William Boyles - cp's OEIS frontend

A353201 a(n) = smallest m such that f(m,x) is divisible by g(n,x), where f(m,x) = U(m-1,x/2), and U(k,x) is the k-th Chebyshev polynomial of the second kind over the field GF(2); g(n,x) is the polynomial over GF(2) whose coefficients correspond to the binary digits of n.

1, 2, 3, 4, 3, 6, 5, 4, 15, 6, 9, 12, 7, 10, 6, 8, 6, 30, 17, 12, 5, 18, 21, 12, 15, 14, 15, 20, 9, 6, 17, 8, 51, 6, 35, 60, 31, 34, 9, 12, 31, 10, 15, 36, 30, 42, 33, 24, 45, 30, 12, 28, 51, 30, 11, 20, 21, 18, 33, 12, 31, 34, 15, 8, 15, 102, 65, 12, 9, 70, 93, 60, 63, 62, 42, 68

Offset: 1

William Boyles, Jun 22 2022

nonn

g(n,x) divides f(m,x) if and only if m is a multiple of a(n).

			For n=13, g(13,x) = 1*x^3 + 1*x^2 + 0*x + 1 because 13 is 1101 in binary. f(7,x) is the smallest that is divisible by g(13,x), so a(13) = 7.

William Boyles, Table of n, a(n) for n = 1..2500

A354463 a(n) is the number of trailing zeros in (2^n)!.

Original entry on oeis.org

0, 0, 0, 1, 3, 7, 14, 31, 63, 126, 253, 509, 1021, 2045, 4094, 8189, 16380, 32763, 65531, 131067, 262140, 524285, 1048571, 2097146, 4194297, 8388603, 16777208, 33554424, 67108858, 134217720, 268435446, 536870902, 1073741816, 2147483642, 4294967289, 8589934584, 17179869176, 34359738358, 68719476729

Offset: 0

William Boyles, May 31 2022

			For n = 4, (2^4)! = 20922789888000, which has a(4) = 3 trailing zeros.

William Boyles, Table of n, a(n) for n = 0..650

Cf. A000079, A027868.

seq n = aux (2 ^ n) 0
  where
    aux x acc
      | x < 5 = acc
      | otherwise = aux y (acc + y)
      where
        y = x `div` 5

a[n_]:=IntegerExponent[(2^n)!]; Array[a,38,0] (* Stefano Spezia, Jun 01 2022 *)

a(n) = val(1<David A. Corneth, Jun 01 2022

from sympy import factorial, multiplicity
def a(n): return multiplicity(5, factorial(2**n, evaluate=False))
print([a(n) for n in range(39)]) # Michael S. Branicky, Jun 01 2022

def A354463(n):
    c, m = 0, 2**n
    while m >= 5:
        m //= 5
        c += m
    return c # Chai Wah Wu, Jun 02 2022

a(n) = A027868(A000079(n)). - Michel Marcus, Jun 01 2022

a(n) = 2^(n-2) - A055223(n) for n >= 2. - John Keith, Jun 06 2022

A353071 Maximum number of clicks needed to solve any solvable Lights Out problem on an n X n grid.

Original entry on oeis.org

1, 4, 9, 7, 15, 36, 49, 64, 37, 100, 65, 144, 169, 123, 225, 124, 199, 324, 197, 400, 441, 484

Offset: 1

William Boyles, Apr 21 2022

a(n) = n^2 if and only if A159257(n) = 0.
a(n) >= A075464(n).
If n = 6k-1 for some integer k, then a(n) <= 26k^2 - 12k + 1. This upper bound is equal to a(n) when A159257(n) = 2. Further, it is conjectured that if A159257(n) = 2, then n = 6k-1 for some integer k.
It is conjectured that if A159257(n) = 4, then n = 5k-1 for some integer k, and a(n) = 17k^2 - 10k.
It is conjectured that if A159257(n) = 6, then n = 12k-1 for some integer k, and a(n) = 88k^2 - 24k + 1
It is conjectured that if A159257(n) = 8, then either n = 10k-1 or n = 17k-1 for some integer k. If n = 10k-1, then a(n) = 60k^2 - 20k - 3. If n = 17k-1, then a(n) = 161k^2 - 34k - 3.
It is conjectured that if A159257(n) = 10, then n = 30k-1 for some integer k, and a(n) = 506k^2 - 60k - 3.
239 <= a(23) <= 305.

