Thomas Ordowski - cp's OEIS frontend

A387016 Permutation of the odd integers >= 3 formed by ordering them first by odd k >= 3 and then by integer m >= 1 in their unique representation (k - 2^m)*2^m + 1.

3, 7, 5, 11, 13, 15, 21, 9, 19, 29, 25, 23, 37, 41, 27, 45, 57, 31, 53, 73, 17, 35, 61, 89, 49, 39, 69, 105, 81, 43, 77, 121, 113, 47, 85, 137, 145, 51, 93, 153, 177, 55, 101, 169, 209, 59, 109, 185, 241, 63, 117, 201, 273, 33, 67, 125, 217, 305, 97

Offset: 1

Thomas Ordowski, Aug 13 2025

A term t must have m = A007814(t-1), and k follows from that so that the representation is unique.
For given k, successive terms have m in the range 1 <= m <= floor(log_2(k)) and this regularity permits a(n) to be calculated from the index n.
The terms where m is the maximum for each k are A369901 (in order) and are a permutation of the Proth numbers A080075.

Amiram Eldar, Table of n, a(n) for n = 1..11703 (terms corresponding to k < 2500)

Cf. A080075 (Proth numbers in ascending order), A369901 (permutation of Proth numbers).

Table[(k - 2^m)*2^m + 1,{k, 3, 35, 2}, {m, 1, Log2[k-1]}] // Flatten (* Amiram Eldar, Aug 13 2025 *)

P(n,m) = (2n+1 - 2^m)*2^m + 1 = (2n+1)*2^m - 4^m + 1, where m > 0 with 2^m < 2n+1, for n > 0.

More terms from Amiram Eldar, Aug 13 2025

A385959 a(0) = 1; a(n) = a(n-1)*(b(n)+1)/(b(n)-1), where b(n) = A385958(n) is the largest prime p such that a(n) is an integer.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 14, 15, 16, 18, 19, 38, 57, 76, 114, 115, 120, 121, 132, 135, 136, 138, 139, 278, 279, 310, 312, 314, 471, 628, 942, 1099, 2198, 2199, 2932, 4398, 5131, 10262, 10995, 10996, 16494, 19243, 38486, 41235, 41236, 41358, 41471, 41838, 41841, 46490, 55788, 55789, 111578, 167367, 168554, 252831, 252832, 252864

Offset: 0

Thomas Ordowski, Jul 13 2025

a(0) = 1; a(n) is the smallest k such that (k + a(n-1))/(k - a(n-1)) is a prime (A385958).

Note that a(n-1)+1 <= a(n) <= 2*a(n-1).

Martin Fuller, Table of n, a(n) for n = 0..3460

Cf. A385958.

a(n) = Product_{k=1..n} (b(k)+1)/(b(k)-1), where b(n) = A385958(n).

a(n) = (1+t(n))/(1-t(n)) with t(n) = tanh(Sum_{k=1..n} arctanh(1/b(k))).

More terms from Morné Louw and Martin Fuller, Jul 15 2025

A385958 a(n) is the largest prime p such that b(n) = b(n-1)*(p+1)/(p-1) is an integer (A385959), where b(0) = 1.

Original entry on oeis.org

3, 5, 7, 5, 13, 3, 29, 31, 17, 37, 3, 5, 7, 5, 229, 47, 241, 23, 89, 271, 137, 277, 3, 557, 19, 311, 313, 5, 7, 5, 13, 3, 4397, 7, 5, 13, 3, 29, 21991, 5, 13, 3, 29, 82471, 677, 733, 227, 27893, 19, 11, 111577, 3, 5, 283, 5, 505663, 15803

Offset: 1

Thomas Ordowski, Jul 13 2025

a(n) = (b(n)+b(n-1))/(b(n)-b(n-1)), where b(n) = A385959(n) is the smallest k such that a(n) is a prime, where b(0) = 1.
a(n) is the largest prime p such that p-1 divides 2*b(n-1).
Note that 3 <= a(n) <= 2*b(n-1)+1.
Does this sequence contain all odd primes?

Martin Fuller, Table of n, a(n) for n = 1..3460

Cf. A065091, A073409, A385959.

allocatemem(2^30);
default(factor_add_primes, 1);
{
my(a,b=1);
for(n=1,100,
  removeprimes(select(p->b%p, addprimes()));
  fordiv(2*b, d, a=2*b/d+1; if(isprime(a),break));
  b+=b*2/(a-1);
  print1(a, ", ");
);
} \\ Martin Fuller, Jul 16 2025

a(n) = A073409(b(n-1)), where b(n) = A385959(n) = Product_{k=1..n} (a(k)+1)/(a(k)-1).

Also tanh(Sum_{k=1..n} arctanh(1/a(k))) = (b(n)-1)/(b(n)+1).

