Sebastian F. Orellana - cp's OEIS frontend

A362138 a(n) = gpf(a(n-1) + prime(n)) where gpf is the greatest prime factor and a(1)=2.

2, 5, 5, 3, 7, 5, 11, 5, 7, 3, 17, 3, 11, 3, 5, 29, 11, 3, 7, 13, 43, 61, 3, 23, 5, 53, 13, 5, 19, 11, 23, 11, 37, 11, 5, 13, 17, 5, 43, 3, 13, 97, 3, 7, 17, 3, 107, 11, 17, 41, 137, 47, 3, 127, 3, 19, 3, 137, 23, 19, 151, 37, 43, 59, 31, 29, 5, 19, 61, 41, 197

Offset: 1

Sebastian F. Orellana, Jun 12 2023

nonn

			a(3) = gpf(a(2) + prime(3)) = gpf(5+5) = 5.

Sebastian F. Orellana, Table of n, a(n) for n = 1..10000

Cf. A006530 (gpf), A000040.

Cf. A076560, A076561.

gpf[n_] := FactorInteger[n][[-1, 1]]; a[1] = 2; a[n_] := a[n] = gpf[a[n - 1] + Prime[n]]; Array[a, 100] (* Amiram Eldar, Jun 15 2023 *)

from sympy import factorint, prime
list=[2]
num=1
k=50
for i in range(0, k):
  list.append(max(factorint(list[i]+prime(i+1))))
print(list)

A362289 a(n) is the largest denominator when the greedy algorithm for Egyptian fractions is applied to 1/n + 1/(n+1).

Original entry on oeis.org

2, 3, 12, 180, 30, 1428, 56, 2520, 90, 2310, 132, 100292556, 182, 9240, 240, 119952, 306, 614444040, 380, 23100, 462, 42190274940, 552, 77390453400, 650, 201474, 756, 23370247110, 870, 200880, 992, 14523137084239067683872, 1122, 2206260, 1260, 104845560637757648698080

Offset: 1

Sebastian F. Orellana, Apr 14 2023

nonn

			For n=16, 1/16 + 1/17 = 33/272 which written in Egyptian fractions is 1/9 + 1/98 + 1/119952 and the largest denominator is 119952.

Cf. A050210.

egyptFraction[f_] := Ceiling[1/Most[NestWhileList[# - 1/Ceiling[1/#] &, f, # != 0 &]]]; a[n_] := egyptFraction[1/n + 1/(n + 1)][[-1]]; Array[a, 40] (* Amiram Eldar, Apr 14 2023 *)

a(n) = A050210(n*(n+1), 2*n+1). - Michel Marcus, Apr 14 2023

A361520 a(n) is the greatest prime factor of a(n-2)^2 + a(n-1)^2 where a(1)=2 and a(2)=3.

Original entry on oeis.org

2, 3, 13, 89, 809, 349, 409, 144541, 10446133981, 1361264878245241, 4398505263882824939701, 17847523009215848981, 512996953133650208042047593649109478833

Offset: 1

Sebastian F. Orellana, Mar 14 2023

nonn

Sebastian F. Orellana, Table of n, a(n) for n = 1..15

Cf. A006530, A113494, A175723.

A[1]:= 2: A[2]:= 3:
for n from 3 to 15 do A[n]:= max(numtheory:-factorset(A[n-2]^2 + A[n-1]^2)) od:
seq(A[],n=1..15); # Robert Israel, Mar 17 2023

a[1] = 2; a[2] = 3; a[n_] := a[n] = FactorInteger[a[n - 1]^2 + a[n - 2]^2][[-1, 1]]; Array[a, 14] (* Amiram Eldar, Mar 14 2023 *)

A360825 a(n) is the remainder after dividing n! by its least nondivisor.

Original entry on oeis.org

1, 1, 2, 2, 4, 1, 6, 2, 5, 1, 10, 1, 12, 3, 8, 1, 16, 1, 18, 4, 11, 1, 22, 22, 6, 5, 14, 1, 28, 1, 30, 33, 20, 31, 18, 1, 36, 7, 20, 1, 40, 1, 42, 8, 23, 1, 46, 19, 11, 9, 26, 1, 52, 30, 27, 10, 29, 1, 58, 1, 60, 43, 53, 56, 33, 1, 66, 12, 35, 1, 70, 1, 72, 27, 23

Offset: 0

Sebastian F. Orellana, Feb 22 2023

nonn

For every term besides a(3), the least nondivisor is the next prime after n.

