A273366 -id:A273366 - cp's OEIS frontend

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

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)

A273368 Numbers k such that 10*k+9 is a perfect square.

Original entry on oeis.org

0, 4, 16, 28, 52, 72, 108, 136, 184, 220, 280, 324, 396, 448, 532, 592, 688, 756, 864, 940, 1060, 1144, 1276, 1368, 1512, 1612, 1768, 1876, 2044, 2160, 2340, 2464, 2656, 2788, 2992, 3132, 3348, 3496, 3724, 3880, 4120, 4284, 4536

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, A273367.

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

CoefficientList[Series[4*x*(x^2+3x+1)/((1-x)^3*(1+x)^2), {x,0,50}], x] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {0, 4, 16, 28, 52}, 50] (* G. C. Greubel, May 20 2016 *)

is(n)=issquare(10*n+9) \\ Charles R Greathouse IV, Jan 31 2017

a(2n) = 10*n^2 + 6*n, n>=0.
a(2n-1) = 10*n^2 - 6*n, n>=1.
G.f.: 4*x*(x^2+3x+1)/((1-x)^3*(1+x)^2).
From G. C. Greubel, May 21 2016: (Start)
E.g.f.: (1/2)*((5*x^2 + 9*x)*cosh(x) + (5*x^2 + 11*x -1)*sinh(x)).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)
a(n) = 4*A085787(n). - R. J. Mathar, Jun 03 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.

A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

Original entry on oeis.org

0, 2, 7, 15, 25, 37, 52, 70, 90, 112, 137, 165, 195, 227, 262, 300, 340, 382, 427, 475, 525, 577, 632, 690, 750, 812, 877, 945, 1015, 1087, 1162, 1240, 1320, 1402, 1487, 1575, 1665, 1757, 1852, 1950, 2050, 2152, 2257

Offset: 0

Paul Curtz, Feb 14 2020

a(-2)=2, a(-1)=0. 4 evens followed by 4 odds.
Last digit is only 0, 2, 5, 7.
The vertical spoke S-N of the pentagonal spiral for A004526.
                          37
                      37  25  25
                  36  24  15  15  26
              36  24  14   7   8  16  26
          35  23  14   7   2   3   8  16  27
      35  23  13   6   2   0   0   3   9  17  27
        34  22  13   6   1   1   4   9  17  28
          34  22  12   5   5   4  10  18  28
            33  21  12  11  11  10  18  29
              33  21  20  20  19  19  29
                32  32  31  31  30  30
Rank of multiples of 10: 0, 7, 8, 15, 16, ... = A047521. Compare to A154260 in the formula.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A004526, A033429, A062786, A168668, A135706, A147874, 2*A147875 (all in the spiral).

Cf. A005408, A047521, A154260

CoefficientList[Series[x (2 + x + 2 x^2)/((1 - x)^3*(1 + x^2)), {x, 0, 42}], x] (* Michael De Vlieger, Feb 14 2020 *)

concat(0, Vec(x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Feb 14 2020

a(-1-n) = a(n).
a(2*n) + a(1+2*n) = 2, 22, 62, ... = A273366(n).
Second differences give the sequence of period 4: repeat [3, 3, 2, 2].
From Colin Barker, Feb 14 2020: (Start)
G.f.: x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4.
(End)
Multiples of 10: 10*(0, 7, 9, 30, 34, ... = A154260).
4*a(n) = A087960(n) +5*n -1 +5*n^2. - R. J. Mathar, Feb 28 2020

A350760 Decimal expansion of Pi/(2sqrt(5)) tan(Pi/(2*sqrt(5))).

Original entry on oeis.org

5, 9, 4, 6, 7, 8, 1, 2, 3, 5, 3, 5, 2, 7, 8, 5, 1, 9, 1, 6, 8, 1, 1, 4, 2, 6, 9, 7, 6, 0, 5, 5, 4, 9, 3, 7, 6, 0, 3, 6, 3, 9, 4, 6, 9, 6, 1, 0, 2, 4, 2, 4, 3, 7, 9, 0, 5, 1, 1, 2, 5, 6, 8, 9, 5, 7, 9, 4, 5, 2, 5, 6, 3, 2, 6, 6, 1, 9, 0, 1, 5, 8, 8, 8, 4, 5, 2, 7, 3, 8, 9, 2, 6, 1, 2, 6, 0, 2, 5, 5, 2, 4, 3, 1, 0

Offset: 0

Amiram Eldar, Jan 14 2022

			0.59467812353527851916811426976055493760363946961024...

Robert Frontczak, Problem B-1267, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; The Zeta Riemann Function and the Cotangent Function, Solution to Problem B-1267 by Brian Bradie, ibid., Vol. 59, No. 2 (2021), pp. 179-180.

Cf. A000032, A000045, A062786, A244979.

RealDigits[(Pi/(2*Sqrt[5]))*Tan[Pi/(2*Sqrt[5])], 10, 100][[1]]

Equals Sum_{n>=1} zeta(2*n)*Fibonacci(2*n)/5^n (Frontczak, 2020).
Equals -1 + Sum_{n>=1} zeta(2*n)*Lucas(2*n)/5^n (Frontczak, 2020).
Equals Sum_{n>=0} 1/A273366(n).

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

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

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A273368 Numbers k such that 10*k+9 is a perfect square.

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

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A332495 a(n-2) = a(n-6) + 5*(1+2*n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

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A350760 Decimal expansion of Pi/(2*sqrt(5)) * tan(Pi/(2*sqrt(5))).

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A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

A350760 Decimal expansion of Pi/(2sqrt(5)) tan(Pi/(2*sqrt(5))).