Darrell Minor - cp's OEIS frontend

User: Darrell Minor

Darrell Minor has authored 7 sequences.

A191093 [Squarefree part of (ABC)]/C for A=2, C=A+B, as a function of B, rounded to nearest integer.

2, 1, 6, 1, 10, 1, 5, 1, 6, 3, 22, 3, 26, 1, 30, 0, 34, 2, 38, 5, 42, 3, 9, 3, 1, 7, 6, 7, 58, 1, 62, 1, 66, 3, 70, 3, 74, 5, 78, 5, 82, 11, 29, 11, 30, 3, 13, 1, 14, 3, 102, 1, 106, 1, 110, 7, 114, 15, 118, 15, 41, 1, 42, 1, 130, 17, 134, 17, 138, 3, 142, 3, 29, 19, 30, 19, 154

Offset: 1

Darrell Minor, May 25 2011

			For B=10, we have C=12 so SQP(ABC)=SQP(240)=2*3*5=30, so SQP(ABC)/C=30/12=2.5, which rounds off to 3.
For B=16, we have C=18 so SQP(ABC)=SQP(576)=2*3=6, so SQP(ABC)/C=6/18=0.33, which rounds off to 0.

Wikipedia, abc Conjecture

Cf. A190846, A191100, A120498.

SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A191093:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 2 >; [ A191093(n): n in [1..80] ]; // Klaus Brockhaus, May 27 2011

rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
a(n)=rad(2*n^2+4*n)\/(n+2) \\ Charles R Greathouse IV, Mar 11 2014

from operator import mul
from sympy import primefactors
def rad(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a(n): return int(round(rad(2*n**2 + 4*n)/(n + 2))) # Indranil Ghosh, May 24 2017

A191100 [Squarefree part of (ABC)]/C for A=3, C=A+B, as a function of B, rounded to nearest integer.

Original entry on oeis.org

2, 6, 1, 6, 4, 1, 21, 6, 1, 30, 33, 2, 5, 42, 2, 6, 26, 2, 57, 30, 2, 13, 69, 0, 8, 78, 1, 42, 5, 10, 93, 6, 2, 102, 105, 2, 28, 114, 13, 30, 62, 5, 129, 66, 1, 20, 28, 2, 11, 30, 2, 78, 40, 2, 165, 42, 10, 174, 177, 3, 6, 186, 7, 6, 98, 22, 201, 102, 2, 210, 213, 0, 110, 222, 5, 114

Offset: 1

Darrell Minor, May 25 2011

Given A,B natural numbers, and C=A+B. The ratio [squarefree part of (ABC)]/C (notation: SQP(ABC)/C) can get arbitrarily small, while the unsolved ABC conjecture (i.e., Oesterle-Masser conjecture) is equivalent to the statement that [SQP(ABC)]^n/C has a minimum value if n>1 (because there are conjectured to be finitely many instances of [SQP(ABC)^(1+epsilon)]

			For B=5, we have C=8 so SQP(ABC)=SQP(120)=2*3*5=30, so SQP(ABC)/C=30/8=3.75, which rounds off to 4.
For B=15, we have C=18 so SQP(ABC)=SQP(810)=2*3*5=30, so SQP(ABC)/C=30/18=1.67, which rounds off to 2.

Eric Weisstein's World of Mathematics, abc Conjecture

Cf. A190846, A191093, A120498.

SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A191100:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 3 >; [ A191100(n): n in [1..80] ]; // Klaus Brockhaus, May 26 2011

rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
a(n)=rad(3*n^2+9*n)\/(n+3) \\ Charles R Greathouse IV, Mar 11 2014

from operator import mul
from sympy import primefactors
def rad(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a(n): return int(round(rad(3*n**2 + 9*n)/(n + 3))) # Indranil Ghosh, May 24 2017

A190846 (Squarefree part of (ABC))/C for A=1, C=A+B, as a function of B, rounded to the nearest integer.

Original entry on oeis.org

1, 2, 2, 2, 5, 6, 2, 1, 3, 10, 6, 6, 13, 14, 2, 2, 6, 6, 10, 10, 21, 22, 6, 1, 5, 3, 2, 14, 29, 30, 2, 2, 33, 34, 6, 6, 37, 38, 10, 10, 41, 42, 22, 7, 15, 46, 6, 1, 1, 10, 26, 26, 6, 6, 14, 14, 57, 58, 30, 30, 61, 21, 1, 2, 65, 66, 34, 34, 69, 70, 6, 6, 73, 15, 8, 38, 77, 78, 10, 0, 3, 82, 42

Offset: 1

Darrell Minor, May 25 2011

Given A, B natural numbers, and C=A+B, the ABC conjecture deals with the ratio of the squarefree part of the product A*B*C, divided by C. Here, B plays the role of the OEIS index n.

			For B=14, we have C=15, so SQP(ABC)=SQP(210)=2*3*5*7=210, so SQP(ABC)/C=210/15=14.
For B=19, we have C=20, so SQP(ABC)=SQP(380)=2*5*19=190, so SQP(ABC)/C=190/20=9.5, which rounds to 10.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Abderrahmane Nitaj, The ABC conjecture homepage
Eric Weisstein's World of Mathematics, abc Conjecture
Wikipedia, ABC conjecture

Cf. A191093, A191100, A120498.

SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A190846:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 1 >; [ A190846(n): n in [1..85] ]; // Klaus Brockhaus, May 27 2011

A190846 := proc(n) c := 1+n ; round(A007947(n*c)/c) ; end proc:
seq(A190846(n),n=1..80) ; # R. J. Mathar, Jun 10 2011

Array[Round[SelectFirst[Reverse@ Divisors[#1 #2], SquareFreeQ]/#2] & @@ {#, # + 1} &, 83] (* Michael De Vlieger, Feb 19 2019 *)

rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
a(n)=rad(n^2+n)\/(n+1) \\ Charles R Greathouse IV, Mar 11 2014

from operator import mul
from sympy import primefactors
def rad(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a(n): return int(round(rad(n**2 + n)/(n + 1))) # Indranil Ghosh, May 24 2017

A141029 Nearest integer to the space diagonal of the smallest (measured by the longest edge) primitive (gcd(a,b,c)=1) Euler bricks (a, b, c, sqrt(a^2 + b^2), sqrt(b^2 + c^2), sqrt(a^2 + c^2) are integers). If the space diagonal is an integer then the Euler brick is called a "perfect cuboid". There are no known perfect cuboids.

Original entry on oeis.org

271, 444, 855, 737, 840, 1887, 1893, 2537, 2897, 3961, 3816, 6596, 8595, 6383, 9260, 8327, 9525, 9405, 13454, 16525, 12122, 12167, 15336, 14721, 22943, 20988, 22444, 25844, 28443, 26336, 30382, 29714, 35079, 31094, 31700, 38989, 32965

Offset: 1

Darrell Minor, Jul 29 2008

nonn

			a(1)=271 because sqrt(240^2 + 117^2 + 44^2) = 270.60, where 240 is the longest edge, 117 the intermediate edge and 44 the smallest edge, of the smallest primitive Euler brick.

Eric Weisstein's World of Mathematics, Euler Brick.

Cf. A031173, A031174, A031175.

A099741 a(1) = a(2) = 1; a(n) = a([n/2])+a([n/3]) (n >= 3).

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21

Offset: 1

Darrell Minor, Nov 09 2004

Let f = f[x,y] be a Fibonacci variant with recurrence f(1) = f(2) = 1; f(n) = f(ceiling((n-1)/x))+f(ceiling((n-2)/y)). This sequence is f[2,3].
Nondecreasing. Increases only when n is of the form 2^x*3^y.
By the Akra-Bazzi theorem, we have a(n) = Theta(n^e), where e ~ 0.78788491102586978 is the root of the equation (1/2)^e + (1/3)^e = 1. - Jeffrey Shallit, Mar 15 2018

			a(19) = a([19/2])+a([19/3]) = a(9)+a(6) = 4+3 = 7.

Robert Israel, Table of n, a(n) for n = 1..10000
Wikipedia, Akra-Bazzi method

Cf. A088468.

f:= proc(n) option remember; procname(floor(n/2))+procname(floor(n/3)) end proc:
f(1):= 1: f(2):= 1:
map(f, [$1..100]); # Robert Israel, Mar 15 2018

a[1] = a[2] = 1;
a[n_] := a[n] = a[Floor[n/2]] + a[Floor[n/3]];
Array[a, 100] (* Jean-François Alcover, Aug 28 2020 *)

G.f. g(x) satisfies g(x) = x + (1+x)*g(x^2) + (1+x+x^2)*g(x^3). - Robert Israel, Mar 15 2018

Name corrected by Robert Israel, Mar 15 2018

A069100 n-th n-digit prime number.

Original entry on oeis.org

2, 13, 107, 1021, 10061, 100069, 1000117, 10000189, 100000193, 1000000181, 10000000259, 100000000193, 1000000000303, 10000000000411, 100000000000487, 1000000000000513, 10000000000000931, 100000000000000591, 1000000000000000861, 10000000000000000667

Offset: 1

Darrell Minor, Apr 05 2002

Eduard Roure Perdices, Table of n, a(n) for n = 1..1000

Cf. A003617, A107108, A107109.

a[n_] := NextPrime[10^(n - 1), n]; (* Eduard Roure Perdices, May 17 2021 *)

a(n)=prime(primepi(10^(n-1))+n) \\ Franklin T. Adams-Watters, Mar 07 2014

from sympy import nextprime
def a(n):  return nextprime(10**(n-1), ith=n)
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Jun 16 2021

More terms from Vladeta Jovovic, Apr 07 2002

Extended by Eduard Roure Perdices, May 17 2021

A068983 a(n) = Sum_{k=0..n} (k^k-k!).

Original entry on oeis.org

0, 2, 23, 255, 3260, 49196, 867699, 17604595, 404662204, 10401033404, 295672787215, 9211294233871, 312080173805324, 11423999821072140, 449316582527563515, 18896039733447227131, 846135945932355895308, 40192537618855187742732, 2018612071634068368034711

Offset: 1

Darrell Minor, Apr 02 2002

a(n) = number of non-injective functions [k]->[k] for 1<=k<=n.

			a(4) = 255 because (1^1-1!)+(2^2-2!)+(3^3-3!)+(4^4-4!) = 255.

Cf. A003422, A062970, A036679.

Accumulate[Table[n^n-n!,{n,20}]] (* Harvey P. Dale, Aug 21 2011 *)

a(n) = Sum_{k=0..n} (k^k-k!).

a(n) = A062970(n) - A003422(n+1). - Alois P. Heinz, Aug 10 2021

cp's OEIS Frontend

User: Darrell Minor

A191093 [Squarefree part of (ABC)]/C for A=2, C=A+B, as a function of B, rounded to nearest integer.

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A191100 [Squarefree part of (ABC)]/C for A=3, C=A+B, as a function of B, rounded to nearest integer.

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A190846 (Squarefree part of (ABC))/C for A=1, C=A+B, as a function of B, rounded to the nearest integer.

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A099741 a(1) = a(2) = 1; a(n) = a([n/2])+a([n/3]) (n >= 3).

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A069100 n-th n-digit prime number.

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Python

Extensions

A068983 a(n) = Sum_{k=0..n} (k^k-k!).

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