Benjamin Lombardo - cp's OEIS frontend

User: Benjamin Lombardo

Benjamin Lombardo has authored 2 sequences.

A325904 Generator sequence for A100982.

1, 0, -3, -8, 15, -91, -54, 2531, -17021, 43035, -66258, 1958757, -24572453, 146991979, -287482322, -3148566077, 35506973089, -198639977241, 1006345648929, -8250266425561, 76832268802555, -517564939540551, 1890772860334557, 3323588929061820, -104547561696315008, 907385094824827328, -6313246535826877248

Offset: 0

Benjamin Lombardo, Sep 08 2019

sign

The name of this sequence is derived from its main purpose as a formula for A100982 (see link). Both formulas below stem from Mike Winkler's 2017 paper on the 3x+1 problem (see below), in which a recursive definition of A100982 and A076227 is created in 2-D space. These formulas redefine the sequences in terms of this 1-D recursive sequence.

Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A020914, A076227, A100982.

import math
numberOfTerms = 20
L6 = [1,0]
def c(n):
    return math.floor(n/(math.log2(3)-1))
def p(a,b):
    return math.factorial(a)/(math.factorial(a-b)*math.factorial(b))
def anotherTerm(newTermCount):
    global L6
    for a in range(newTermCount+1-len(L6)):
        y = len(L6)
        newElement = 0
        for k in range(y):
            newElement -= int(L6[k]*p(c(y)+y-k-2, c(y)-2))
        L6.append(newElement)
anotherTerm(numberOfTerms)
print("A325904")
for a in range(numberOfTerms+1):
    print(a, "|", L6[a])

@cached_function
def a(n):
    if n < 2: return 0^n
    A = floor(n/(log(3, 2) - 1)) - 2
    return -sum(a(k)*binomial(A + n - k, A) for k in (0..n-1))
[a(n) for n in range(100)] # Peter Luschny, Sep 10 2019

a(0)=1, a(1)=0, a(n) = -Sum_{k=0..n-1} a(k)*binomial(A325913(n)+n-k-2, A325913(n)-2) for n>1.

A325913 Integers m such that there are exactly two powers of 2 between 3^m and 3^(m+1).

Original entry on oeis.org

1, 3, 5, 6, 8, 10, 11, 13, 15, 17, 18, 20, 22, 23, 25, 27, 29, 30, 32, 34, 35, 37, 39, 41, 42, 44, 46, 47, 49, 51, 52, 54, 56, 58, 59, 61, 63, 64, 66, 68, 70, 71, 73, 75, 76, 78, 80, 82, 83, 85, 87, 88, 90, 92, 94, 95, 97, 99, 100

Offset: 1

Benjamin Lombardo, Sep 08 2019

nonn

Or m such that A022921(m) = 2.

Also largest m such that 2^(m+n) > 3^m. - Bob Selcoe, Dec 19 2021

			For m=3, there are exactly two powers of 2 between 3^3 = 27 and 3^(3+1) = 81: 32 and 64, since 27 < 32 < 64 < 81. Therefore, m=3 is an element of the sequence (at n=2).

Benjamin Lombardo, Table of n, a(n) for n = 1..1000

Cf. A022921, A054414.

import math
def a(n):
    return math.floor(n/(math.log2(3)-1))
for n in range(1, 101):
    print("a(" + str(n) + ") = " + str(a(n)))

a(n) = floor(n/(log_2(3)-1)).

a(n) = A054414(n) - n - 1.

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User: Benjamin Lombardo

A325904 Generator sequence for A100982.

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