William Boyles, Most Clicks Problem in Lights Out, arXiv:2201.03452 [math.CO], 2022.

Cf. A159257, A075464.

A307826 The number of integers r such that all primes above a certain value have the form primorial(n)*q +- r.

Original entry on oeis.org

1, 1, 4, 24, 240, 2880, 46080, 829440, 18247680, 510935040, 15328051200, 551809843200, 22072393728000, 927040536576000, 42643864682496000, 2217480963489792000, 128613895882407936000, 7716833752944476160000, 509311027694335426560000

Offset: 1

William Boyles, Apr 30 2019

nonn

			For n=3, the third primorial is 2*3*5=30, and all primes at least 17 have the form 30n +- (1,7,11,13). So, a(3) = 4.

William Boyles, Table of n, a(n) for n = 1..350

Cf. A000010, A002110, A005867, A156037.

a[1]=1; a[n_] := EulerPhi[Product[Prime[i], {i, 1, n}]]/2; Array[a, 20] (* Amiram Eldar, Jul 08 2019 *)

import sympy
def A307826(n):
    sympy.sieve.extend_to_no(n)
    s = list(sympy.sieve._list)
    prod = s[0]
    print("1")
    for i in range(1,n):
        prod*=s[i]
        print(sympy.ntheory.factor_.totient(prod)//2)

a(n) = Product_{k=1..n} A156037(k).
a(n) = A000010(A002110(n))/2 for n > 1.
a(n) = A005867(n)/2 for n > 1. - Alexandre Herrera, Apr 16 2023

A275534 Number of primes of the form x^2 + y^2 less than or equal to 2*n^2.

Original entry on oeis.org

1, 2, 4, 5, 7, 9, 12, 15, 18, 22, 25, 29, 33, 37, 43, 46, 51, 56, 62, 68, 75, 79, 86, 93, 102, 107, 114, 119, 127, 136, 143, 150, 160, 169, 179, 184, 195, 206, 215, 223, 233, 242, 254, 264, 274, 285, 297, 307, 318, 330, 339, 350, 362, 376, 386, 400, 415, 428, 441, 452, 465, 483, 498, 510, 525, 541

Offset: 1

William Boyles, Jul 31 2016

nonn

Number of terms in A002313 that are less than A001105(n).

William Boyles, Table of n, a(n) for n = 1..1000

Cf. A001105, A002313.

nn = 66; Table[Count[Take[#, PrimePi[2 n^2]], k_ /; k > 0], {n, nn}] &@
SquaresR[2, Prime@ Range@ PrimePi[2 nn^2]] (* Michael De Vlieger, Aug 01 2016 *)

a(n)=my(s); forprime(p=2,2*n^2, if(p%4<3, s++)); s \\ Charles R Greathouse IV, Sep 02 2016

cp's OEIS Frontend

User: William Boyles

A353201 a(n) = smallest m such that f(m,x) is divisible by g(n,x), where f(m,x) = U(m-1,x/2), and U(k,x) is the k-th Chebyshev polynomial of the second kind over the field GF(2); g(n,x) is the polynomial over GF(2) whose coefficients correspond to the binary digits of n.

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A354463 a(n) is the number of trailing zeros in (2^n)!.

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A353071 Maximum number of clicks needed to solve any solvable Lights Out problem on an n X n grid.

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A307826 The number of integers r such that all primes above a certain value have the form primorial(n)*q +- r.

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Programs

Mathematica

Python

Formula

A275534 Number of primes of the form x^2 + y^2 less than or equal to 2*n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI