A080663 -id:A080663 - cp's OEIS frontend

A157872 a(n) = 9*n^2 - 3.

6, 33, 78, 141, 222, 321, 438, 573, 726, 897, 1086, 1293, 1518, 1761, 2022, 2301, 2598, 2913, 3246, 3597, 3966, 4353, 4758, 5181, 5622, 6081, 6558, 7053, 7566, 8097, 8646, 9213, 9798, 10401, 11022, 11661, 12318, 12993, 13686, 14397, 15126, 15873, 16638, 17421, 18222

Offset: 1

Vincenzo Librandi, Mar 08 2009

The identity (6*n^2 - 1)^2 - (9*n^2 - 3)*(2*n)^2 = 1 can be written as A140811(n-1)^2 - a(n)*A005843(n+1)^2 = 1. - Vincenzo Librandi, Feb 05 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1. [broken link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A080663, A140811.

I:=[6, 33, 78]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 05 2012

LinearRecurrence[{3, -3, 1}, {6, 33, 78}, 50] (* Vincenzo Librandi, Feb 05 2012 *)
9*Range[50]^2-3 (* Harvey P. Dale, Aug 04 2024 *)

for(n=1, 40, print1(9*n^2 - 3", ")); \\ Vincenzo Librandi, Feb 05 2012

From Vincenzo Librandi, Feb 05 2012: (Start)
G.f.: -3*x*(2 + 5*x - x^2)/(x - 1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = a(n-1) + 18*n - 9. (End)
From Amiram Eldar, May 28 2022: (Start)
a(n) = 3*A080663(n).
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(3)*cot(Pi/sqrt(3))))/6.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(3))*csc(Pi/sqrt(3)) - 1)/6. (End)
E.g.f.: 3*(exp(x)*(3*x^2 + 3*x - 1) + 1). - Elmo R. Oliveira, Jan 25 2025

A271675 Numbers m such that 3*m + 4 is a square.

Original entry on oeis.org

0, 4, 7, 15, 20, 32, 39, 55, 64, 84, 95, 119, 132, 160, 175, 207, 224, 260, 279, 319, 340, 384, 407, 455, 480, 532, 559, 615, 644, 704, 735, 799, 832, 900, 935, 1007, 1044, 1120, 1159, 1239, 1280, 1364, 1407, 1495, 1540, 1632, 1679, 1775, 1824, 1924, 1975, 2079, 2132, 2240, 2295, 2407

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2016

7 is the unique prime in this sequence. If m is in this sequence, then 3*m + 4 = k^2 for k is nonzero integer, that is, m = (k^2 - 4)/3 = (k-2)*(k+2)/3. So m can be only prime if one of divisors is prime and another one is 1. Otherwise there should be more than 1 prime divisors, that is n must be composite. - Altug Alkan, Apr 12 2016
From Ray Chandler, Apr 12 2016: (Start)
Square roots of resulting squares gives A001651 (with a different starting point).
Sequence is the union of (positive terms) in A140676 and A270710. (End)
The sequence terms are the exponents in the expansion of Sum_{n >= 0} q^n*(1 - q)*(1 - q^3)*...*(1 - q^(2*n+1)) = 1 - q^4 - q^7 + q^15 + q^20 - q^32 - q^50 + + - - .... - Peter Bala, Dec 19 2024

			a(4) = 32 because 3*32 + 4 = 100 = 10*10.

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001651, A140676, A270710.

Cf. numbers n such that 3*n + k is a square: A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), this sequence (k=4), A100536 (k=6).

