A177735 -id:A177735 - cp's OEIS frontend

A047222 Numbers that are congruent to {0, 2, 3} mod 5.

0, 2, 3, 5, 7, 8, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 27, 28, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 82, 83, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 102, 103, 105, 107

Offset: 1

N. J. A. Sloane

Row sum of a triangle where the top value is 2 and every elementary triangle or triple is required to have the values 1,2,2 (see link below). Compare with A008854 where the triple contains 1,2,2 with 1 at the top. - Craig Knecht, Oct 18 2015
Also, numbers k such that k*(k^2+1)/5 is a nonnegative integer. - Bruno Berselli, Jan 16 2016
Conjecture: Apart from 0, the sequence gives the values for c/6, such that an infinite number of primes, p, result in both p^2-c and p^2+c being positive primes, except when c is a square. When c is square solutions exist for c (both within and outside of the a(n) set), but occur at only a single prime p. See A274609. Other c values with only one prime providing a solution occur when p^2-c=3. See A274610. The only remaining c values with single p solutions are: c=2 (with p=3) and c=6 (with p=5). - Richard R. Forberg, Jun 26 2016
See A047363 for case of p^3 +- c. See A005097 and A177735 for observations on the general case p^q +- c. - Richard R. Forberg, Aug 11 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Craig Knecht, Row sum for the 1,2,2 triangle with 2 at the top.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A008854, A028738, A042965, A047363, A005097.

[n : n in [0..150] | n mod 5 in [0, 2, 3]]; // Wesley Ivan Hurt, Jun 14 2016

A047222 := n -> 2*n - 2 - iquo(n,3): seq(A047222(n), n=1..100); # Wesley Ivan Hurt, Nov 07 2013

Floor[(Range[5, 305, 5] - 4)/3] (* Vladimir Joseph Stephan Orlovsky, Jan 26 2012 *)
Flatten[Table[5n + {0, 2, 3}, {n, 0, 19}]] (* Alonso del Arte, Nov 07 2013 *)
LinearRecurrence[{1, 0, 1, -1}, {0, 2, 3, 5}, 100] (* Vincenzo Librandi, Jun 15 2016 *)

a(n)=(5*n-4)\3 \\ Charles R Greathouse IV, Oct 28 2011

concat(0, Vec(x^2*(2+x+2*x^2)/((1-x)^2*(1+x+x^2)) + O(x^100))) \\ Altug Alkan, Oct 26 2015

From R. J. Mathar, Oct 18 2008: (Start)
G.f.: x^2*(2 + x + 2*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = A028738(n-2), 1 < n < 16. (End)
a(n) = floor((5*n-4)/3). - Gary Detlefs, Oct 28 2011
a(n) = 2*n - 2 - floor(n/3). - Wesley Ivan Hurt, Nov 07 2013
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (15*n-15-3*cos(2*n*Pi/3)-sqrt(3)*sin(2*n*Pi/3))/9.
a(3k) = 5k-2, a(3k-1) = 5k-3, a(3k-2) = 5k-5. (End)
a(n) = n - 1 + floor((2n-1)/3). - Wesley Ivan Hurt, Dec 27 2016
Sum_{n>=2} (-1)^n/a(n) = arccoth(3/sqrt(5))/sqrt(5) - log(2)/5. - Amiram Eldar, Dec 10 2021
From Peter Bala, Aug 04 2022: (Start)
a(n) = a(floor(n/2)) + a(1 + ceiling(n/2)) for n >= 4 with a(1) = 0, a(2) = 2 and a(3) = 3.
a(2*n) = a(n) + a(n+1); a(2*n+1) = a(n) + a(n+2). Cf. A008854 and A042965. (End)

A256675 Denominators of the inverse binomial transform of Bernoulli(n+2).

Original entry on oeis.org

6, 6, 15, 15, 105, 21, 105, 15, 165, 33, 15015, 1365, 1365, 3, 255, 255, 33915, 399, 21945, 165, 3795, 69, 31395, 1365, 1365, 3, 435, 435, 1038345, 7161, 608685, 255, 255, 3, 959595, 959595, 959595, 3, 6765, 6765, 2036265, 903, 103845, 345, 16215, 141, 1090635

Offset: 0

Paul Curtz, Apr 07 2015

nonn

Difference table of B(n+2):
1/6,        0,  -1/30,     0,   1/42,     0, -1/30, ...
-1/6,   -1/30,   1/30,  1/42,  -1/42, -1/30,   ...
2/15,    1/15, -1/105, -1/21, -1/105,   ...
-1/15, -8/105, -4/105, 4/105,    ...
-1/105, 4/105,  8/105,   ...
1/21,   4/105,    ...
-1/105,   ...
...
a(n) is the denominator of the n-th term of the first column.
a(n+2) is the denominator of the n-th term of the third row.
See A239315(n), which is the table without the first two rows.
Inverse binomial transform: 1/6, -1/6, 2/15, -1/15, -1/105, 1/21, -1/105, -1/15, 7/165, 5/33, -2663/15015, ... .

Cf. A190339, A239315, A029765, A001897, A141459, A172087, A168516, A177735, A256595.

max = 42; bb = Table[BernoulliB[n+2], {n, 0, max}]; dd = Table[Differences[bb, n], {n, 0, max}]; dd[[All, 1]] // Denominator (* Jean-François Alcover, Apr 09 2015 *)

lista(nn) = {A = vector(nn, n, bernfrac(n+1)); for (i=1, #A-1, for(j=0,i-1,A[i+1]-=binomial(i,j)*A[j+1])); for (i=1, #A, print1(denominator(A[i]), ", "));} \\ Michel Marcus, Apr 08 2015

a(2n) = A029765(n).
a(2n+3) = A001897(n+2).
a(2n)/a(2n+1) = A177735(n).
a(2n+4)/a(2n+3) = A177735(n+3).

cp's OEIS Frontend

A047222 Numbers that are congruent to {0, 2, 3} mod 5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

A256675 Denominators of the inverse binomial transform of Bernoulli(n+2).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Formula