A273368 -id:A273368 - cp's OEIS frontend

A273366 a(n) = 10n^2 + 10n + 2.

2, 22, 62, 122, 202, 302, 422, 562, 722, 902, 1102, 1322, 1562, 1822, 2102, 2402, 2722, 3062, 3422, 3802, 4202, 4622, 5062, 5522, 6002, 6502, 7022, 7562, 8122, 8702, 9302, 9922, 10562, 11222, 11902, 12602, 13322, 14062, 14822, 15602

Offset: 0

Nathan Fox, Brooke Logan, and N. J. A. Sloane, May 20 2016

These are the numbers k such that 10*k+5 is a perfect square.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A062786, A132356, A273365, A273367, A273368, A350760.

Cf. A033583 (perfect squares ending in 0 in base 10 with final 0 removed).

LinearRecurrence[{3,-3,1}, {2, 22, 62}, 50] (* G. C. Greubel, May 20 2016 *)
Table[10n^2+10n+2,{n,0,40}] (* Harvey P. Dale, May 21 2024 *)

a(n)=10*n^2+10*n+2 \\ Charles R Greathouse IV, Jan 31 2017

G.f.: 2*(x^2+8x+1)/(1-x)^3.
From G. C. Greubel, May 20 2016: (Start)
E.g.f.: 2*(1 + 10*x + 5*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = 2*A062786(n+1). - R. J. Mathar, Jun 03 2016
Sum_{n>=0} 1/a(n) = Pi/(2*sqrt(5)) * tan(Pi/(2*sqrt(5))) (A350760). - Amiram Eldar, Jan 20 2022

A273367 Numbers k such that 10*k+6 is a perfect square.

Original entry on oeis.org

1, 3, 19, 25, 57, 67, 115, 129, 193, 211, 291, 313, 409, 435, 547, 577, 705, 739, 883, 921, 1081, 1123, 1299, 1345, 1537, 1587, 1795, 1849, 2073, 2131, 2371, 2433, 2689, 2755, 3027, 3097, 3385, 3459, 3763, 3841, 4161, 4243, 4579

Offset: 0

Nathan Fox, Brooke Logan, and N. J. A. Sloane, May 20 2016

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

Cf. A132356, A273365, A273366, A273368.

Cf. A033583 (perfect squares ending in 0 in base 10 with final 0 removed).

LinearRecurrence[{1, 2, -2, -1, 1}, {1, 3, 19, 25, 57}, 50] (* G. C. Greubel, May 20 2016 *)

is(n)=issquare(10*n+6) \\ Charles R Greathouse IV, Jan 31 2017

a(2n) = 10*n^2 - 8*n + 1.
a(2n+1) = 10*n^2 + 8*n + 1.
G.f.: (x^4+2x^3+14x^2+2x+1)/((1-x)^3*(1+x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - G. C. Greubel, May 20 2016

A273365 Numbers k such that 10*k+4 is a perfect square.

Original entry on oeis.org

0, 6, 14, 32, 48, 78, 102, 144, 176, 230, 270, 336, 384, 462, 518, 608, 672, 774, 846, 960, 1040, 1166, 1254, 1392, 1488, 1638, 1742, 1904, 2016, 2190, 2310, 2496, 2624, 2822, 2958, 3168, 3312, 3534, 3686, 3920, 4080, 4326, 4494

Offset: 0

Nathan Fox, Brooke Logan, and N. J. A. Sloane, May 20 2016

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

Cf. A132356, A273366, A273367, A273368.

Cf. A033583 (perfect squares ending in 0 in base 10 with final 0 removed).

LinearRecurrence[{1, 2, -2, -1, 1}, {0, 6, 14, 32, 48}, 50] (* G. C. Greubel, May 21 2016 *)
Select[Range[0,5000],IntegerQ[Sqrt[10#+4]]&] (* Harvey P. Dale, Apr 19 2019 *)

is(n)=issquare(10*n+4) \\ Charles R Greathouse IV, Jan 31 2017

a(2n) = 10*n^2 + 4*n, n>=0.
a(2n-1) = 10*n^2 - 4*n, n>=1.
G.f.: 2*x*(3*x^2+4x+3)/((1-x)^3*(1+x)^2).
From G. C. Greubel, May 21 2016: (Start)
E.g.f.: (1/2)*((5*x^2 + 11*x)*cosh(x) + (5*x^2 + 9*x + 1)*sinh(x)).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)

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.

cp's OEIS Frontend

A273366 a(n) = 10*n^2 + 10*n + 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A273367 Numbers k such that 10*k+6 is a perfect square.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A273365 Numbers k such that 10*k+4 is a perfect square.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Julia

Magma

Mathematica

Maxima

PARI

PARI

Python

Sage

Formula

A273366 a(n) = 10n^2 + 10n + 2.