Pablo Hueso Merino - cp's OEIS frontend

A343515 a(n) is the number of real solutions to the equation sin(x) = x/n.

1, 3, 3, 3, 3, 3, 3, 7, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 15, 15, 15, 15, 15, 15, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 27, 27, 27, 27, 27, 27, 31, 31, 31, 31, 31, 31, 35, 35, 35, 35, 35, 35, 35, 39, 39, 39, 39, 39, 39, 43, 43, 43, 43, 43

Offset: 1

Pablo Hueso Merino, Apr 17 2021

nonn

All terms are odd.

All terms are congruent to 3 modulo 4 after the first term. Proof: define sin(x)/x to be 1 at x = 0. If a(n) == 1 (mod 4), then the horizontal line y = 1/n is tangent to the curve y = sin(x)/x at (x_n, sin(x_n)/x_n) for some x_n >= 0. We have tan(x_n) = x_n and sin(x_n)/x_n = 1/n, so cos(x_n) = 1/n. By the Lindemann-Weierstrass theorem, we have either x_n = 0 or x_n must be transcendental (if x is a nonzero algebraic number, then exp(x) is transcendental). On the other hand, x_n = tan(x_n) = sqrt(n^2-1) is algebraic, so the only possibility is n = 1. - Jianing Song, Jul 13 2021

			a(3) = 3 because the equation sin(x) = x/3 has 3 real solutions: {-2.27886..., 0, 2.27886...}.

Join[{1}, Table[CountRoots[n*Sin[x] - x, {x, -n, n}], {n, 2, 100}]] (* Vaclav Kotesovec, Jun 25 2021 *)

a(n) ~ 4*(floor((n-Pi/2)/(2*Pi))+1)-1.
For n > 1, a(n) = 4*(floor((n-Pi/2)/(2*Pi))+1)-1 + r(n), where r(n) is an error term defined as follows: let E be the system of equations given by cos(sqrt(n^2-1)) = 1/n and sin(sqrt(n^2-1)) = sqrt(n^2-1)/n; r(n) = 4 if the closest solution of E from the left to Pi/2 + 2*Pi*(floor((n-Pi/2)/2*Pi)+1) is smaller than n; r(n) = 0 otherwise.
From Jianing Song, Jul 13 2021: (Start)
Define x_k to be root of tan(x) = x in [k*Pi, (k+1)*Pi), k >= 0. For n > 1, if sec(x_(2*k)) < n < sec(x_(2*k+2)) (or equivalently, x_(2*k) < sqrt(n^2-1) < x_(2*k+2)), then a(n) = 4*k + 3.
For n >= 2, a(n+1) - a(n) is either 0 or 4. a(n+1) - a(n) = 4 if n is of the form floor(sec(x_(2*k))) = floor(sqrt((x_(2*k))^2+1)) for some k > 0. (End)

A339310 a(n) = a(n-1-a(n-1)) + a(n-a(n-2)) for n>2; starting with a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 2, 3, 3, 3, 5, 4, 6, 5, 6, 8, 8, 6, 9, 8, 8, 11, 11, 10, 10, 14, 12, 11, 16, 15, 12, 17, 14, 17, 16, 18, 14, 19, 19, 16, 21, 22, 19, 21, 25, 18, 22, 25, 23, 24, 25, 25, 23, 31, 28, 22, 33, 28, 29, 32, 28, 29, 30, 30, 33, 35, 29, 33, 32, 28, 41, 36, 35

Offset: 1

Pablo Hueso Merino, Dec 02 2020

nonn

{a(n)} is the Pinn F 1,0(n) sequence (see link section).

			a(3)=2 because a(3) = a(3-1-a(3-1))+a(3-a(3-2)) = a(2-1)+a(3-1) = 1+1 = 2.

K. Pinn, A Chaotic Cousin Of Conway's Recursive Sequence, arXiv:cond-mat/9808031 [cond-mat.stat-mech], 1998.

Cf. A005185, A070867, A046699.

a[1] = a[2] = 1; a[n_] := a[n] = a[n - 1 - a[n - 1]] + a[n - a[n - 2]]; Table[ a[n], {n, 1, 40}]

lista(nn) = {my(va = vector(nn)); va[1] = 1; va[2] = 1; for (n=3, nn, va[n]=va[n-1-va[n-1]]+va[n-va[n-2]];); va;} \\ Michel Marcus, Dec 07 2020

a=[1,1]
for n in range(100):
    i1=len(a)-1-a[len(a)-1]
    i2=len(a)-a[len(a)-2]
    if i1>=0 and i2>=0 :
        a.append(a[i1]+a[i2])
    else :
        print("Sequence dies. Contains ", n+2, " terms.")
        break
print(a)

A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

Original entry on oeis.org

1, 3, 8, 22, 60, 163, 443, 1204, 3273, 8897, 24184, 65739, 178698, 485751, 1320408, 3589241, 9756569, 26521104, 72091835, 195965925, 532690613, 1448003214, 3936080824, 10699376979, 29083922018, 79058296722, 214902731368, 584166189564, 1587928337892, 4316436745787

Offset: 1

Alejandro Argüelles Trujillo and Pablo Hueso Merino, Jan 07 2020

nonn

a(n) is equal to A024581(n) through a(10), and grows very similarly for n > 10.

Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Cf. A331030. - Jon E. Schoenfield, Jan 14 2020

			a(1)=1 because 1 >= 1,
a(2)=3 because 1/2 + 1/3 + 1/4 = 1.0833... >= 1, etc.

Cf. A004080, A024581, A136616, A331030.

Some sequences in the same spirit as this: A002387, A004080, A055980, A115515.

default(realprecision, 10^5); e=exp(1);
lista(nn) = {my(r=1); print1(r); for(n=2, nn, print1(", ", -r+(r=floor(e*r+(e+1)/2+(e-1/e)/(24*(r+1/2)))))); } \\ Jinyuan Wang, Mar 31 2020

x = 0.0
y = 0.0
for i in range(1,100000000000000000000000):
  y += 1
  x = x + 1/i
  if x >= 1:
    print(y)
    y = 0
    x = 0

a(n) = min(p): Sum_{b=r+1..p+r} 1/b >= 1, r = Sum_{k=1..n-1} a(k), a(1) = 1.

a(20)-a(21) from Giovanni Resta, Jan 14 2020

More terms from Jinyuan Wang, Mar 31 2020

A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

Original entry on oeis.org

2, 5, 13, 36, 98, 266, 723, 1965, 5342, 14521, 39472, 107296, 291661, 792817, 2155100, 5858169, 15924154, 43286339, 117664468, 319845186, 869429357, 2363354022, 6424262292, 17462955450, 47469234471, 129034757473, 350752836478, 953445061679, 2591732385596

Offset: 1

Pablo Hueso Merino and Alejandro Argüelles Trujillo, Jan 07 2020

nonn

a(n) = A046171(n+1) through a(5), and grows similarly for n > 5.

Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Verified through n = 10000 (using the approximation Sum_{j=1..k} 1/j = log(k) + gamma + 1/(2*k) - 1/(12*k^2) + 1/(120*k^4) - 1/(252*k^6) + 1/(240*k^8) - ..., where gamma is the Euler-Mascheroni constant, A001620). Cf. A081881. - Jon E. Schoenfield, Jan 10 2020

			a(1)=2 because 1 + 1/2 = 1.5 > 1,
a(2)=5 because 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 1.0928... > 1,
etc.

Cf. A002387, A046171, A081881, A096005, A180224, A295572, A331028.

lista(lim=oo)={my(s=0, p=0); for(i=1, lim, s+=1/i; if(s>1, print1(i-p, ", "); s=0; p=i))} \\ Andrew Howroyd, Jan 08 2020

x = 0.0
y = 0.0
z = 0.0
for i in range(1,100000000000000000000000):
  y += 1
  x = x + 1/i
  z = z + 1/i
  if x > 1:
    print(y)
    y = 0
    x = 0

a(1)=2, a(n) = (min(p) : Sum_{s=r..p} 1/s > 1)-r+1, r=Sum_{k=1..n-1} a(k).

a(25)-a(29) from Jon E. Schoenfield, Jan 10 2020

A329547 Number of natural numbers k <= n such that k^k is a square.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 38, 38, 39, 39, 40, 40

Offset: 1

Pablo Hueso Merino, Nov 16 2019

For even k, k^k is always a square. For odd k, k^k is a square if and only if k is a square.
It seems the unrepeated terms form A266304 \ {0}. - Ivan N. Ianakiev, Nov 21 2019
Indices of unrepeated terms are A081349. - Rémy Sigrist, Dec 07 2019

			a(5) = 3 because among 1^1, 2^2, ..., 5^5 there are 3 squares: 1^1, 2^2, and 4^4.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A081349, A176693, A266304.

Table[Floor[n/2] + Ceiling[Floor[Sqrt[n]]/2], {n, 1, 100}]

a(n) = sum(k=1, n, issquare(k^k)); \\ Michel Marcus, Nov 17 2019

first(n) = my(res=vector(n), inc); res[1] = 1; for(i=2, n, inc = (1-(i%2)) || issquare(i); res[i] = res[i-1] + inc); res \\ David A. Corneth, Dec 07 2019

a(n) = n\2 + (sqrtint(n)+1)\2 \\ David A. Corneth, Dec 07 2019

from math import isqrt
def A329547(n): return (n>>1)+(isqrt(n)+1>>1) # Chai Wah Wu, Sep 18 2024

a(n) = floor(n/2) + ceiling(floor(sqrt(n))/2).

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User: Pablo Hueso Merino

A343515 a(n) is the number of real solutions to the equation sin(x) = x/n.

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A339310 a(n) = a(n-1-a(n-1)) + a(n-a(n-2)) for n>2; starting with a(1) = a(2) = 1.

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A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

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A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

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Formula

Extensions

A329547 Number of natural numbers k <= n such that k^k is a square.

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Programs

Mathematica

PARI

PARI

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Python

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