Magma
```
[n: n in [0..4000] | IsSquare(3*n+4)];
```

Select[Range[0,2500], IntegerQ@ Sqrt[3 # + 4] &] (* Michael De Vlieger, Apr 12 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{0,4,7,15,20},60] (* Harvey P. Dale, Dec 09 2016 *)

from gmpy2 import is_square
for n in range(0,10**5):
    if(is_square(3*n+4)):print(n)
# Soumil Mandal, Apr 12 2016

O.g.f.: x^2*(4 + 3*x - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: 1 + (1 - 2*x)*exp(-x)/8 - 3*(3 - 4*x - 2*x^2)*exp(x)/8.
a(n) = A001082(n+1) - 1 = (6*n*(n+1) + (2*n + 1)*(-1)^n - 1)/8 - 1. Therefore: a(2*k+1) = k*(3*k+4), a(2*k) = (k+1)*(3*k-1).
Sum_{n>=2} 1/a(n) = 19/16 - Pi/(4*sqrt(3)). - Amiram Eldar, Jul 26 2024

Edited and extended by Bruno Berselli, Apr 12 2016

A378066 Array read by ascending antidiagonals: A(n, k) = (-2n)^k Euler(k, (n - 1)/(2*n)) for n >= 1 and A(0, k) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -3, -2, 1, 1, 1, -8, -11, 0, 1, 1, 1, -15, -26, 57, 16, 1, 1, 1, -24, -47, 352, 361, 0, 1, 1, 1, -35, -74, 1185, 1936, -2763, -272, 1, 1, 1, -48, -107, 2976, 6241, -38528, -24611, 0, 1

Offset: 0

Peter Luschny, Nov 15 2024

This is the counterpart of A377666, where A(1, n) are the secant numbers A122045(n). Here A(1, n) are the tangent numbers A155585(n).

			Array starts:
  [0]  1, 1,   1,    1,     1,     1,        1, ...  A000012
  [1]  1, 1,   0,   -2,     0,    16,        0, ...  A155585
  [2]  1, 1,  -3,  -11,    57,   361,    -2763, ...  A188458
  [3]  1, 1,  -8,  -26,   352,  1936,   -38528, ...  A000810
  [4]  1, 1, -15,  -47,  1185,  6241,  -230895, ...  A000813
  [5]  1, 1, -24,  -74,  2976, 15376,  -906624, ...  A378065
  [6]  1, 1, -35, -107,  6265, 32041, -2749355, ...
  [7]  1, 1, -48, -146, 11712, 59536, -6997248, ...

Rows: A000012, A155585, A188458, A000810, A000813, A378065.
Columns: A005563 (k=2), A080663 (k=3), A378064 (k=4).
Cf. A378063 (main diagonal), A377666 (secant), A081658 (column generating polynomials).

A := (n, k) -> ifelse(n = 0, 1, (-2*n)^k * euler(k, (n - 1) / (2*n))):
for n from 0 to 7 do seq(A(n, k), k = 0..9) od; # row by row
# Alternative:
A := proc(n, k) local j; add(binomial(k, j)*euler(j, 1/2)*(-2*n)^j, j = 0..k) end: seq(seq(A(n - k, k), k = 0..n), n = 0..10);
# Using generating functions:
egf := n -> exp(x)/cosh(n*x): ser := n -> series(egf(n), x, 14):
row := n -> local k; seq(k!*coeff(ser(n), x, k), k = 0..7):
seq(lprint(row(n)), n = 0..7);

A(n, k) = k! * [x^k] exp(x)/cosh(n*x).

A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 1/2) *(-2*n)^j.

A271723 Numbers k such that 3*k - 8 is a square.

Original entry on oeis.org

3, 4, 8, 11, 19, 24, 36, 43, 59, 68, 88, 99, 123, 136, 164, 179, 211, 228, 264, 283, 323, 344, 388, 411, 459, 484, 536, 563, 619, 648, 708, 739, 803, 836, 904, 939, 1011, 1048, 1124, 1163, 1243, 1284, 1368, 1411, 1499, 1544, 1636, 1683, 1779, 1828, 1928, 1979, 2083, 2136, 2244, 2299

Offset: 1

Juri-Stepan Gerasimov, Apr 13 2016

Square roots of resulting squares gives A001651. - Ray Chandler, Apr 14 2016

			a(1) = 3 because 3*3 - 8 = 1^2.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001651.

Cf. numbers n such that 3*n + k is a square: this sequence (k=-8), A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6), A271741 (k=7), A067725 (k=9).

Magma
```
[n: n in [1..2400] | IsSquare(3*n-8)];
```

seq(seq(((3*m+k)^2+8)/3, k=1..2),m=0..50); # Robert Israel, Dec 05 2016

Select[Range@ 2400, IntegerQ@ Sqrt[3 # - 8] &] (* Bruno Berselli, Apr 14 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{3,4,8,11,19},60] (* Harvey P. Dale, Oct 02 2020 *)

from gmpy2 import is_square
[n for n in range(3000) if is_square(3*n-8)] # Bruno Berselli, Dec 05 2016

[(6*(n-1)*n-(2*n-1)*(-1)**n+23)/8 for n in range(1, 60)] # Bruno Berselli, Dec 05 2016

From Ilya Gutkovskiy, Apr 13 2016: (Start)
G.f.: x*(3 + x - 2*x^2 + x^3 + 3*x^4)/((1 - x)^3*(1 + x)^2).
a(n) = (6*(n - 1)*n - (2*n - 1)*(-1)^n + 23)/8. (End)

A173202 Solutions y of the Mordell equation y^2 = x^3 - 3a^2 + 1 for a = 0,1,2, ... (solutions x are given by the sequence A000466).

Original entry on oeis.org

0, 5, 58, 207, 500, 985, 1710, 2723, 4072, 5805, 7970, 10615, 13788, 17537, 21910, 26955, 32720, 39253, 46602, 54815, 63940, 74025, 85118, 97267, 110520, 124925, 140530, 157383, 175532, 195025, 215910, 238235, 262048, 287397, 314330, 342895

Offset: 1

Michel Lagneau, Feb 12 2010

For many values of k for the equation y^2 = x^3 + k, all the solutions are known. For example, we have solutions for k=-2: (x,y) = (3,-5) and (3,5). A complete resolution for all integers k is unknown. Theorem: Let k be < -1, free of square factors, with k == 2 or 3 (mod 4). Suppose that the number of classes h(Q(sqrt(k))) is not divisible by 3. Then the equation y^2 = x^3 + k admits integer solutions if and only if k = 1 - 3a^2 or 1 - 3a^2 where a is an integer. In this case, the solutions are x = a^2 - k, y = a(a^2 + 3k) or -a(a^2 + 3k) (the first reference gives the proof of this theorem). With k = -1 - 3a^2, we obtain the solutions x = 4a^2 + 1, y = a(8a^2 + 3) or -a(8a^2 + 3). For the case k = 1 - 3a^2, we obtain the solution x = 4a^2 - 1 given by the sequence A000466.

			With a=3, x = 35 and y = 207, and then 207^2 = 35^2 - 26.

T. Apostol, Introduction to Analytic Number Theory, Springer, 1976
D. Duverney, Theorie des nombres (2e edition), Dunod, 2007, p.151

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
W. J. Ellison, F. Ellison, J. Pesek, C. E. Stall & D. S. Stall, The diophantine equation y^2 + k = x^3, J. Number Theory 4 (1972), 107-117.
Helmut Richter, Solutions of Mordell's equation y^2 = x^3 + k (solutions for 0
School of Mathematics and Statistics, University of St Andrews, Louis Joel Mordell.
Eric Weisstein's World of Mathematics, Mordell Curve.
D. J. Wright, Mordell's Equation.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Diophantine equations: see also Pellian equation: (A081233, A081234), (A081231, A082394), (A081232, A082393); Mordell equation: A053755, A173200; Diophantine equations: A006452, A006451, A006454.

I:=[0, 5, 58, 207]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 02 2012

for a from 0 to 100 do : z := evalf(a*(8*a^2 - 3)) : print (z) :od :

CoefficientList[Series[x*(5+38*x+5*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jul 02 2012 *)
CoefficientList[Series[E^x (5 x + 24 x^2 + 8 x^3), {x, 0, 40}], x]*Table[n!, {n, 0, 40}] (* Stefano Spezia, Dec 04 2018 *)

y = a*(8*a^2 - 3).
a(n) = sqrt(A000466(n)^3 - A080663(n)). - Artur Jasinski, Nov 26 2011
From Colin Barker, Apr 26 2012: (Start)
a(n) = 8*n^3 - 24*n^2 + 21*n - 5.
G.f.: x^2*(5 + 38*x + 5*x^2)/(1 - x)^4. (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jul 02 2012
E.g.f.: exp(x)*(5*x + 24*x^2 + 8*x^3). - Stefano Spezia, Dec 04 2018

A239064 Positive integers whose square can be expressed as kx^2+(k+1)x+(k+2) with k and x positive integers.

