A145979 -id:A145979 - cp's OEIS frontend

A288994 a(n) = n(n+3) when n is congruent to 0 or 3 (mod 4), and n(n+3)/2 otherwise.

0, 2, 5, 18, 28, 20, 27, 70, 88, 54, 65, 154, 180, 104, 119, 270, 304, 170, 189, 418, 460, 252, 275, 598, 648, 350, 377, 810, 868, 464, 495, 1054, 1120, 594, 629, 1330, 1404, 740, 779, 1638, 1720, 902, 945, 1978, 2068, 1080, 1127, 2350, 2448, 1274, 1325

Offset: 0

Jean-François Alcover and Paul Curtz, Jun 21 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A000096, A014695, A028552, A060819, A130658, A145979, A227316.

a[n_] := n (n+3) Switch[Mod[n, 4], 0|3, 1, _, 1/2]; Table[a[n], {n, 0, 50}]
Table[If[MemberQ[{0,3},Mod[n,4]],n(n+3),(n(n+3))/2],{n,0,50}] (* or *) LinearRecurrence[{3,-6,10,-12,12,-10,6,-3,1},{0,2,5,18,28,20,27,70,88},60] (* Harvey P. Dale, Jun 05 2021 *)

concat(0, Vec(x*(2 - x + 15*x^2 - 16*x^3 + 18*x^4 - 9*x^5 + 5*x^6 - 2*x^7) / ((1 - x)^3*(1 + x^2)^3) + O(x^60))) \\ Colin Barker, Jun 21 2017

i=I; a(n) = (1/8 + i/8)*(((3 - 3*i) - i*(-i)^n + i^n)*n*(3 + n)) \\ Colin Barker, Jun 21 2017

a(n) = n*(n+3)/2 * (2 - floor((n+1)/2) mod 2), where n*(n+3)/2 is A000096(n).
a(n) = A060819(n+3)*A145979(n-2).
a(n) = (2*n*(n+3))/(GCD(4, n+2)*GCD(4, n+3)).
a(n) = A227316(n+1) - (period 4 repeat 2,1,1,2).
From Colin Barker, Jun 21 2017: (Start)
G.f.: x*(2 - x + 15*x^2 - 16*x^3 + 18*x^4 - 9*x^5 + 5*x^6 - 2*x^7) / ((1 - x)^3*(1 + x^2)^3).
a(n) = (1/8 + i/8)*(((3 - 3*i) - i*(-i)^n + i^n)*n*(3 + n)), where i=sqrt(-1). (End)
Sum_{n>=1} 1/a(n) = 17/18 + log(2)/6. - Amiram Eldar, Aug 12 2022

A306591 a(n) is the denominator of 1/2 - 1/(prime(n)+1), where prime(n) is the n-th prime.

Original entry on oeis.org

6, 4, 3, 8, 12, 7, 9, 20, 24, 15, 32, 19, 21, 44, 48, 27, 60, 31, 68, 72, 37, 80, 84, 45, 49, 51, 104, 108, 55, 57, 128, 132, 69, 140, 75, 152, 79, 164, 168, 87, 180, 91, 192, 97, 99, 200, 212, 224, 228, 115, 117, 240, 121, 252, 129, 264, 135, 272, 139, 141, 284, 147, 308, 312, 157

Offset: 1

Michel Marcus, Feb 26 2019

Nitya Mani and Simon Rubinstein-Salzedo, Diophantine Tuples over Zp, arXiv:1902.09048 [math.NT], 2019. See Lemma 3.3. p. 4.

Cf. A236965 (numerators).

Cf. A145979 (denominator of 1/2 - 1/(n+1)).

1/2-1/(#+1)&/@Prime[Range[70]]//Denominator (* Harvey P. Dale, Feb 19 2023 *)

a(n) = denominator(1/2 - 1/(prime(n)+1));

from sympy import prime
from fractions import Fraction
def a(n): return (Fraction(1, 2) - Fraction(1, (prime(n)+1))).denominator
print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Jun 04 2021

cp's OEIS Frontend

A288994 a(n) = n(n+3) when n is congruent to 0 or 3 (mod 4), and n(n+3)/2 otherwise.

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A306591 a(n) is the denominator of 1/2 - 1/(prime(n)+1), where prime(n) is the n-th prime.

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cp's OEIS Frontend

A288994 a(n) = n*(n+3) when n is congruent to 0 or 3 (mod 4), and n*(n+3)/2 otherwise.

Views

Author

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Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A306591 a(n) is the denominator of 1/2 - 1/(prime(n)+1), where prime(n) is the n-th prime.

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Author

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Links

Crossrefs

Programs

Mathematica

PARI

Python

A288994 a(n) = n(n+3) when n is congruent to 0 or 3 (mod 4), and n(n+3)/2 otherwise.