A132356 -id:A132356 - cp's OEIS frontend

A273372 Squares ending in digit 1.

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

A321383 Numbers k such that the concatenation k21 is a square.

Original entry on oeis.org

1, 15, 37, 79, 123, 193, 259, 357, 445, 571, 681, 835, 967, 1149, 1303, 1513, 1689, 1927, 2125, 2391, 2611, 2905, 3147, 3469, 3733, 4083, 4369, 4747, 5055, 5461, 5791, 6225, 6577, 7039, 7413, 7903, 8299, 8817, 9235, 9781, 10221, 10795, 11257, 11859, 12343, 12973, 13479

Offset: 1

Bruno Berselli, Nov 08 2018

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

Cf. A008805.

Numbers k such that the concatenation km is a square: A132356 (m = 1), A273365 (m = 4), A273366 (m = 5), A273367 (m = 6), A273368 (m = 9); missing sequence for m = 16; this sequence for m = 21; missing sequence for m = 24; A002378 (m = 25).

List([1..50], n -> (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8);

[div((50*(n-1)*n+3*(2*n-1)*(-1)^n+11), 8) for n in 1:50] |> println

[(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8: n in [1..50]];

Table[(50 (n - 1) n + 3 (2 n - 1) (-1)^n + 11)/8, {n, 1, 50}]

makelist((50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8, n, 1, 50);

vector(50, n, nn; (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8)

Vec(x*(1 + 14*x + 20*x^2 + 14*x^3 + x^4) / ((1 - x)^3*(1 + x)^2) + O(x^50)) \\ Colin Barker, Nov 12 2018

[(50*(n-1)*n+3*(2*n-1)*(-1)**n+11)/8 for n in range(1, 50)]

[(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8 for n in (1..50)]

G.f.: x*(1 + 14*x + 20*x^2 + 14*x^3 + x^4)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = 2*a(n-2) - a(n-4) + 50.
a(n) = (50*(n - 1)*n + 3*(2*n - 1)*(-1)^n + 11)/8. Therefore:
a(n) = (25*n^2 - 22*n + 4)/4 for even n;
a(n) = (25*n^2 - 28*n + 7)/4 for odd n.

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A273372 Squares ending in digit 1.

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A321383 Numbers k such that the concatenation k21 is a square.

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