0,3

Table of n, a(n) for n=0..24.

a(n) ~ 2^(n+2) * n! / Pi^(n+1). - Vaclav Kotesovec, May 06 2021

seq(add(A109449(n-k, k), k = 0..n/2), n = 0..25);

Table[Sum[Binomial[n-k, k] * 2^(n-2*k) * Abs[EulerE[n-2*k, 1/2] + EulerE[n-2*k, 1]], {k, 0, Floor[n/2]}] - (1 + (-1)^n)/2, {n, 0, 25}] (* Vaclav Kotesovec, May 06 2021 *)

Cf. A109449.

nonn,new

Peter Luschny, May 06 2021

approved

0,2

Table of n, a(n) for n=0..7.

Cf. A014577, A337580, A343181, A343182.

nonn,new

N. J. A. Sloane, May 06 2021

approved

0,4

Table of n, a(n) for n=0..21.

a(n) = (-2)^n*Sum_{k=0..n} A109449(n, k)*(-1/2)^k.

From Vaclav Kotesovec, May 06 2021: (Start)

a(n) ~ (-1)^n * exp(-Pi/4) * 4^(n+1) * n! / Pi^(n+1).

E.g.f.: exp(x)*(1 - tan(x))/(1 + tan(x)). (End)

a := n -> add((-1)^k*binomial(n, k)*A000831(k), k=0..n):

seq(a(n), n = 0..21);

Table[-1 + Sum[(-1)^k * Binomial[n, k] * 4^k * Abs[EulerE[k, 1/2] + EulerE[k, 1]], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 06 2021 *)

Cf. A000831, A000834, A109449.

sign,new

Peter Luschny, May 06 2021

approved

0,2

Table of n, a(n) for n=0..19.

a(n) = binomial(2*n, n) * |Euler(n) - 2^n*Euler(n, 0)|.

a(n) = A109449(2*n, n) for n >= 1.

a(n) = (-1)^binomial(n, 2) * binomial(2*n, n) * A163747(n).

a(n) ~ 2^(3*n + 5/2) * n^n / (Pi^(n+1) * exp(n)). - Vaclav Kotesovec, May 06 2021

a := n -> binomial(2*n, n)*abs(euler(n) - 2^n*euler(n, 0)):

seq(a(n), n=0..19);

Table[Binomial[2*n, n]*Abs[EulerE[n] - 2^n*EulerE[n, 0]], {n, 0, 20}] (* Vaclav Kotesovec, May 06 2021 *)

Cf. A109449, A163747.

nonn,new

Peter Luschny, May 06 2021

approved

0,2

Take a sheet of paper, and fold the right edge up and onto the left edge. Do this n times. and unfold. Write a 0 for every valley and a 1 for every ridge, and read the sequence backwards.

a(7) is too large to include in the DATA section.

Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.

Sunggye Lee, Jinsoo Kim, and Won Choi, Relation between folding and un-folding paper of rectangle and (0,1)-pattern [Korean], J. Korean Soc. Math. Ed. Ser. E, 23(3) (2009), 507-522.

Rémy Sigrist and N. J. A. Sloane, Two-Dimensional Paper-Folding, Manuscript in preparation, May 2021.

Table of n, a(n) for n=0..5.

When converted to base 10 we get A343183.

Cf. A014577. A variant of A343181.

nonn,new

N. J. A. Sloane, May 06 2021

approved

1,2

This sequence is infinite as it contains 7^n. - David A. Corneth, May 06 2021

Table of n, a(n) for n=1..37.

q:= k-> is(0=11&^k+7&^k+5&^k+3&^k+2&^k mod k):

select(q, [$1..100000])[];  # Alois P. Heinz, May 06 2021

q[k_] := Divisible[Plus @@ (PowerMod[#, k, k] & /@ {2, 3, 5, 7, 11}), k]; Select[Range[10^6], q] (* Amiram Eldar, May 06 2021 *)

(PARI) is(n) = lift(Mod(11, n)^n + Mod(7, n)^n + Mod(5, n)^n + Mod(3, n)^n + Mod(2, n)^n) == 0 \\ David A. Corneth, May 06 2021

Cf. A000420, A045576, A220170.

nonn,new

Zak Seidov, May 06 2021

a(24)-a(37) from Alois P. Heinz, May 06 2021

approved

1,8

Conjecture 1: a(n) > 0 for all n > 7.

We have verified a(n) > 0 for all n = 8..50000. Clearly, a(2*n) > 0 if a(n) > 0.

Conjecture 2: For any integer n > 7, we can write n as x + y + z with x,y,z positive integers such that x^2 + 2*y^2 + 3*z^2 is a square.

Conjecture 3: For any integer n > 4, we can write n as x + y + z with x,y,z positive integers such that 3*x^2 + 4*y^2 + 5*z^2 (or x^2 + 3*y^2 + 5*z^2) is a square.