Original entry on oeis.org

3, 5, 6, 9, 10, 12, 15, 16, 18, 19, 21, 23, 24, 26, 27, 30, 33, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 51, 52, 54, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 78, 79, 81, 82, 83, 84, 86, 87, 89, 90, 92, 93, 96, 99, 100, 102, 103, 105, 107, 108, 110, 111, 114, 117, 120, 121, 123, 124, 126, 128, 129, 130, 131, 132, 134, 135, 138, 141, 142, 144, 145, 147, 149, 150, 152, 153, 156, 159, 162, 163, 165, 166, 168, 170, 171, 173, 174, 176, 177, 180, 183, 184, 186, 187, 189, 191, 192, 194, 195, 196, 198, 201, 204

Offset: 1

Carmine Suriano, Mar 28 2014

nonn

Some can be expressed in more than one way. E.g., a(46) = 81 for x = 1, k = 26 and for x = 2 and k = 11.
x=1 for k=3*m^2-1 that is A080663, with a = 3*m.
From Wolfdieter Lang, Apr 26 2014: (Start)
The solution of this problem can be found as follows. Consider k*x^2 + (k+1)*x + (k+2) - a^2 = 0. Solve for x (positive) as function of k: x = (-(k+1) + sqrt((2*a)^2*k - c(k)))/(2*k), where c(k) = 3*k(k+2) - 1 is the sequence [8, 23, 44, 71, 104, 143, 188, 239, 296, 359,...]. A necessary condition for a solution is (2*a)^2*k - c(k) = b^2, or k*X^2 - Y^2 = c(k), with X = 2*a and Y = b. This is a binary indefinite quadratic form with discriminant D = (2^2)*k > 0. If k is a square then there is no solution with even X because (K*X)^2 - Y^2 = c(K^2) has only the solution with K=1, X=3, Y=1. For k not a square there are either no solutions for given k (e.g., k = 5, 6, 10, ...) or countable infinitely many ones of the representation k*X^2 - Y^2 = c(k). From the solutions one has to pick first the ones with even X (a = X/2) and then to test whether 2*k divides Y - (k+1). The solution for x is then x = (Y - (k+1))/(2*k).
For representations of integers by indefinite binary quadratic forms see, e.g., the D. A. Buell reference, and also a W. Lang link (with further references) with an on-line program.
(End)

			a(10)=19 from 19^2=361=51*2^2+(51+1)*2+(51+2).
From _Wolfdieter Lang_, Apr 21 2014: (Start)
k = 2: There are two proper solutions of 2*X^2 - Y^2 = 23, namely [4, 3], [6, 7]. Both generate infinitely many new solutions, all with even X, namely [4, 3], [18, 25], [104, 147], [606, 857],[3532, 4995], [20586, 29113], ... and [6, 7], [32, 45], [186, 263], [1084, 1533],[6318, 8935], [36824, 52077], ... . Only every other solution has 2*k = 4 dividing Y - (2+1) = Y-3, giving the positive solutions for (a=X/2, b=Y; x), starting with the second proper solution and then alternating between the two sets of solutions (3, 7; 1), (52, 147; 36), (93, 263; 65), (1766, 4995; 1248), (3159, 8935; 2233), ... . Thus the positive x solutions for k = 2 are 1, 36, 65, 1248, 2233, ..., with a = 3, 52, 93, 1766, 3159, ... .
k = 3: the positive solutions for x are 2, 5, 38, 79, 538, 1109, ..., with a = 5, 10, 67, 138, 933, 1922, ..., coming from the even X solutions of 3*X^2 - Y^2 = 44, [4, 2], [10, 16], [36, 62], [134, 232], [500, 866], [1866, 3232], ... and [6, 8], [20, 34], [74, 128], [276, 478], [1030, 1784], [3844, 6658], ... . Then 2*k = 6 has to divide Y - 4, leaving every other of these solutions with (a = X/2, b=Y; x) given by (5, 16; 2), (10, 34; 5), (67, 232; 38), (138, 478; 79), (933, 3232; 538), (1922, 6658; 1109), ... .
k = 5: there are no solutions of 5*X^2 - Y2 = 104.
k = 6: there are no solutions of 6*X^2 - Y2 = 143.
(End)

D. A. Buell, Binary Quadratic Forms. Springer-Verlag, NY, 1989.

Wolfdieter Lang, Binary Quadratic Forms (indefinite case).

for k=1 to z
  for x=1 to z
    a=k*x*x+(k+1)*x+(k+2)
    if sqr(a)-int(sqr(a))=0 then
      j=j+1
      a_n(j)=sqr(a)
    endif
   next x
next k

Increasingly ordered set {a from the positive integers: a^2 = k*x^2 + (k+1)*x + (k+2), with k and x positive integers}. - Wolfdieter Lang, Apr 21 2014

Edited, name specified, comment changed, reference and link added by Wolfdieter Lang, Apr 21 2014

A292717 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. -log(1 - x)/(1 - x)^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 5, 11, 6, 0, 1, 7, 26, 50, 24, 0, 1, 9, 47, 154, 274, 120, 0, 1, 11, 74, 342, 1044, 1764, 720, 0, 1, 13, 107, 638, 2754, 8028, 13068, 5040, 0, 1, 15, 146, 1066, 5944, 24552, 69264, 109584, 40320, 0, 1, 17, 191, 1650, 11274, 60216, 241128, 663696, 1026576, 362880

Offset: 0

Ilya Gutkovskiy, Sep 21 2017

			E.g.f. of column k: A_k(x) = x/1! + (2*k + 1)*x^2/2! + (3*k^2 + 6*k + 2)*x^3/3! + 2*(2*k^3 + 9*k^2 + 11*k + 3)*x^4/4! + ...
Square array begins:
   0,    0,     0,     0,     0,      0,  ...
   1,    1,     1,     1,     1,      1,  ...
   1,    3,     5,     7,     9,     11,  ...
   2,   11,    26,    47,    74,    107,  ...
   6,   50,   154,   342,   638,   1066,  ...
  24,  274,  1044,  2754,  5944,  11274,  ...

Columns k=0..11 give A104150, A000254, A001705, A001711 (with offset 1), A001716 (with offset 1),  A001721 (with offset 1), A051524, A051545, A051560, A051562, A051564, A203147.
Rows n=0..3 give  A000004, A000012, A005408, A080663 (with offset 0).
Main diagonal gives A058806.

Table[Function[k, n! SeriesCoefficient[-Log[1 - x]/(1 - x)^k, {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

E.g.f. of column k: -log(1 - x)/(1 - x)^k.

cp's OEIS Frontend

A157872 a(n) = 9*n^2 - 3.

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A271675 Numbers m such that 3*m + 4 is a square.

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A378066 Array read by ascending antidiagonals: A(n, k) = (-2*n)^k * Euler(k, (n - 1)/(2*n)) for n >= 1 and A(0, k) = 1.

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A271723 Numbers k such that 3*k - 8 is a square.

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A173202 Solutions y of the Mordell equation y^2 = x^3 - 3a^2 + 1 for a = 0,1,2, ... (solutions x are given by the sequence A000466).

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A239064 Positive integers whose square can be expressed as k*x^2+(k+1)*x+(k+2) with k and x positive integers.

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A292717 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. -log(1 - x)/(1 - x)^k.

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A378066 Array read by ascending antidiagonals: A(n, k) = (-2n)^k Euler(k, (n - 1)/(2*n)) for n >= 1 and A(0, k) = 1.

A239064 Positive integers whose square can be expressed as kx^2+(k+1)x+(k+2) with k and x positive integers.