A047205 -id:A047205 - cp's OEIS frontend

A226096 Squares with doubled (4*n+2)^2.

1, 4, 4, 9, 16, 25, 36, 36, 49, 64, 81, 100, 100, 121, 144, 169, 196, 196, 225, 256, 289, 324, 324, 361, 400, 441, 484, 484, 529, 576, 625, 676, 676, 729, 784, 841, 900, 900, 961, 1024, 1089, 1156, 1156, 1225, 1296, 1369

Offset: 0

Paul Curtz, May 26 2013

Also nondecreasing ordered values of A226008 (except 0).
Consider A225948/A226008 ordered according to a(n): 0/1, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = b(n)/a(n), and consider the sequence with period 5: 1, 64, 16, 1, 4, ... = t(n); then a(n) = 4*b(n) + t(n).
The recurrences in Formula lines are also valid for b(n).
Note that the fractions b(n)/a(n) of rank 0, 3,4,5, 8,9,10, ... = A047205:
0,  2/9, 3/16, 6/25, 12/49, 15/64, 20/81, ... are all in A226023(n).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2,-2,0,0,0,-1,1).

Cf. A016826, A226008.

MapIndexed[ If [Mod[First[#2], 4] == 2, Sequence @@ {#1, #1}, #1] &, Range[40]]^2 (* Jean-François Alcover, May 28 2013 *)

a(n+5) - a(n) = 8*A090223(n+4).
a(n) = 1 followed by (A090223(n) + 2)^2.
a(n) = 3*a(n-5) -3*a(n-10) +a(n-15).
G.f.: (x^9 + 3*x^8 + 5*x^6 + 7*x^5 + 7*x^4 + 5*x^3 + 3*x + 1)/((1 - x)*(1 - x^5)^2). [Ralf Stephan, May 30 2013]
a(n) = a(n-1) +2*a(n-5) -2*a(n-6) -a(n-10) +a(n-11). [Bruno Berselli, May 30 2013]
a(n) = (24*(16*floor(n/5)^2 + 8*floor(n/5) + 1) - (11 + 24*floor(n/5))*(n - 5*floor(n/5))^4 + 2*(49 + 104*floor(n/5))*(n - 5*floor(n/5))^3 - 23*(11 + 24*floor(n/5))*(n - 5*floor(n/5))^2 + 2*(119 + 280*floor(n/5))*(n - 5*floor(n/5)))/24. - Luce ETIENNE, May 08 2017

A317657 Numbers congruent to {15, 75, 95} mod 100.

Original entry on oeis.org

15, 75, 95, 115, 175, 195, 215, 275, 295, 315, 375, 395, 415, 475, 495, 515, 575, 595, 615, 675, 695, 715, 775, 795, 815, 875, 895, 915, 975, 995, 1015, 1075, 1095, 1115, 1175, 1195, 1215, 1275, 1295, 1315, 1375, 1395, 1415, 1475, 1495, 1515

Offset: 1

Paul Curtz, Aug 03 2018

Numbers written in French ending in "quinze".

a(n) = 5 * (3, 15, 19, 23, 35, 39, 43, 55, 59, ... ).

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1)

Cf. A008587, A008854, A047205, A090772, A290781, A317633.

Filtered([0..1520], n->n mod 100=15 or n mod 100=75 or n mod 100=95); # Muniru A Asiru, Aug 29 2018

select(n->modp(n,100)=15 or modp(n,100)=75 or modp(n,100)=95,[$0..1520]); # Muniru A Asiru, Aug 29 2018

Rest@ CoefficientList[Series[(5 x (x^3 + 4 x^2 + 12 x + 3))/((x^2 + x + 1) (x - 1)^2), {x, 0, 46}], x] (* Michael De Vlieger, Aug 05 2018 *)
Table[100*n/3 - 80*Sin[2*n*Pi/3]/(3*Sqrt[3]) - 5,{n,1,46}] (* Stefano Spezia, Aug 29 2018 *)

a(n) = 10*A317633(n) + 5.
a(n) = a(n-3) + 100, a(1) = 15, a(2) = 75, a(3) = 95.
From Franck Maminirina Ramaharo, Aug 05 2018: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4), n>4.
a(n) = A290781(A047205(n)).
a(n) = 20*A008854(n+1) - 5.
a(n) = 100*n/3 - 80*sin(2*n*Pi/3)/(3*sqrt(3)) - 5.
G.f.: (5*x*(x^3 + 4*x^2 + 12*x + 3))/((x^2 + x + 1)*(x - 1)^2).
E.g.f.: 100*x*exp(x)/3 - 80*sin(sqrt(3)*x/2)/(exp(x/2)*(3*sqrt(3)))-5*exp(x).
(End)

cp's OEIS Frontend

A226096 Squares with doubled (4*n+2)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula