cp's OEIS Frontend

In a tree with binary nodes (0, 1 children only), the maximum number of unique child nodes at level n.

Number of binary trees (each vertex has 0, or 1 left, or 1 right, or 2 children) such that all leaves are at level n. Example: a(1) = 3 because we have (i) root with a left child, (ii) root with a right child and (iii) root with two children. a(n) = A000215(n) - 2. - Emeric Deutsch, Jan 20 2004

Similarly, this is also the number of full balanced binary trees of height n. (There is an obvious 1-to-1 correspondence between the two sets of trees.) - David Hobby (hobbyd(AT)newpaltz.edu), May 02 2010

Partial products of A000215.

The first 5 terms n (only) have the property that phi(n)=(n+1)/2, where phi(n) = A000010(n) is Euler's totient function. - Lekraj Beedassy, Feb 12 2007

If A003558(n) is of the form 2^n and A179480(n+1) is even, then (2^(A003558(n)) - 1) is in A051179. Example: A003558(25) = 8 with A179480(25) = 4, even. Then (2^8 - 1) = 255. - Gary W. Adamson, Aug 20 2012

For any odd positive a(0), the sequence defined by a(n) = a(n-1) * (a(n-1) + 2) gives a constructive proof that there exist integers with at least n distinct prime factors, e.g., a(n), since omega(a(n)) >= n. As a corollary, this gives a constructive proof of Euclid's theorem stating that there are infinitely many primes. - Daniel Forgues, Mar 07 2017

From Sergey Pavlov, Apr 24 2017: (Start)

I conjecture that, for n > 7, omega(a(n)) > omega(a(n-1)) > n.

It seems that the largest prime divisor p(n+1) of a(n+1) is always bigger than the largest prime divisor of a(n): p(n+1) > p(n). For 3 < n < 8, p(n+1) > 100 * p(n).

(End)

It appears that a(n) is the integer whose bits indicate the possible subset sums of the first n powers of two. For another example, see the calculation for primes at A368491 - Yigit Oktar, Mar 20 2025

Arises in the approximation of 14-fold quasipatterns by 14 Fourier modes.

Let DTS(n^c) denote the set of languages accepted by a deterministic Turing machine with space n^(o(1)) and time n^(c+o(1)), and let SAT denote the Boolean satisfiability problem. Then (1) SAT is not in DTS(n^c) for any c < 2*cos(Pi/7), and (2) the Williams inference rules cannot prove that SAT is not in DTS(n^c) for any c >= 2*cos(Pi/7). These results also apply to the Boolean satisfiability problem mod m where m is in A085971 except possibly for one prime. - Charles R Greathouse IV, Jul 19 2012

rho(7):= 2*cos(Pi/7) is the length ratio (smallest diagonal)/side in the regular 7-gon (heptagon). The algebraic number field Q(rho(7)) of degree 3 is fundamental for the 7-gon. See A187360 for the minimal polynomial C(7, x) of rho(7). The other (larger) diagonal/side ratio in the heptagon is sigma(7) = -1 + rho(7)^2, approx. 2.2469796. (see the decimal expansion in A231187). sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences like A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon. See the P. Steinbach reference. - Wolfdieter Lang, Nov 21 2013

An algebraic integer of degree 3 with minimal polynomial x^3 - x^2 - 2x + 1. - Charles R Greathouse IV, Nov 12 2014

The other two solutions of the minimal polynomial of rho(7) = 2*cos(Pi/7) are 2*cos(3*Pi/7) and 2*cos(5*Pi/7). See eq. (20) of the W. Lang link. - Wolfdieter Lang, Feb 11 2015

The constant is the square root of 3.24697... (cf. A116425). It is the fifth-longest diagonal in the regular 14-gon with unit radius, which equals 2*sin(5*Pi/14). - Gary W. Adamson, Feb 14 2022

Row sums are odd-indexed Fibonacci numbers.