More terms from Morné Louw and Martin Fuller, Jul 15 2025

A386311 a(1) = 2, a(n+1) is the largest prime p such that b(n+1) = b(n)*(p + a(n))/(p - a(n)) is a positive integer, where b(1) = 1.

Original entry on oeis.org

2, 3, 13, 29, 71, 73, 3673, 3677, 1970327, 8879341, 30578677

Offset: 1

Thomas Ordowski, Jul 18 2025

This sequence is finite and full.
Note that a(n) < a(n+1) <= 2*b(n) + a(n).
b(1) = 1, b(n+1) is the smallest k such that a(n+1) = a(n)*(k + b(n))/(k - b(n)) is a prime, where a(1) = 2.
b(n) = 1, 5, 8, 21, 50, 3600, 3746, 6883275, 6909014, 10849668, and 19729009.
Conjecture: a'(n) = prime(n) for "the smallest prime p" and b'(n) = A352743(n-1) for "the largest k".
If a(n+1) <= 2*b(n) + a(n), then a(11) = 30578677 is the last term. - M. F. Hasler, Jul 19 2025

Cf. A000040, A352743 (see author's conjecture).

{a=List(2); b=List(1); for(n=1,oo, print1(a[n]", "); my(an=a[n], bn=b[n], p=precprime(2*bn+an)); iferr(while(bn*(p+an)%(p-an), p=precprime(p-1)), E, break); listput(a, p); listput(b, bn*(p+an)\(p-an))); print("that's all."); a=Vec(a)} \\ M. F. Hasler, Jul 19 2025

Product_{k=1..n} (a(k+1) + a(k))/(a(k+1) - a(k)) = b(n+1).
a(n+1)/a(n) = (b(n+1) + b(n))/(b(n+1) - b(n)).
b(n+1)/b(n) = (a(n+1) + a(n))/(a(n+1) - a(n)).

a(7)-a(10) from M. F. Hasler, Jul 18 2025

a(11) = 2*b(10)+a(10) from Thomas Ordowski, Jul 19 2025

A384438 Composite numbers k such that ((2^k+1)/3)^k == 1 (mod k^2).

Original entry on oeis.org

341, 1105, 1387, 1729, 1771, 2047, 2465, 2485, 2701, 2821, 3277, 3445, 4033, 4369, 4681, 5185, 5461, 6601, 7957, 8321, 8911, 9361, 10261, 10585, 11305, 11713, 11891, 13741, 13747, 13981, 14491, 15709, 15841, 16105, 16705, 18145, 18721, 19951, 23377, 28441, 29341

Offset: 1

Thomas Ordowski, May 29 2025

nonn

If p > 3 is prime, then ((2^p+1)/3)^p == 1 (mod p^2).
Fermat pseudoprimes to base 2 not divisible by 3 (A066488) are a proper subsequence.
The terms k that are not 2^(k-1) == 1 (mod k) are 1771, 2485, 3445, 5185, 9361, ...

Cf. A001567, A066488 (subsequence), A384148.

isok(k) = (k>1) && (k%2) && !isprime(k) && (Mod((2^k+1)/3, k^2)^k == 1); \\ Michel Marcus, May 29 2025

a(18)-a(41) from Jinyuan Wang, May 29 2025

A384456 Positive integers k such that (2^k - 1)^k + 2 is prime.

Original entry on oeis.org

1, 2, 4, 8, 16, 40

Offset: 1

Thomas Ordowski, May 30 2025

If it exists, a(7) > 800. - Hugo Pfoertner, Jun 03 2025

Cf. A019434.

Select[Range[100],PrimeQ[(2^#-1)^#+2] &] (* Stefano Spezia, Jun 01 2025 *)

isok(k) = ispseudoprime((2^k - 1)^k + 2); \\ Michel Marcus, May 30 2025

A384148 Numbers k such that (2^k-1)^k == 1 (mod (2^k+1)*k^2) and 2^(k-1) != 1 (mod k).

Original entry on oeis.org

30457, 33865, 80185, 82621, 86785, 104845, 212401, 250705

Offset: 1

Thomas Ordowski, May 20 2025

If p > 3 is prime, then (2^p-1)^p == 1 (mod (2^p+1)*p^2).
Generally, if m is not divisible by 3 and 2^(m-1) == 1 (mod m), then (2^m-1)^m == 1 (mod (2^m+1)*m^2).
However, there are composite numbers satisfying this congruence that are not Fermat pseudoprimes to base 2. These exceptions constitute this sequence.

Cf. A001567, A066488 (Fermat pseudoprimes to base 2 that are not divisible by 3).

isok(k) = if (!isprime(k) && (Mod(2, k)^(k-1) != 1), Mod((2^k-1),(2^k+1)*k^2)^k == 1); \\ Michel Marcus, May 20 2025

a(3)-a(6) from Michel Marcus, May 21 2025

a(7)-a(8) from Michael S. Branicky, May 28 2025

A383431 a(n) is the denominator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.

