harvey P. Dale, Apr 22 2025 *)

isA363491(k) = bigomega(2^k-5) == 2; \\ Hugo Pfoertner, Jun 25 2023

A361803 Least k > 1 such that k^n - n > 1 is semiprime, or 0 if no such k exists.

Original entry on oeis.org

5, 4, 5, 3, 6, 2, 2, 5, 8, 3, 4, 11, 15, 5, 2, 0, 4, 2, 14, 7, 48, 42, 6, 35, 2, 7, 602, 3, 16, 13, 2, 3, 2, 6, 37, 3185, 6, 9, 2, 33, 28, 2, 20, 9, 2, 135, 6, 5, 2, 49, 100, 5, 166, 5, 4, 9, 98, 15, 4, 27, 24, 2, 4, 17343, 34, 19, 24, 15, 56, 6, 90, 5, 2, 85

Offset: 1

Kevin P. Thompson, Jun 12 2023

nonn

For n = 16, k^16 - 16 = (k^8 - 4)(k^8 + 4) = (k^4 - 2)(k^4 + 2)(k^8 + 4) always has at least three factors, so a(16) = 0. Similarly for any n of the form (2m)^4, so a(A016744(n)) = 0.

			For n = 3:
k = 1: 1^3 - 3 = -2 < 0 so ignore.
k = 2: 2^3 - 3 = 5 which is not semiprime.
k = 3: 3^3 - 3 = 24 = 2 * 2 * 2 * 3 which is not semiprime.
k = 4: 4^3 - 3 = 61 which is not semiprime.
k = 5: 5^3 - 3 = 122 = 2 * 61 which is semiprime.
Therefore, a(3) = 5 since k = 5 is the first value for which k^3 - 3 is semiprime.

Kevin P. Thompson, Table of n, a(n) for n = 1..115
Kevin P. Thompson, Table of n, a(n) for n = 1..154 with uncertain values

Cf. A001358, A016744, A130827.

A363374 Numbers k such that 2^k - 3 is a semiprime.

Original entry on oeis.org

8, 11, 13, 15, 17, 18, 21, 23, 25, 30, 32, 33, 34, 35, 36, 37, 40, 44, 54, 58, 60, 61, 71, 73, 92, 95, 101, 102, 106, 144, 160, 164, 183, 200, 209, 210, 216, 241, 244, 270, 273, 274, 281, 293, 309, 313, 344, 365, 422, 430, 461, 475, 477, 480, 504, 509, 556, 579, 597, 609, 612, 631, 650

Offset: 1

Kevin P. Thompson, May 29 2023

nonn

The numbers 717, 720, 759 are also terms with 713 being the only remaining unknown below them.

			11 is a member because 2^11 - 3 = 2045 = 5 * 409 is a semiprime.

Cf. A085724.

Select[Range[700],PrimeOmega[2^#-3]==2&] (* Harvey P. Dale, Dec 14 2024 *)

A359070 Smallest k > 1 such that k^n - 1 is the product of n distinct primes.

Original entry on oeis.org

3, 4, 15, 12, 39, 54, 79, 86, 144, 318, 1591, 144, 20131, 2014, 1764, 1308, 46656, 1296

Offset: 1

Kevin P. Thompson, Dec 15 2022

a(19) > 60000 and a(20) = 3940.
a(19) > 5 * 10^5; a(21) = 132023; a(22) = 229430; a(24) = 4842. - Daniel Suteu, Dec 16 2022
Because of the algebraic factorization of x^n-1 (via cyclotomic polynomials), there is good reason to expect (on average) that prime values of n will have larger solutions than other numbers. That is, those values of n with many factors already get a head start by having many algebraic factors. - Sean A. Irvine, Jan 06 2023

			a(3) = 15 since 15^3 - 1 = 3374 = 2*7*241 is the product of 3 distinct primes and 15 is the smallest number with this property.

Cf. A001597, A005117, A045542, A219019, A281940, A359069.

isok(k, n) = my(f=factor(k^n - 1)); issquarefree(f) && (omega(f) == n);
a(n) = my(k=2); while (!isok(k, n), k++); k; \\ Michel Marcus, Dec 15 2022

a(n) >= A219019(n). - Daniel Suteu, Dec 16 2022

A359069 Smallest prime p such that p^(2n-1) - 1 is the product of 2n-1 distinct primes.