T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k double rises. Mirror image of A085478. - Emeric Deutsch, May 31 2004

Diagonal sums are A052535. - Paul Barry, Jan 21 2005

Matrix inverse is the triangle of Salie numbers A098435. - Paul Barry, Jan 21 2005

Coefficients of Morgan-Voyce polynomial b(n,x); e.g., b(3,x)=x^3+5x^2+6x+1. See A172431 for coefficients of Morgan-Voyce polynomial B(n,x). - Clark Kimberling, Feb 13 2010

T(n,k) is the number of stack polyominoes of perimeter 2n+4 with k+1 columns. - Emanuele Munarini, Apr 07 2011

Roots of signed n-th polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths A003558(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697... = A116415; we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...; the 3 roots to the polynomial with cycle length 3 matching A003558(3) = 3. The operation (-2, x^2) is the reversal of the well known chaotic operation (x^2 - 2) [Kappraff, Adamson, 2004] starting with seed 2*cos(2*Pi/N). Check: given 2*cos(2*Pi/7) = 1.24697..., we obtain the 3-cycle using (x^2 - 2): (1.24697...-> -0.445041...-> 1.801937...; where the terms in either set are intermediate terms in the other, irrespective of sign. - Gary W. Adamson, Sep 22 2011

A054142 is jointly generated with A172431 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section of A172431. - Clark Kimberling, Mar 09 2012

Subtriangle of the triangle given by (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 01 2012

The o.g.f. for row n of the array A(n, k) = binomial(2*n-k,k), k >= 0, n >= 0 is G(n,x) = Sum_{k=0..n} T(n, k)*x^k + (-x)^(2*n+1) * c(-x)^(2*n+1) / sqrt(1-4*(-x)), for n >= 0. Here c(x) is the o.g.f. of A000108 (Catalan). For powers of c(x) see the W. Lang link in A115139. For the alternating sign case replace x by -x. - Wolfdieter Lang, Sep 12 2016

Multiplying the n-th diagonal by A001147(n) generates A001497. - Tom Copeland, Oct 04 2016

Pseudoprimes for the primality test from [Schick]: n odd is probably prime if (n-1) | A003558((n-1)/2). (Succeeds for 99.9975% of odd natural numbers less than 10^8.) - Jonathan Skowera, Jun 29 2013

Equivalently, these are composites n such that ((n-1)/2)^((n-1)/2) == +-1 (mod n). - Thomas Ordowski, Nov 28 2023

The period of the expansion of 1/p, base N (where N=4), is equivalent to determining for base integer 4, the period of the sequence 1, 4, 4^2, 4^3, ... mod p. Thus the cycle length for base 4, 1/7 = 0.021021021... (cycle length 3).

The cycle length, base 4, mod p, is equivalent to "clock cycles", given angle A, then the algebraic identity for the doubling angle, 2A.

Examples: Given cos A, f(x) for 2A = 2x^2 - 1, seed 2 Pi/7, i.e., (.623489801 == (arrow), -.222520934... == -.900968867...== .623489801...(cycle length 3). Given 2 cos A, the algebraic identity for 2 cos 2A, f(x) = x^2 - 2; e.g., given seed 2 cos A = 2 Pi/7, the 3 cycle is 1.246979604...== .445041867...== -1.801937736...== back to 1.24697... Likewise, the doubling function given sin^2 A, f(x) for sin^2 2A = 4x(1 - x), the logistic equation; getting cycle length of 3 using the seed sin^2 2 Pi/7. Similarly, the doubling function for tan 2A given tan A, where A = 2 Pi/7 gives 2x/(1 - x^2), cycle length of 3. The doubling function for cot 2A given cot A, with A = 2 Pi/7 gives (x^2 - 1)/2x, cycle length of 3. Note that (x^2 - 1)/2x = sinh(log(x)), and is also generated from using Newton's method on x^2 + 1 = 0.

Consider the odd pseudoprimes, composite numbers x such that 2^(x-1) = 1 mod x, that have prime(n) as a factor. It appears that all such x can be factored as prime(n) * (2 a(n) k + 1) for some integer k. For example, the first few pseudoprimes having the factor 31 are 31*11, 31*91, 31*141 and 3*151. The 11th prime is 31 and a(11) = 5. Therefore all the cofactors of 31 should have the form 10k+1, which is clearly true. - T. D. Noe, Jun 10 2003

A dual sequence to A179382.

Let b = (2*n-1) and k = A003558(n-1). If a(n) is odd, b divides (2^k + 1); but if a(n) is even, b divides (2^k - 1). Examples: a(14) = 5, odd; with b = 27 and A003558(13) = 9. Then 27 divides (2^9 + 1) or 513 = 27 * 19. a(18) = 6, even. b = 35, with k= A003558(17) = 12. Then 35 divides (2^12 - 1). - Gary W. Adamson, Aug 20 2012.