Original entry on oeis.org

1, 2, 11, 18, 127, 463, 1717, 3218, 24311, 92379, 352717, 1352079, 5200301, 20058301, 77558761, 150270098, 1166803111, 4537567651, 17672631901, 68923264411, 269128937221, 1052049481861, 4116715363801, 16123801841551, 63205303218877, 247959266474053, 973469712824057, 3824345300380221, 15033633249770521

Offset: 1

Thomas Ordowski, Apr 27 2025

a(2^m) is even for m > 0.

			Denominators of 0, 1/2, 9/11, 17/18, 125/127, 461/463, 1715/1717, 3217/3218, ...

Cf. A001700, A382257 (numerators).

a(n) = (binomial(2n-1, n-1) + 1)/2 if n = 2^m or a(n) = binomial(2n-1, n-1) + 1 otherwise, because tanh(Sum_{k=1..n-1} artanh(k/n)) = (binomial(2n-1, n-1) - 1)/(binomial(2n-1, n-1) + 1) reduced.

a(n) = A382257(n) + 1 if n = 2^m or a(n) = A382257(n) + 2 otherwise.

A377026 Numbers k such that 2^(2^k+1) == -1 (mod k^2).

Original entry on oeis.org

1, 3, 9, 171, 3249, 97641, 1855179, 29884473, 55753011, 567804987, 892145817, 1059307209

Offset: 1

Thomas Ordowski, Oct 13 2024

It seems that these are cubefree terms of A006521.

Subsequence of A006521.

A376473 Numbers k such that 2^(2^(k-1)-1) == 1 (mod k^2) and 2^(k-1) =/= 1 (mod k).

Original entry on oeis.org

951481, 2215441, 28758601, 81844921, 1221936841, 10370479321, 16287076081, 26946809137, 33663998161, 35094800881, 134619011281, 188455112353, 299226038833, 314240366881, 383116075201, 594981050401, 1230227375833, 1572186445201, 2096189123113, 2377714473001

Offset: 1

Thomas Ordowski, Sep 24 2024

nonn

The terms k of A374953 for which A002326((k-1)/2) is odd.
Numbers k in A376253 that are not strong pseudoprimes to base 2.
Every term of this sequence must have a Wieferich prime factor (for example, 951481 = 271 * 3511). The Wieferich prime 1093 cannot divide such a number (see A374953).

Subsequence of A374953.

Cf. A001220, A002326, A001262, A001567, A376253.

q[k_] := Module[{m = MultiplicativeOrder[2, k^2]}, PowerMod[2, k - 1, m] == 1]; Select[Range[1, 2300000, 2], PowerMod[2, # - 1, #] != 1 && q[#] &] (* Amiram Eldar, Sep 24 2024 *)

is(k) = (k > 1) && k % 2 && !isprime(k) && Mod(2, k)^(k-1) != 1 && Mod(2, znorder(Mod(2, k^2)))^(k-1) == 1; \\ Amiram Eldar, Sep 24 2024

list(lim)=my(v=List()); if(lim>3<<64, warning("May miss multiples of Wieferich primes > 2^64.")); forstep(n=10533,lim,7022, if(Mod(2, znorder(Mod(2, n^2)))^(n-1) == 1 && Mod(2,n)^n != 2, listput(v,n))); Vec(v) \\ Charles R Greathouse IV, Sep 24 2024

More terms from Amiram Eldar, Sep 24 2024

cp's OEIS Frontend

User: Thomas Ordowski

A387016 Permutation of the odd integers >= 3 formed by ordering them first by odd k >= 3 and then by integer m >= 1 in their unique representation (k - 2^m)*2^m + 1.

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A385959 a(0) = 1; a(n) = a(n-1)*(b(n)+1)/(b(n)-1), where b(n) = A385958(n) is the largest prime p such that a(n) is an integer.

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A385958 a(n) is the largest prime p such that b(n) = b(n-1)*(p+1)/(p-1) is an integer (A385959), where b(0) = 1.

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A386311 a(1) = 2, a(n+1) is the largest prime p such that b(n+1) = b(n)*(p + a(n))/(p - a(n)) is a positive integer, where b(1) = 1.

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A384438 Composite numbers k such that ((2^k+1)/3)^k == 1 (mod k^2).

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A384456 Positive integers k such that (2^k - 1)^k + 2 is prime.

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A384148 Numbers k such that (2^k-1)^k == 1 (mod (2^k+1)*k^2) and 2^(k-1) != 1 (mod k).

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A383431 a(n) is the denominator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.

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A377026 Numbers k such that 2^(2^k+1) == -1 (mod k^2).

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