A090771 -id:A090771 - cp's OEIS frontend

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

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

A273372 Squares ending in digit 1.

Original entry on oeis.org

1, 81, 121, 361, 441, 841, 961, 1521, 1681, 2401, 2601, 3481, 3721, 4761, 5041, 6241, 6561, 7921, 8281, 9801, 10201, 11881, 12321, 14161, 14641, 16641, 17161, 19321, 19881, 22201, 22801, 25281, 25921, 28561, 29241, 32041, 32761, 35721, 36481, 39601, 40401

Offset: 1

Vincenzo Librandi, May 21 2016

Intersection of A000290 and A017281; also, union of A017282 and A017378. The square roots are in A017281 or in A017377 (numbers ending in 1 or 9, respectively). - David A. Corneth, May 22 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, A090771, A132356.
Cf. A017281 (numbers ending in 1), A017283 (cubes ending in 1).
Cf. similar sequences listed in A273373.

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 1];

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

Table[5 (10 n + (-1)^n - 5) (2 n + (-1)^n - 1)/4 + 1, {n, 1, 50}]

A273372_list = [(10*n+m)**2 for n in range(10**3) for m in (1,9)] # Chai Wah Wu, May 24 2016

p (1..(n + 1) / 2).inject([]){|s, i| s + [(10 * i - 9) ** 2, (10 * i - 1) ** 2]}[0..n - 1] # Seiichi Manyama, May 24 2016

G.f.: x*(1 + 80*x + 38*x^2 + 80*x^3 + x^4) / ((1 + x)^2*(1 - x)^3).
a(n) = 10*A132356(n-1) + 1 = 5*(10*n+(-1)^n-5)*(2*n+(-1)^n-1)/4+1.
a(n) = (5*n - 5/2 + (3/2)*(-1)^n)^2 = 25*n^2 - 25*n + 17/2 + 15*n*(-1)^n - (15/2)*(-1)^n. - David A. Corneth, May 21 2016
a(n) = A090771(n)^2. - Michel Marcus, May 25 2016
Sum_{n>=1} 1/a(n) = Pi^2*(3+sqrt(5))/50. - Amiram Eldar, Feb 16 2023

Edited by Bruno Berselli, May 24 2016

A266297 Numbers whose last digit is a square.

Original entry on oeis.org

0, 1, 4, 9, 10, 11, 14, 19, 20, 21, 24, 29, 30, 31, 34, 39, 40, 41, 44, 49, 50, 51, 54, 59, 60, 61, 64, 69, 70, 71, 74, 79, 80, 81, 84, 89, 90, 91, 94, 99, 100, 101, 104, 109, 110, 111, 114, 119, 120, 121, 124, 129, 130, 131, 134, 139, 140, 141, 144, 149

Offset: 1

Wesley Ivan Hurt, Dec 26 2015

Numbers ending in 0, 1, 4 and 9.
Union of A008592, A017281, A017317 and A017377. - Hurt
None of these numbers are prime in Z[phi] (where phi = 1/2 + sqrt(5)/2 is the golden ratio), since the numbers in this sequence that are prime in Z can be expressed in the form (a - b sqrt(5))(a + b sqrt(5)). - Alonso del Arte, Dec 30 2015
Union of A197652 and A016897. - Wesley Ivan Hurt, Dec 31 2015
Union of A146763 and A090771. - Wesley Ivan Hurt, Jan 01 2016

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A008592, A016897, A017281, A017317, A017377, A090771, A091084, A141158, A146763, A197652.

[(10*n-11+(-1)^n+(4+2*(-1)^n)*(-1)^((2*n-1+(-1)^n) div 4))/4: n in [1..60]]; // Vincenzo Librandi, Dec 27 2015

A266297:=n->(10*n-11+(-1)^n+(4+2*(-1)^n)*(-1)^((2*n-1+(-1)^n)/4))/4: seq(A266297(n), n=1..100);

Table[(10 n - 11 + (-1)^n + (4 + 2 (-1)^n)*(-1)^((2 n - 1 + (-1)^n)/4))/4, {n, 50}] (* G. C. Greubel, Dec 27 2015 *)
LinearRecurrence[{1, 0, 0, 1, -1}, {0, 1, 4, 9, 10}, 60] (* Vincenzo Librandi, Dec 27 2015 *)
CoefficientList[Series[x*(1 + 3*x + 5*x^2 + x^3)/((x - 1)^2*(1 + x + x^2 + x^3)), {x, 0, 100}], x] (* Wesley Ivan Hurt, Dec 30 2015 *)
Flatten[Table[10n + {0, 1, 4, 9}, {n, 0, 19}]] (* Alonso del Arte, Dec 30 2015 *)
Select[Range[0,150],MemberQ[{0,1,4,9},Mod[#,10]]&] (* Harvey P. Dale, Jul 30 2019 *)

is(n) = issquare(n%10); \\ Altug Alkan, Dec 29 2015

G.f.: x^2*(1 + 3*x + 5*x^2 + x^3)/((x - 1)^2*(1 + x + x^2 + x^3)).
a(n) = a(n - 1) + (n - 4) - a(n - 5) for n > 5.
a(n) = (10n - 11 + (-1)^n + (4 + 2(-1)^n) * (-1)^((2n - 1 + (-1)^n)/4))/4.
a(n+1) - a(n) = A091084(n+1) for n>0.
Sum_{n>=2} (-1)^n/a(n) = (14*sqrt(5)*arccoth(sqrt(5)) - 2*Pi*sqrt(1-2/sqrt(5)) + 16*log(2) + 5*log(5))/40. - Amiram Eldar, Jul 30 2024

A299250 Numbers congruent to {9, 11, 21, 29} mod 30.

Original entry on oeis.org

9, 11, 21, 29, 39, 41, 51, 59, 69, 71, 81, 89, 99, 101, 111, 119, 129, 131, 141, 149, 159, 161, 171, 179, 189, 191, 201, 209, 219, 221, 231, 239, 249, 251, 261, 269, 279, 281, 291, 299, 309, 311, 321, 329, 339, 341, 351, 359, 369, 371, 381, 389, 399, 401, 411

Offset: 1

Arkadiusz Wesolowski, Feb 05 2018

For any m >= 0, if F(m) = 2^(2^m) + 1 has a factor of the form b = a(n)*2^k + 1 with odd k >= m + 2 and n >= 1, then the cofactor of F(m) is equal to F(m)/b = j*2^k + 1, where j is congruent to 1 mod 10 if n == 0 or 1 mod 4, or j is congruent to 9 mod 10 if n == 2 or 3 mod 4. That is, the integer a(n) + j must be divisible by 10.

			39 belongs to this sequence and d = 39*2^13 + 1 is a divisor of F(11) = 2^(2^11) + 1, so 10 | (39 + (F(11)/d - 1)/2^13).

Wikipedia, Fermat number
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Subsequence of A090771.

Cf. A000215, A298360.

[n: n in [0..411] | n mod 30 in {9, 11, 21, 29}];

LinearRecurrence[{1, 0, 0, 1, -1}, {9, 11, 21, 29, 39}, 60]
CoefficientList[ Series[(9 + 2x + 10x^2 + 8x^3 + x^4)/((-1 + x)^2 (1 + x + x^2 + x^3)), {x, 0, 54}], x] (* Robert G. Wilson v, Feb 08 2018 *)

Vec(x*(9 + 2*x + 10*x^2 + 8*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2 + O(x^55)))

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
a(n) = a(n-4) + 30.
G.f.: x*(9 + 2*x + 10*x^2 + 8*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2).

A237128 Angles n expressed in degrees such that 2*cos(n) = phi where phi is the golden ratio (A001622).

Original entry on oeis.org

36, 324, 396, 684, 756, 1044, 1116, 1404, 1476, 1764, 1836, 2124, 2196, 2484, 2556, 2844, 2916, 3204, 3276, 3564, 3636, 3924, 3996, 4284, 4356, 4644, 4716, 5004, 5076, 5364, 5436, 5724, 5796, 6084, 6156, 6444, 6516, 6804, 6876, 7164, 7236, 7524, 7596, 7884

Offset: 1

Michel Lagneau, Feb 04 2014

a(n) == 36, 324 mod 360 and a(n)/36 is congruent to {1,9} mod 10 (A090771).
See A019863 = half of the golden ratio (A001622) => a(1) = 90 - 54 degrees and a(2) = 360 - a(1) = 324 degrees.
The squares in the sequence are 36, 324, 1764, 2916, 4356, 6084, 10404, 12996, 15876, 19044, 26244, 30276, 34596, 39204, 49284, 54756, 60516, 66564, 79524,... with the following properties:
If a(n) == 36 mod 360 is a perfect square, sqrt(36+360*n)/6 = A090771 (numbers that are congruent to {1, 9} mod 10).
If a(n) == 324 mod 360 is a perfect square, sqrt(324+360*n)/6 = A063226 (numbers that are congruent to {3, 7} mod 10).

			1476 is in the sequence because 2*cos(1476°) = 2*cos(1476*Pi/180) = 1.61803398... = phi.

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

Cf. A001622, A019863.

***first program***
with(numtheory):err:=1/10^10:Digits:=20:for n from 1 to 20000 do:x:=evalf(2*cos(n*Pi/180)):ph:=evalf((1+sqrt(5)))/2:if abs(ph-x)

Select[Range[8000],2*Cos[# Degree]==GoldenRatio&] (* or *) LinearRecurrence[ {1,1,-1},{36,324,396},50] (* Harvey P. Dale, Aug 14 2015 *)

Vec(36*x*(x^2+8*x+1)/((x-1)^2*(x+1)) + O(x^100)) \\ Colin Barker, Feb 04 2014

a(n) = 18*(-5+3*(-1)^n+10*n). a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: 36*x*(x^2+8*x+1) / ((x-1)^2*(x+1)). - Colin Barker, Feb 04 2014

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

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A273372 Squares ending in digit 1.

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A266297 Numbers whose last digit is a square.

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A299250 Numbers congruent to {9, 11, 21, 29} mod 30.

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A237128 Angles n expressed in degrees such that 2*cos(n) = phi where phi is the golden ratio (A001622).

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