Iff a(n) = n/2 or (n-1)/2, then 2*n - 1 is a prime with one coach and is in A216371. Examples: a(19) = 9, so 37 is in A216371. a(12) = 6, so 23 is in A216371. - _Gary W. Adamson, Sep 08 2012.

The troubadour Arnaut Daniel composed sestinas based on the permutation 123456 -> 615243, which cycles after 6 iterations.

Roubaud quotes the number 141, but the corresponding Queneau-Daniel permutation is only of order 47 = 141/3.

This appears to coincide with the numbers n such that a type-2 optimal normal basis exists for GF(2^n) over GF(2). But are these two sequences really the same? - Joerg Arndt, Feb 11 2008

The answer is Yes - see Theorem 2 of the Dumas reference. [Jean-Guillaume Dumas (Jean-Guillaume.Dumas(AT)imag.fr), Mar 20 2008]

From Peter R. J. Asveld, Aug 17 2009: (Start)

a(n) is the n-th T-prime (Twist prime). For N >= 2, the family of twist permutations is defined by

p(m,N) == +2m (mod 2N+1) if 1 <= m < k = ceiling((N+1)/2),

p(m,N) == -2m (mod 2N+1) if k <= m < N.

N is T-prime if p(m,N) consists of a single cycle of length N.

The twist permutation is the inverse of the Queneau-Daniel permutation.

N is T-prime iff p=2N+1 is a prime number and exactly one of the following three conditions holds;

(1) N == 1 (mod 4) and +2 generates Z_p^* (the multiplicative group of Z_p) but -2 does not,

(2) N == 2 (mod 4) and both +2 and -2 generate Z_p^*,

(3) N == 3 (mod 4) and -2 generate Z_p^* but +2 does not. (End)

The sequence name says the permutation is of order n, but P. R. J. Asveld's comment says it's an n-cycle. Is there a proof that those conditions are equivalent for the Queneau-Daniel permutation? (They are not equivalent for any arbitrary permutation; e.g., (123)(45)(6) has order 6 but isn't a 6-cycle.) More generally, I have found that for all n <= 9450, (order of Queneau-Daniel permutation) = (length of orbit of 1) = A003558(n). Does this hold for all n? - David Wasserman, Aug 30 2011

Given that prime p has only one coach, the corresponding value of k in A003558 must be (p-1)/2, and vice versa. Using the Coach theorem of Jean Pedersen et al., phi(b) = 2 * c * k, with b odd. Let b = p, prime. Then phi(p) = (p-1), and k must be (p-1)/2 iff c = 1. Or, phi(p) = (p-1) = 2 * 1 * (p-1)/2.

Conjecture relating to odd integers: iff an integer is in the set A216371 and is either of the form 4q - 1 or 4q + 1, (q>0); then the top row of its coach (cf. A003558) is composed of a permutation of the first q odd integers. Examples: 11 is of the form 4q - 1, q = 3; with the top row of its coach [1, 5, 3]. 13 is of the form 4q + 1, q = 3; so has a coach of [1, 3, 5]. 37 is of the form 4q + 1, q = 9; so has a coach with the top row composed of a permutation of the first 9 odd integers: [1, 9, 7, 15, 11, 13, 3, 17, 5]. - Gary W. Adamson, Sep 08 2012

Odd primes p such that 2^m is not congruent to 1 or -1 (mod p) for 0 < m < (p-1)/2. - Charles R Greathouse IV, Sep 15 2012

These are also the odd primes a(n) for which there is only one periodic Schick sequence (see the reference, and also the Brändli and Beyne link, eq. (2) for the recurrence but using various inputs. See also a comment in A332439). This sequence has primitive period length (named pes in Schick's book) A003558((a(n)-1)/2) = A005034(a(n)) = A000010(a(n))/2 = (a(n) - 1)/2, for n >= 1. - Wolfdieter Lang, Apr 09 2020

From Jianing Song, Dec 24 2022: (Start)

Primes p such that the multiplicative order of 4 modulo p is (p-1)/2. Proof of equivalence: let ord(a,k) be the multiplicative of a modulo k.

If 2^m is not 1 or -1 (mod p) for 0 < m < (p-1)/2, then ord(2,p) is either p-1 or (p-1)/2. If ord(2,p) = p-1, then ord(4,p) = (p-1)/2. If ord(2,p) = (p-1)/2, then p == 3 (mod 4), otherwise 2^((p-1)/4) == -1 (mod p), so ord(4,p) = (p-1)/2.