Original entry on oeis.org

3, 59, 47, 79, 347, 6343, 56711, 4523

Offset: 1

Kevin P. Thompson, Dec 15 2022

a(9) > 113500.
a(9) > 1000000, a(10) > 237000, a(11) > 209021. - Sean A. Irvine, Jan 10 2023
a(n)-1 is squarefree for all n. - Chai Wah Wu, Jan 30 2023

			a(3) = 47 since 47^(2*3-1) - 1 = 229345006 = 2*11*23*31*14621 is the product of 5 distinct primes and 47 is the smallest prime number with this property.

Cf. A001597, A005117, A045542, A280005, A359070.

isok(p, n) = my(f=factor(p^(2*n-1)-1)); issquarefree(f) && (omega(f) == 2*n-1);
a(n) = my(p=2); while (!isok(p, n), p=nextprime(p+1)); p; \\ Michel Marcus, Dec 15 2022

A352597 a(n) is the smallest k > 1 such that k^n + 1 has all prime divisors p == 1 (mod n).

Original entry on oeis.org

2, 2, 6, 2, 10, 6, 28, 2, 18, 10, 22, 6, 52, 14, 60, 2, 102, 36, 190, 20, 756, 66, 46, 18, 2550, 26, 2970, 28, 58, 120, 310, 2, 330, 170, 11550, 6, 148, 570, 156, 140, 82, 2184, 172, 88, 3040020, 184, 282, 42, 7252, 110, 7548, 312, 106, 1440, 41800, 42, 11172

Offset: 1

Kevin P. Thompson, Mar 21 2022

Equivalently, a(n) is the smallest k > 1 such that for all divisors d of k^n + 1, d == 1 (mod n).
A298299 is a subsequence.
All terms in this sequence are even since for odd k the expression k^n + 1 is divisible by 2 which is not congruent to 1 (mod n) for any n > 1.
If n is odd, a(n)^n + 1 is divisible by a(n) + 1. Therefore, a(n) + 1 == 1 (mod n) and so n | a(n) for odd n.
Theorem: a(n) = 2 if and only if n is a power of 2.

			a(3) = 6 since 6^3 + 1 = 217 = 7 * 31 and both factors are congruent to 1 (mod 3).

Cf. A298076, A298299 (bisection), A298310, A298398.

isok(k,n) = my(f=factor(k^n+1)); for (i=1, #f~, if (Mod(f[i,1], n) != 1, return(0))); return(1);
a(n) = my(k=2); while (!isok(k, n), k+=2); k; \\ Michel Marcus, Mar 22 2022

a(2n) = A298299(n).

A350800 Numbers k such that k and k+1 have the same number and sum of divisors but a different number of distinct prime factors.

Original entry on oeis.org

64665, 109214, 2305557, 4701537, 6444873, 10118654, 32225337, 33876117, 70282053, 105967784, 149205914, 187434621, 268890218, 279113505, 334925577, 357340922, 391392134, 424942604, 575712494, 610752933, 612863198, 641703842, 701792234, 743194142, 800679495

Offset: 1

Kevin P. Thompson, Jan 16 2022

nonn

Subsequence of A054004. Most members of A054004 are not a part of this subsequence, so consecutive numbers with equal tau and sigma most often achieve this with an equal count of distinct prime factors.

			64665 is a term of this sequence since tau(64665) = tau(64666) = 8 and sigma(64665) = sigma(64666) = 2160, but omega(64665) = 4 and omega(64666) = 3.

Amiram Eldar, Table of n, a(n) for n = 1..223 (terms 1..106 from Kevin P. Thompson)

Cf. A000005, A000203, A001221, A054004.

Select[Range[10^7], DivisorSigma[{0, 1}, #] ==  DivisorSigma[{0, 1}, # + 1] && PrimeNu[#] != PrimeNu[# + 1] &] (* Amiram Eldar, Jan 20 2022 *)

A350370 a(n) is the smallest k such that the Collatz sequence for k includes a record number of consecutive tripling steps.

Original entry on oeis.org

1, 3, 7, 15, 27, 127, 255, 511, 1023, 1819, 4095, 4255, 16383, 32767, 65535, 77671, 262143, 459759, 1048575, 2097151, 4194303, 7456539, 16777215, 33554431, 67108863, 125687199, 1073741823, 2147483647, 4294967295, 8589934591, 17179869183, 20361326439, 68719476735

Offset: 1

Kevin P. Thompson, Dec 27 2021

nonn

See A350369 for a description of "consecutive tripling steps."
Records for A350369, recorded by the Collatz sequence starting value.
Differs from A213215 in that repeated values are removed, i.e., if a gap in the number of consecutive tripling steps occurs, A213215 will report the starting value multiple times but this sequence will not. Example: The Collatz sequence for 15 has 4 tripling steps but the sequence for 27 has 6, so 27 is reported by A213215 for n=5 and n=6. This sequence only reports 27 once as having set a new record.
Differs from A222598 in that certain consecutive tripling step lengths will not be represented here when a gap in the record number of consecutive tripling steps occurs. Example: Since the consecutive tripling step record moves from 4 in the Collatz sequence for 15 to 6 in the Collatz sequence for 27, this sequence will not report the Collatz sequence for 159 with 5 consecutive tripling steps like A222598 does.

			a(5) = 27 since the Collatz sequence for 27 is the 5th sequence to set a record for the most consecutive tripling steps, i.e., A350369(27) = 6 is the first occurrence of 6 in A350369.