Zhi-Wei Sun, Table of n, a(n) for n = 1..3200

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].

a(4) = 1, and 4 = 1 + 1 + 2 with 1^2 + 4*1^2 + 5*2^2 = 5^2.

a(5) = 1, and 5 = 2 + 2 + 1 with 2^2 + 4*2^2 + 5*1^2 = 5^2.

a(9) = 1, and 9 = 4 + 1 + 4 with 4^2 + 4*1^2 + 5*4^2 = 10^2.

a(14) = 1, and 14 = 7 + 5 + 2 with 7^2 + 4*5^2 + 5*2^2 = 13^2.

a(23) = 1, and 23 = 7 + 8 + 8 with 7^2 + 4*8^2 + 5*8^2 = 25^2.

a(46) = 1, and 46 = 14 + 16 + 16 with 14^2 + 4*16^2 + 5*16^2 = 50^2.

a(71) = 1, and 71 = 42 + 8 + 21 with 42^2 + 4*8^2 + 5*21^2 = 65^2.

a(92) = 1, and 92 = 28 + 32 + 32 with 28^2 + 4*32^2 + 5*32^2 = 100^2.

a(142) = 1, and 142 = 84 + 16 + 42 with 84^2 + 4*16^2 + 5*42^2 = 130^2.

PowQ[n_]:=PowQ[n]=n>1&&IntegerQ[Log[2, n]];

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

tab={}; Do[r=0; Do[If[(PowQ[y]||PowQ[n-x-y])&&SQ[x^2+4*y^2+5*(n-x-y)^2], r=r+1], {x, 1, n-3}, {y, 1, n-1-x}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Cf. A000079, A000290, A271510, A271513, A271518, A230121, A230747.

nonn,new

Zhi-Wei Sun, May 05 2021

approved

1,1

Primes p such that p-1 and p+1 each have at most one odd prime factor (counted by multiplicity).

Terms > 3 must be either of the form 3*2^k+1 with 3*2^(k-1)+1 prime, or of the form 3*2^k-1 with 3*2^(k-1)-1 prime.

There are no more terms up to 3*2^5000+1.

Conjecture: these are all the terms.

Table of n, a(n) for n=1..10.

a(5) = 11 is a term because 11-1=5*2^1 and 11+1=3*2^2 with 11, 5 and 3 prime.

{3, 7} union select(t -> isprime(t) and isprime((t+1)/2), {seq(3*2^k+1, k=1..3000)})

union select(t -> isprime(t) and isprime((t-1)/2), {seq(3*2^k-1, k=1..3000)});

Cf. A093641.

nonn,more,new

J. M. Bergot and Robert Israel, May 05 2021

approved

1,1

The triples (a, b, c) are displayed in increasing order of side a, and if sides a coincide then in increasing order of the side b.

When sides satisfy 2/a = 1/b + 1/c, or a = 2*b*c/(b+c) then a is always the middle side with b < a < c.

Equivalent relations: the heights and sines satisfy 2*h_a = h_b + h_c and 2/sin(A) = 1/sin(B) + 1/sin(C).

Inequalities between sides: a/2 < b < a < c < b*(1+sqrt(2)).

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-337 p. 179, André Desvigne.

Table of n, a(n) for n=1..69.

(4, 3, 6) is the first triple with 2/4 = 1/3 + 1/6 and 6-4 < 3 < 6+4.

The table begins:

   4,  3,  6;

  12, 10, 15;

  15, 12, 20;

  21, 15, 35;

  24, 21, 28;

  35, 30, 42;

  ...

for a from 4 to 200 do

for b from floor(a/2)+1 to a-1 do

c := a*b/(2*b-a);

if c=floor(c) and igcd(a, b, c)=1 and c-b<a then print(a, b, c); end if;

end do;

end do;

Cf. A020883 (side a), A343892 (side b), A343893 (side c), A343894 (perimeter).

nonn,tabf,new

Bernard Schott, May 03 2021

approved

1,2

Row n starts with n, then the lowest power of 2 dividing n is subtracted to produce the next entry in the row.

n first appears at position A000788(n)+1.

Peter Kagey, Rows n = 1..1023, flattened

The triangle begins

1

2

3 2

4

5 4

6 4

7 6 4

Table[Most @ NestWhileList[# - 2^IntegerExponent[#, 2] &, n, # > 0 &], {n, 1, 30}] // Flatten (* Amiram Eldar, May 05 2021 *)

(Python)

def gen_a():

    for n in range(1, 100):

        k = n

        while k>0:

            yield k

            k = k & (k-1)

a = gen_a()

Cf. A000120 (row widths), A000788, A006519, A129760.

nonn,easy,tabf,new

Christian Perfect, May 04 2021

approved