Conversely, if ord(4,p) = (p-1)/2, then ord(2,p) = p-1, or ord(2,p) = (p-1)/2 and p == 3 (mod 4) (otherwise ord(4,p) = (p-1)/4). In the first case, (p-1)/2 is the smallest m > 0 such that 2^m == +-1 (mod p); in the second case, since (p-1)/2 is odd, 2^m == -1 (mod p) has no solution. In either case, so 2^m is not 1 or -1 (mod p) for 0 < m < (p-1)/2.

{(a(n)-1)/2} is the sequence of indices of fixed points of A053447.

A prime p is a term if and only if one of the two following conditions holds: (a) 2 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 2 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 8) since 2 is a quadratic residue modulo p). (End)

From Jianing Song, Aug 11 2023: (Start)

Primes p such that 2 or -2 (or both) is a primitive root modulo p. Proof of equivalence: if ord(2,p) = p-1, then clearly ord(4,p) = (p-1)/2. If ord(-2,p) = p-1, then we also have ord(4,p) = (p-1)/2. Conversely, suppose that ord(4,p) = (p-1)/2, then ord(2,p) = p-1 or (p-1)/2, and ord(-2,p) = p-1 or (p-1)/2. If ord(2,p) = ord(-2,p) = (p-1)/2, then we have that (p-1)/2 is odd and (-1)^((p-1)/2) == 1 (mod p), a contradiction.

A prime p is a term if and only if one of the two following conditions holds: (a) -2 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of -2 modulo p is (p-1)/2 (in this case, we have p == 3 (mod 8) since -2 is a quadratic residue modulo p). (End)

No terms are congruent to 1 modulo 8, since otherwise we would have 4^((p-1)/4) = (+-2)^((p-1)/2) == 1 (mod p). - Jianing Song, May 14 2024

The n-th prime A000040(n) is a term iff A376010(n) = 2. - Max Alekseyev, Sep 05 2024

Since for some a < n, 2^a == 1 (mod n) (a consequence of Euler's Theorem), searching up to k=n is sufficient to determine whether an integer is in the sequence. - Michael B. Porter, Dec 06 2009

A195470(a(n)) > 0; A195610(n) gives the smallest k such that a(n) divides 2^k + 1. - Reinhard Zumkeller, Sep 21 2011

This sequence is the subset of odd integers > 1 as (2*n - 1) in A179480, such that the corresponding entry in A179480 is odd. Example: A179480(14) = 5, odd, with (2*14 - 1) = 27; and 5 is a term of this sequence. A014659 (odd and does not divide (2^k + 1) for any k >= 1) represents the subset of odd terms >1 corresponding to A179480 entries that are even. - Gary W. Adamson, Aug 20 2012

All prime factors of a(n) are in A091317. Sequence has asymptotic density 0. - Robert Israel, Aug 12 2014

This sequence, for m>2, is those m for which, for some e, (m-1)(2^e-1)/m is a term of A253608. Moreover, e(n) is 2*A195610(n) when m is a(n). - Donald M Davis, Jan 12 2018

From Wolfdieter Lang, Aug 22 2020: (Start)

Without a(2) = 2 this is the complement of A014659 relative to the odd positive integers A005408.

For the least nonnegative integer k(n) with 2^k(n) + 1 == d(n)*a(n), for n >= 1, see k(n) = A195610(n) and d(n) = A337220(n).

Starting with a(3) = 3 these numbers are the odd moduli, named 2*n+1 in the definition of A003558, for which the minus signs applies (see A332433(m) for the signs applying for A003558(m)). (End)

Given the quasi-order and coach theorems of Hilton and Pedersen (cf. A003558): phi(b) = 2 * k * c with b odd. Dividing through by 2 we obtain phi(b)/2 = k * c which is the same as a(n) = A003558(n-1) * A135303(n-1). Here k refers to the "least exponent" (cf. A003558), while c is the number of coaches for odd b (cf. A135303). - Gary W. Adamson, Sep 25 2012

After the initial term, this is the totient function phi(2n-1)/2 (A037225(n-1)/2). Also a(n) is the number of partitions of the odd number (2n+1) into two ordered relatively prime parts. If p and q are such parts where p > q and p+q = 2n+1 then they can generate primitive Pythagorean triples of the form (p^2 - q^2, 2*p*q, p^2 + q^2). - Frank M Jackson, Oct 30 2012