			a(5) = 5! mod 7 = 120 mod 7 = 1.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A000040, A000142, A006093, A040976, A151800, A213636, A275111, A360805.

a[n_] := Module[{f = n!, m = n + 1}, While[Divisible[f, m], m++]; Mod[f, m]]; Array[a, 100, 0] (* Amiram Eldar, Feb 22 2023 *)

a(n) = my(k=1, r); while(!(r=(n! % (n+k))), k++); r; \\ Michel Marcus, Feb 22 2023

from functools import reduce
from sympy import nextprime
def A360825(n):
    if n == 3: return 2
    m = nextprime(n)
    return reduce(lambda i, j: i*j%m,range(2,n+1),1)%m # Chai Wah Wu, Feb 22 2023

from functools import reduce
from sympy import nextprime
def A360825(n):
    if n == 3: return 2
    m = nextprime(n)
    return (m-1)*pow(reduce(lambda i,j:i*j%m,range(n+1,m),1),-1,m)%m # Chai Wah Wu, Feb 23 2023

a(n) = 1 <=> n in { A040976 } \ { 3 }.
a(n) = n <=> n in { A006093 }.
a(n) = n! mod A151800(n) for n > 3.
a(n) = A213636(n!) = A213636(A000142(n)).
a(A000040(n)) = A275111(n) for n >= 3.
a(n) > n <=> n in { A360805 }.

A360496 a(n) is the remainder after dividing n by its largest prime factor plus 1, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 1, 5, 2, 7, 2, 1, 4, 11, 0, 13, 6, 3, 1, 17, 2, 19, 2, 5, 10, 23, 0, 1, 12, 3, 4, 29, 0, 31, 2, 9, 16, 3, 0, 37, 18, 11, 4, 41, 2, 43, 8, 3, 22, 47, 0, 1, 2, 15, 10, 53, 2, 7, 0, 17, 28, 59, 0, 61, 30, 7, 1, 9, 6, 67, 14, 21, 6, 71, 0, 73, 36, 3, 16, 5, 8, 79, 2, 1, 40, 83, 4, 13

Offset: 1

Sebastian F. Orellana, Feb 09 2023

nonn

			a(15) = 15 mod (5+1) = 15 mod 6 = 3.

Sebastian F. Orellana, Table of n, a(n) for n = 1..6999

Cf. A006530.

a:= n-> irem(n, max(numtheory[factorset](n))+1):
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 10 2023

a[n_] := Mod[n, FactorInteger[n][[-1, 1]] + 1]; Array[a, 100] (* Amiram Eldar, Feb 10 2023 *)

a(n) = if (n==1, 1, n % (vecmax(factor(n)[, 1])+1)); \\ Michel Marcus, Feb 10 2023

a(n) = n mod (1 + A006530(n)).

A359950 a(n) is the greatest prime factor of n^n - n!.

Original entry on oeis.org

2, 7, 29, 601, 29, 116929, 11887, 4778489, 82207, 296987, 2767, 464089, 36922117, 71722471217, 10219277051, 9406703479, 2040247819, 122450719, 1265072927, 18353142818474353, 21514105057, 46999724987, 29693667067, 5684341885088084044195811037649, 692132186353, 12114317049616531

Offset: 2

Sebastian F. Orellana, Jan 19 2023

nonn

			a(5) = greatest prime factor of 5^5 - 5! = greatest prime factor of 3125 - 120 = greatest prime factor of 3005 = 3005/5 = 601.

Amiram Eldar, Table of n, a(n) for n = 2..106

Cf. A006530, A020639, A036679, A126130.

Table[Max[First/@FactorInteger[n^n-n!]],{n,2,27}] (* Stefano Spezia, Jan 22 2023 *)

a(n) = vecmax(factor(n^n - n!)[,1]); \\ Michel Marcus, Jan 22 2023

a(n) = A006530(A036679(n)) = A006530(n*A126130(n-1)).

More terms from Michel Marcus, Jan 22 2023

A352798 a(n) = 1/(cf[0;n,n,n,...,n] - cf[0;n,n,...,n]) where the first continued fraction has n+1 terms and the second has n terms.

Original entry on oeis.org

1, -10, 330, -21960, 2551640, -461930274, 120572270007, -42930583856160, 20008932768992430, -11825788272679695050, 8643081649999714376976, -7654102744143874729100040, 8076084821027629176909996013, -10010473694454865001226770534530, 14402393216408406872433735669683370

Offset: 1

Sebastian F. Orellana, Apr 03 2022

			a(2) = -10 because the two continued fractions are cf[0;2,2] = 0 + 1/(2 + 1/2) = 2/5 and cf[0;2] = 0 + 1/2 = 1/2 and the reciprocal of their difference is 1/(2/5 - 1/2) = -10.
a(3) = 330 because the two continued fractions are cf[0;3,3,3] = 0 + 1/(3 + 1/(3 + 1/3)) = 10/33 and cf[0;3,3] = 0 + 1/(3 + 1/3) = 3/10, and 1/(10/33 - 3/10) = 330.

Cf. A084844, A084845.

a:= n-> (f-> -(-1)^n*f(n,n)*f(n+1,n))(combinat[fibonacci]):
seq(a(n), n=1..15);  # Alois P. Heinz, Jul 06 2022

a(n) = (-1)^(n+1) * vecprod(Vec(lift(Mod('x,'x^2-n*'x-1)^(n+1)))); \\ Kevin Ryde, Apr 18 2022

from sympy.ntheory.continued_fraction import continued_fraction_reduce
def A352798(n): return int(1/(continued_fraction_reduce([0]+[n]*n)-continued_fraction_reduce([0]+[n]*(n-1)))) # Chai Wah Wu, Jul 06 2022

a(n) = A084844(n)*A084845(n)*(-1)^(n+1).

More terms from Kevin Ryde, Apr 18 2022

A352285 a(n) is the number of steps in John Conway's game of life that it takes for the smallest square checkerboard pattern with a diagonal of n living cells to either die out or enter a cycle; or -1 if it never cycles.

Original entry on oeis.org

1, 1, 1, 1, 3, 4, 4, 4, 40, 7, 58, 9, 38, 8, 37, 29, 71, 55, 51, 41, 49, 70, 60, 93, 102, 79, 333, 123, 181, 69, 200, 279, 372, 117, 188, 212, 122, 137, 263, 576, 96, 149, 225, 169, 150, 276, -1, 304, 281, 106, 215, 160, 206, 197, -1, 359, 221, 355, -1, 447, 178, 314, 431

Offset: 1

Sebastian F. Orellana, Mar 10 2022

sign

a(n) = -1 iff the pattern's extent grows without bound (since a bounded region must eventually repeat). The first a(n) = -1 is at n=47 where the square launches 8 gliders into open space.

			For n = 1:
. . . | . . . |
. o . | . . . |
. . . | . . . |
all cells are dead after one generation, hence a(1)=1.
For n = 2:
. . . . | . . . . |
. o . . | . . . . |
. . o . | . . . . |
. . . . | . . . . |
all cells are dead after one generation, hence a(2)=1.
For n = 3:
. . . . .| . . . . . |
. o . o .| . . o . . |
. . o . .| . o . o . |
. o . o .| . . o . . |
. . . . .| . . . . . |
a pattern repeats after one generation, hence a(3)=1.
For n = 4:
. . . . . . | . . . . . . |
. o . o . . | . . o o . . |
. . o . o . | . o . . o . |
. o . o . . | . o . . o . |
. . o . o . | . . o o . . |
. . . . . . | . . . . . . |
a pattern repeats after one generation, hence a(4) = 1.
For n = 5:
. . . . . . . . . | . . . . . . . . . | . . . . . . . . . | . . . . . . . . . |
. . . . . . . . . | . . . . . . . . . | . . . . o . . . . | . . . o o o . . . |
. . o . o . o . . | . . . o o o . . . | . . . o o o . . . | . . . . . . . . . |
. . . o . o . . . | . . o . . . o . . | . . o . o . o . . | . o . . . . . o . |
. . o . o . o . . | . . o . . . o . . | . o o o . o o o . | . o . . . . . o . |
. . . o . o . . . | . . o . . . o . . | . . o . o . o . . | . o . . . . . o . |
. . o . o . o . . | . . . o o o . . . | . . . o o o . . . | . . . . . . . . . |
. . . . . . . . . | . . . . . . . . . | . . . . o . . . . | . . . o o o . . . |
. . . . . . . . . | . . . . . . . . . | . . . . . . . . . | . . . . . . . . . |
a pattern begins to oscillate between four parallel "blinkers" after one generation, hence a(5) = 3.

Cf. A089520 (filled square).

A352026 a(n) is the nearest integer to 1/(H(k) - n), where H(k) is the smallest harmonic number that exceeds n.

Original entry on oeis.org

1, 2, 12, 50, 37, 483, 229, 785, 2059, 4806, 23251, 56470, 327690, 813351, 734186, 2643630, 10476269, 67340402, 268822185, 102740092, 618260119, 2491694355, 7222972533, 50525424196, 44010188391, 164490666033, 131444704333, 548839044705, 1808874061272, 9913711133738

Offset: 0

Sebastian F. Orellana, Feb 28 2022

nonn

The k-th harmonic number, H(k), is Sum_{j=1..k} 1/j. A002387(n) is the smallest k such that H(k) > n.

			H(2) = 1/1 + 1/2 = 3/2 = 1.5;
H(3) = 1/1 + 1/2 + 1/3 = 11/6 = 1.8333...;
H(4) = 1/1 + 1/2 + 1/3 + 1/4 = 25/12 = 2.08333..., which is the first harmonic number that exceeds 2, so a(2) = round(1/(25/12 - 2)) = round(1/(1/12)) = 12.
H(10) = 7381/2520 = 2.92896...;
H(11) = 83711/27720 = 3.01987..., which is the first harmonic number > 3, and the fractional part of 83711/27720 = 551/27720, so a(3) = round(27720/551) = round(50.30852...) = 50.

Peter Luschny, Table of n, a(n) for n = 0..100

Cf. A001008, A002805, A002387.

a(n)={my(s=0,k=0); while(s<=n, k++;s+=1/k); round(1/(s-n))} \\ Andrew Howroyd, Mar 01 2022

RR = RealField(1000)
def A352026(n):
    g = RR.euler_constant()
    u = exp(RR(n) - g)
    a = u + RR(3/2) - RR(1/(24*u)) + RR(3/(640*u^3))
    h = RR(psi(RR(floor(a)))) + g
    return round(RR(1/(h - RR(n))))
print([A352026(n) for n in range(30)])  # Peter Luschny, Mar 03 2022

a(n) = round(1/(H(A002387(n)) - n)).

More terms from Peter Luschny, Mar 01 2022

A350037 a(n) = n^2 mod round(sqrt(n)).

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 4, 4, 1, 0, 1, 4, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4

Offset: 1

Sebastian F. Orellana, Dec 09 2021

			a(5) = 5^2 mod round(sqrt(5)) = 25 mod 2 = 1.

Winston de Greef, Table of n, a(n) for n = 1..10000

Cf. A336302 (with ceiling instead of round).

import java.util.Arrays;
public class modulus_sequence {
  static int[] solutions = new int[480];
  static int a_of_n (double n) {
    int z = (int)Math.round(Math.sqrt(n));
    int w = (int)(Math.pow(n, 2));
    int k = w%z;
    return k;
  }
  public static void main(String[] args) {
    for (double j = 2; j < 482; j++) {
      int h = a_of_n(j);
      solutions[(int) (j-2)]=h;
    }
    System.out.println(Arrays.toString(solutions));
  }
}

a[n_] := Mod[n^2, Round @ Sqrt[n]]; Array[a, 100, 2] (* Amiram Eldar, Dec 10 2021 *)
Table[PowerMod[n,2,Round[Sqrt[n]]],{n,2,101}] (* Stefano Spezia, Dec 15 2021 *)

a(n) = n^2 % round(sqrt(n)); \\ Michel Marcus, Dec 14 2021

a(n) = lift(Mod(n, ((sqrtint(4*n) + 1)\2))^2); \\ Michel Marcus, Dec 14 2021

from math import isqrt
def A350037(n): return pow(n,2,(m:=isqrt(n))+int(4*n>=(2*m+1)**2)) # Chai Wah Wu, Jan 10 2022

a(n) = A000290(n) mod A000194(n). - Michel Marcus, Dec 14 2021

cp's OEIS Frontend

User: Sebastian F. Orellana

A362138 a(n) = gpf(a(n-1) + prime(n)) where gpf is the greatest prime factor and a(1)=2.

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A362289 a(n) is the largest denominator when the greedy algorithm for Egyptian fractions is applied to 1/n + 1/(n+1).

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A361520 a(n) is the greatest prime factor of a(n-2)^2 + a(n-1)^2 where a(1)=2 and a(2)=3.

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A360825 a(n) is the remainder after dividing n! by its least nondivisor.

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A360496 a(n) is the remainder after dividing n by its largest prime factor plus 1, a(1) = 1.

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A359950 a(n) is the greatest prime factor of n^n - n!.

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A352798 a(n) = 1/(cf[0;n,n,n,...,n] - cf[0;n,n,...,n]) where the first continued fraction has n+1 terms and the second has n terms.

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