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

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

1,2

Links

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).

Formula

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.

Mathematica

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

Prog

(PARI) vector(50, n, nn; (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8)
(PARI) 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
(Sage) [(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8 for n in (1..50)]
(Maxima) makelist((50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8, n, 1, 50);
(GAP) List([1..50], n -> (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8);
(Magma) [(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8: n in [1..50]];
(Python) [(50*(n-1)*n+3*(2*n-1)*(-1)**n+11)/8 for n in range(1, 50)]
(Julia) [div((50*(n-1)*n+3*(2*n-1)*(-1)^n+11), 8) for n in 1:50] |> println

Crossrefs

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).

Sequence in context: A059605 A147221 A051461 * A039605 A032654 A320721

Adjacent sequences: A321380 A321381 A321382 * A321384 A321385 A321386

Keyword

nonn,easy,base

Author

Bruno Berselli, Nov 08 2018

Status

approved