A273372 -id:A273372 - cp's OEIS frontend

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

Vincenzo Librandi, May 21 2016

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

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).

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).

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

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

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

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 *)

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

Corrected and extended by Bruno Berselli, May 23 2016

A348487 Positive numbers whose square starts and ends with exactly one 1.

Original entry on oeis.org

1, 11, 39, 41, 101, 111, 119, 121, 129, 131, 139, 141, 319, 321, 329, 331, 349, 351, 359, 361, 369, 371, 379, 381, 389, 391, 399, 401, 409, 411, 419, 421, 429, 431, 439, 441, 1001, 1009, 1011, 1019, 1021, 1029, 1031, 1039, 1041, 1099, 1101, 1109, 1111, 1119, 1121, 1129, 1131, 1139

Offset: 1

Bernard Schott, Oct 21 2021

When a square ends with 1, this square ends with exactly one 1.

Sequences A000533 and A253213 show that there are an infinity of terms. The square of their terms, for n >= 3, starts and ends with exactly one 1. Also, the numbers 119, 1119, 11119, ..., ((10^k + 71) / 9)^2, (k >= 3) are terms. The squares ((10^k + 71) / 9)^2, have the last digit 1 and because 12*10^(2*k - 3) < ((10^k + 71) / 9)^2 <13*10^(2*k - 3), for k >= 3, the squares ((10^k + 71) / 9)^2, k >= 4, start with 12. - Marius A. Burtea, Oct 21 2021

			39 is a term since 39^2 = 1521.
109 is not a term since 109^2 = 11881.
119 is a term since 119^2 = 14161.

Cf. A045855, A090771, A253213, A273372 (squares ending with 1), A017281, A017377.
Cf. A000533, A253213 for n >= 2 (subsequences).
Subsequence of A305719.

[1] cat [n:n in [2..1200]|Intseq(n*n)[1] eq 1 and Intseq(n*n)[#Intseq(n*n)] eq 1 and Intseq(n*n)[-1+#Intseq(n*n)] ne 1]; // Marius A. Burtea, Oct 21 2021

Join[{1}, Select[Range[11, 1200], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 1 && d[[2]] != 1 &]] (* Amiram Eldar, Oct 21 2021 *)

isok(k) = my(d=digits(sqr(k))); (d[1]==1) && (d[#d]==1) && if (#d>2, (d[2]!=1) && (d[#d-1]!=1), 1); \\ Michel Marcus, Oct 21 2021

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("1")) == len(s.lstrip("1")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+d for i in count(0) for d in [1, 9]))
  return [k for k in r if ok(k)]
print(aupto(1140)) # Michael S. Branicky, Oct 21 2021

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A273373 Squares ending in digit 6.

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A348487 Positive numbers whose square starts and ends with exactly one 1.

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