cp's OEIS Frontend

A brilliant number is a semiprime (products of two primes, A001358) whose two prime factors have the same number of decimal digits. For an n-digit brilliant number, the two prime factors must each have ceiling(n/2) decimal digits.

Since all brilliant numbers are semiprimes, a(n) >= A098449(n), also, a(n) = A098449(n) for n = 1, 2, 4, 16, 78, ..., are there infinitely many n such that a(n) = A098449(n)?

For every n, all sufficiently large primes p such that n*p+1 is a perfect power are of the form ((n+1)^q-1)/n with q prime.

a(34) = (35^313-1)/34 is too large to include; it has 482 decimal digits.

a(35) - a(37) = {37, 61, 1483}.

a(38) = (39^349-1)/38 is too large to include; it has 554 decimal digits.

a(39) - a(100) = {5, 2, 43, 3500201, 5, 71, 43, 3851, 178481, 11, 47, 3221, 5, 178250690949465223, 2971, 127, 53, 3, 7, 3541, 61, 2, 59, 2, 61, 17, 3, 751410597400064602523400427092397, 21700501, 4831, 7, 19, 73, 5, 7, 5701, 73, 6007, 79, 39449441, 6481, 19, 79, 48037081, 6218272796370530483675222621221, 2, 3, 438668366137, 89, 5, 23, 331, 89, 654022685443, 11, 1001523179, 97, 3, 792806586866086631668831, 9901, 97, 10303}.

If n*p+1 = m^k, then n*p = m^k-1 = (m-1)*(m^(k-1) + m^(k-2) + ... + m + 1). If p >= n, then m^k = n*p+1 >= n^2+1 > n^2, and we have these three cases: Case 1: m-1 > n, then p can't be prime. Case 2: m-1 = n, this is A084738. Case 3: m-1 < n. If gcd(n, m-1) != m-1, then because m^(k-1) + m^(k-2) + ... + m + 1 > n, p can't be prime. This implies m-1 | n. The three cases means that we only need to check p < n and numbers m such that m-1 | n.

The first numbers n such that a(n) = 0 are {124, 215, 224, 242, ...}. a(268) is unknown; it is the smallest prime of the form (269^q - 1)/268 with prime q if such a prime exists (in which case it must be greater than (269^63659-1)/268), otherwise 0.

Sequence is finite with 77 terms, the largest being 5*10^30 + 27 (which can be written 5(0_28)27, where 0_28 means the string of 28 0's). See text file for proof (this file also has proofs for bases 2, 3, 4, 5, 6, 8, 12).

Minimal elements for the base b representations of the primes > b for other bases b: (see the text file for 9 <= b <= 16) (all written in base b)

b=2: {11}

b=3: {12, 21, 111}

b=4: {11, 13, 23, 31, 221}

b=5: {12, 21, 23, 32, 34, 43, 104, 111, 131, 133, 313, 401, 414, 3101, 10103, 14444, 30301, 33001, 33331, 44441, 300031, 10^95 + 13}

b=6: {11, 15, 21, 25, 31, 35, 45, 51, 4401, 4441, 40041}

b=7: {14, 16, 23, 25, 32, 41, 43, 52, 56, 61, 65, 113, 115, 131, 133, 155, 212, 221, 304, 313, 335, 344, 346, 364, 445, 515, 533, 535, 544, 551, 553, 1022, 1051, 1112, 1202, 1211, 1222, 2111, 3031, 3055, 3334, 3503, 3505, 3545, 4504, 4555, 5011, 5455, 5545, 5554, 6034, 6634, 11111, 11201, 30011, 30101, 31001, 31111, 33001, 33311, 35555, 40054, 100121, 150001, 300053, 351101, 531101, 1100021, 33333301, 5100000001, 33333333333333331} (conjectured, not proven)

b=8: {13, 15, 21, 23, 27, 35, 37, 45, 51, 53, 57, 65, 73, 75, 107, 111, 117, 141, 147, 161, 177, 225, 255, 301, 343, 361, 401, 407, 417, 431, 433, 463, 467, 471, 631, 643, 661, 667, 701, 711, 717, 747, 767, 3331, 3411, 4043, 4443, 4611, 5205, 6007, 6101, 6441, 6477, 6707, 6777, 7461, 7641, 47777, 60171, 60411, 60741, 444641, 500025, 505525, 3344441, 4444477, 5500525, 5550525, 55555025, 444444441, 744444441, 77774444441, 7777777777771, 555555555555525, (10^220-1)/9*40 + 7}.