Kevin P. Thompson, Table of n, a(n) for n = 1..44

Cf. A006370, A213215, A222598, A347270, A347668, A350369.

k=0;nmax=0;Do[While[t=0;max=0;NestWhileList[If[OddQ@#,t++;If[t>max,max=t];(3#+1)/2,t=0;#/2]&,++k,#!=1&];maxGiorgos Kalogeropoulos, Jan 11 2022 *)

A350369 a(n) is the length of the longest sequence of consecutive tripling steps in the Collatz (3x+1) sequence beginning at n.

Original entry on oeis.org

0, 0, 2, 0, 1, 2, 3, 0, 3, 1, 2, 2, 1, 3, 4, 0, 1, 3, 2, 1, 1, 2, 3, 2, 2, 1, 6, 3, 2, 4, 6, 0, 2, 1, 2, 3, 3, 2, 3, 1, 6, 1, 3, 2, 1, 3, 6, 2, 3, 2, 2, 1, 1, 6, 6, 3, 3, 2, 2, 4, 3, 6, 6, 0, 3, 2, 2, 1, 1, 2, 6, 3, 6, 3, 2, 2, 2, 3, 4, 1, 3, 6, 6, 1, 1, 3, 3

Offset: 1

Kevin P. Thompson, Dec 27 2021

"Consecutive tripling steps" are repeated (3x+1)/2 operations that are not interrupted by a second division by 2.
This sequence attempts to measure the largest upward thrust in each Collatz sequence and so is correlated to some degree with the maximum value (A025586) and length (A006577) of Collatz sequences.
If n = 2^x * (2^y*z - 1), then a(n) >= y. - Charles R Greathouse IV, Oct 25 2022

			The Collatz sequence for n=7 has a streak of 3 consecutive tripling steps (at 7, 11, and 17), so a(7) = 3.
7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
^      ^       ^

Kevin P. Thompson, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A006577, A006667, A025586, A213215, A221469, A347270, A347409.

a(n)=my(c,r); n>>=valuation(n,2); while(n>1, n+=(n+1)/2; if(n%2, c++, r=max(r,c+1); n>>=valuation(n,2); c=0)); max(r,c) \\ Charles R Greathouse IV, Oct 25 2022

A349592 Number of prime lucky numbers k between powers of 2, 2^n < k < 2^(n+1).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 4, 6, 10, 19, 25, 46, 70, 126, 226, 360, 708, 1237, 2215, 3902, 7281, 13243, 24140, 44006, 81479, 150526, 278981, 519383, 966444, 1803380, 3376209, 6330436, 11904432

Offset: 0

Kevin P. Thompson, Nov 22 2021

			a(6) = 4 since there are 4 prime lucky numbers (67, 73, 79, and 127) between 2^6 = 64 and 2^7 = 128.

Cf. A000040, A000959, A031157, A036378, A309399.

a(n) = # { x in A031157 | 2^n < x < 2^(n+1) }. - M. F. Hasler, Nov 24 2021

cp's OEIS Frontend

User: Kevin P. Thompson

A363491 Numbers k such that 2^k - 5 is a semiprime.

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A361803 Least k > 1 such that k^n - n > 1 is semiprime, or 0 if no such k exists.

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A363374 Numbers k such that 2^k - 3 is a semiprime.

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A359070 Smallest k > 1 such that k^n - 1 is the product of n distinct primes.

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A359069 Smallest prime p such that p^(2n-1) - 1 is the product of 2n-1 distinct primes.

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A352597 a(n) is the smallest k > 1 such that k^n + 1 has all prime divisors p == 1 (mod n).

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A350800 Numbers k such that k and k+1 have the same number and sum of divisors but a different number of distinct prime factors.

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A350370 a(n) is the smallest k such that the Collatz sequence for k includes a record number of consecutive tripling steps.

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A350369 a(n) is the length of the longest sequence of consecutive tripling steps in the Collatz (3x+1) sequence beginning at n.

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