A089928 -id:A089928 - cp's OEIS frontend

A084703 Squares k such that 2*k+1 is also a square.

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

A220185 Numbers n such that n^2 + n(n+1) is an oblong number (A002378).

Original entry on oeis.org

0, 10, 348, 11830, 401880, 13652098, 463769460, 15754509550, 535189555248, 18180690368890, 617608282987020, 20980500931189798, 712719423377466120, 24211479893902658290, 822477596969312915748, 27940026817062736477150, 949138434183163727307360

Offset: 1

Alex Ratushnyak, Apr 12 2013

Also numbers n such that the sum of the hexagonal numbers H(n) and H(n+1) is equal to m^2 + (m+1)^2 for some m. - Colin Barker, Dec 10 2014

Also nonnegative integers x in the solutions to 4*x^2-2*y^2+2*x-2*y = 0, the corresponding values of y being A251867. - Colin Barker, Dec 10 2014

			a(3) = A089928(7) = 348.

Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002378, A014105 (n^2 + n(n+1)), A029546, A084703 (numbers n such that n^2 + n(n+1) is a square).

Cf. A000290, A000384, A251867.

#include 
#include 
typedef unsigned long long U64;
U64 rootPronic(U64 a) {
    U64 sr = 1L<<31, s, b;
    if (a < sr*(sr+1)) {
          sr>>=1;
          while (a < sr*(sr+1))  sr>>=1;
    }
    for (b = sr>>1; b; b>>=1) {
            s = sr+b;
            if (a >= s*(s+1))  sr = s;
    }
    return sr;
}
int main() {
  U64 a, n, r, t;
  for (n=0; n < (1L<<31); n++) {
    a = (n*(n+1)) + n*n;
    t = rootPronic(a);
    if (a == t*(t+1))  printf("%llu\n", n);
  }
}

[Floor(((1+Sqrt(2))^(4*n-3)+(1-Sqrt(2))^(4*n-3)-2)/8): n in [1..20]]; // Vincenzo Librandi, Sep 08 2015

f:= gfun:-rectoproc({a(n)=35*(a(n-1)-a(n-2))+a(n-3),a(1)=0,a(2)=10,a(3)=348},a(n),remember):
map(f, [$1..50]); # Robert Israel, Sep 06 2015

LinearRecurrence[{35, - 35, 1}, {0, 10, 348}, 20] (* Vincenzo Librandi, Sep 06 2015 *)

concat(0, Vec(2*x^2*(5-x)/((1-x)*(1-34*x+x^2))+O(x^100))) \\ Colin Barker, Dec 10 2014

For n>1, a(n) = A089928(n*4-5).
From Bruno Berselli, Apr 12 2013: (Start)
G.f.: 2*x^2*(5-x)/((1-x)*(1-34*x+x^2)).
a(n) = ((1+sqrt(2))^(4n-3)+(1-sqrt(2))^(4n-3)-2)/8.
a(n+2) = 10*A029546(n)-2*A029546(n-1). (End)
a(n) = 35*a(n-1)-35*a(n-2)+a(n-3). - Colin Barker, Dec 10 2014
a(n) = A251867(n) - A001542(n-1)^2. - Alexander Samokrutov, Sep 05 2015

More terms from Bruno Berselli, Apr 12 2013

A215099 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is prime.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 18, 24, 25, 29, 34, 38, 39, 41, 44, 48, 53, 55, 56, 58, 71, 73, 78, 84, 85, 89, 94, 102, 103, 109, 120, 124, 131, 133, 138, 144, 145, 149, 162, 164, 169, 173, 178, 180, 181, 187, 192, 196, 197, 201

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

For n>0 and (n mod 4)<2, a(n) is odd.
Same definition, but k+a(n-2) is a
Fibonacci number: A006498 except first two terms,
Lucas number: A000045 except first two terms,
Pell number: A089928(n-1),
Jacobsthal number: A215095,
factorial: A215096,
square: A194274,
cube: A215097,
triangular number: A011848(n+2),
oblong number: A215098.
Example of a related sequence definition: a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a cube.

Iain Fox, Table of n, a(n) for n = 0..10000

Cf. A062042: a(1) = 2, a(n) = least k>a(n-1) such that k+a(n-1) is a prime.

Cf. A073627, A073628.

first(n) = my(res = vector(n, i, i-1), k); for(x=3, n, k=res[x-1]+1; while(!isprime(k+res[x-2]), k++); res[x]=k); res \\ Iain Fox, Apr 22 2019 (corrected by Iain Fox, Apr 25 2019)

from sympy import prime
prpr = 0
prev = 1
for n in range(77):
    print(prpr, end=', ')
    b = c = 0
    while c<=prev:
        c = prime(b+1) - prpr
        b+=1
    prpr = prev
    prev = c

A215095 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a Jacobsthal number.

Original entry on oeis.org

0, 1, 3, 4, 8, 17, 35, 68, 136, 273, 547, 1092, 2184, 4369, 8739, 17476, 34952, 69905, 139811, 279620, 559240, 1118481, 2236963, 4473924, 8947848, 17895697, 35791395, 71582788, 143165576, 286331153, 572662307, 1145324612, 2290649224, 4581298449, 9162596899

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

Same definition, but k+a(n-2) is a
Fibonacci number: A006498 except first two terms,
Lucas number: A000045 except first two terms,
Pell number: A089928(n-1),
factorial: A215096,
square: A194274,
cube: A215097,
triangular number: A011848(n+2),
oblong number: A215098,
prime number: A215099.
Example of a related sequence definition: a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a cube.

Cf. A001045, A006498, A089928, A000045.

prpr = 0
prev = 1
jac = [0]*10000
for n in range(10000):
    jac[n] = prpr
    curr = prpr*2 + prev
    prpr = prev
    prev = curr
prpr, prev = 0, 1
for n in range(1, 44):
    print(prpr, end=', ')
    b = c = 0
    while c<=prev:
        c = jac[b] - prpr
        b+=1
    prpr = prev
    prev = c

Conjecture: G.f. (x+2*x^2)/(1-x-x^2-x^3-2*x^4). - David Scambler, Aug 06 2012

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A084703 Squares k such that 2*k+1 is also a square.

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A220185 Numbers n such that n^2 + n(n+1) is an oblong number (A002378).

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A215099 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is prime.

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A215095 a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a Jacobsthal number.

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