Equivalently: primes > 10 such that no proper substring (i.e., deleting any positive number of digits) is again a prime > 10. - M. F. Hasler, May 03 2022

a(26) = T(5,3) is conjectured to be 0, but this has not been proved.

By Fermat's last theorem, T(n,2) = 0 for n > 2.

Euler's sum of powers conjecture is that T(n,k) = 0 for n > k > 1, but this conjecture is not true: T(4,3) = 422481, T(5,4) = 144, there are no known counterexamples for n >= 6.

There are no known 0s for k > 2.

Conjecture: If T(n,k) = 0, then T(r,k) = T(n,s) = T(r,s) = 0 for all r >= n, 2 <= s <= k.

Numbers not appearing in A047994.

Indices of -1 in A135347.

Unitary version of A007617.

This sequence to A047994 is A007617 to A000010.

This sequence to A135347 is A007617 to A049283 (for the case that no such numbers exist, A135347 uses -1 and A049283 uses 0).

All odd numbers not of the form 2^k-1 (i.e. not in A000225) are in this sequence, since uphi(n) = A047994(n) is an even number unless n is a power of 2 (A000079), in this case uphi(n) = n-1.

The intersection of this sequence and A049225 is empty, since for squarefree numbers, all divisors are unitary divisors, note that the intersection of this sequence and A002202 is not empty, the number 110 is in both sequences.

By definition, this is the union of A000396, A259180, and A122726. However, at present A122726 is not known to be complete. There is no proof that 564 (for example) is missing from this sequence. - N. J. A. Sloane, Sep 17 2021

Numbers m for which there exists k>=1 such that s^k(m) = m, where s is A001065.

Conjecture: There are no aliquot cycles containing even numbers and odd numbers simultaneously, i.e., every aliquot cycle either has only even numbers or has only odd numbers.

This sequence is a permutation of the nonnegative integers iff Catalan's aliquot sequence conjecture (also called Catalan-Dickson conjecture) is true.

a(563) = 276 is the smallest number whose aliquot sequence has not yet been fully determined.

As long as the aliquot sequence of 276 is not known to be finite or eventually periodic, a(563+k) = A008892(k).

Smallest k such that the k-th Fibonacci polynomial evaluated at x=n is prime. (The first few Fibonacci polynomials are 1, x, x^2 + 1, x^3 + 2*x, x^4 + 3*x^2 + 1, x^5 + 4*x^3 + 3*x, ...)

All terms are primes, since if a divides b, then the a-th term of the n-Fibonacci sequence also divides the b-th term of the n-Fibonacci sequence.

Corresponding primes are 2, 2, 3, 17, 5, 37, 7, 4289, 726120289954448054047428229, 101, 11, 21169, 13, 197, 82088569942721142820383601, 257, 17, 34539049, 19, 401, ...

a(n) = 2 if and only if n is prime.

a(n) = 3 if and only if n^2 + 1 is prime (A005574), except n=2 (since 2 is the only prime p such that p^2 + 1 is also prime).

a(34) > 1024, does a(n) exist for all n >= 1? (However, 17 is the only prime in the first 1024 terms of the 4-Fibonacci sequence, and it seems that 17 is the only prime in the 4-Fibonacci sequence.)

a(35)..a(48) = 71, 3, 2, 17, 11, 3, 2, 37, 2, 31, 5, 11, 2, 5, a(50)..a(54) = 11, 11, 23, 2, 3, a(56) = 3, a(58)..a(75) = 5, 2, 47, 2, 5, 311, 13, 233, 3, 2, 5, 11, 5, 2, 7, 2, 3, 5. Unknown terms a(34), a(49), a(55), a(57), exceed 1024, if they exist.

a(49) > 20000, if it exists. - Giovanni Resta, Jun 06 2018