A273373 Squares ending in digit 6.

16, 36, 196, 256, 576, 676, 1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776, 7056, 7396, 8836, 9216, 10816, 11236, 12996, 13456, 15376, 15876, 17956, 18496, 20736, 21316, 23716, 24336, 26896, 27556, 30276, 30976, 33856, 34596, 37636, 38416, 41616

Offset: 1

Author

Vincenzo Librandi, May 21 2016

Keywords

• nonn
• base
• easy

Comments

These are the only squares whose second last digit is odd. This implies that the only squares whose last two digits are the same are those ending with 0 or 4; those ending with 1, 5, and 9 are paired with even second last digits. - Waldemar Puszkarz, May 24 2016

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A000290, A047221, A090773.

Cf. A017341 (numbers ending in 6), A017343 (cubes ending in 6).

Cf. squares with last digit k: A017270 (k=0), A273372 (k=1), A273375 (k=4), A017330 (k=5), this sequence (k=6), A273374 (k=9).

Programs

Magma

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 6];

Magma

[(10*n - 3*(-1)^n - 5)^2/4: n in [1..50]];

Maple

seq(seq((10*i+j)^2,j=[4,6]),i=0..20); # Robert Israel, May 24 2016

Mathematica

Table[(10 n - 3 (-1)^n - 5)^2/4, {n, 1, 50}]
CoefficientList[Series[4 (4 + 5 x + 32 x^2 + 5 x^3 + 4 x^4) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x]
Select[Range[250]^2,Mod[#,10]==6&] (* Harvey P. Dale, May 31 2020 *)

Formula

G.f.: 4*x*(4 + 5*x + 32*x^2 + 5*x^3 + 4*x^4)/((1 + x)^2*(1 - x)^3).

a(n) = 4*A047221(n)^2 = (10*n - 3*(-1)^n - 5)^2/4.

a(n) = A090773(n)^2. - Michel Marcus, May 25 2016

Sum_{n>=1} 1/a(n) = 2*Pi^2/(25*(5+sqrt(5))). - Amiram Eldar, Feb 16 2023

Extensions

Corrected and extended by Bruno Berselli, May 23 2016