A061292 -id:A061292 - cp's OEIS frontend

A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

0, 1, 2, 2, 3, 4, 10, 37, 366, 13532, 4952675, 67019597734, 331926286207224918, 22245566178948766568816183137, 7383888166355511098764350563784314022618210032

Offset: 1

Robert G. Wilson v, Jul 05 2000

Theorem 1.1 of Hare et al. (2010, 2011) involves a shifted version of this sequence and the Fibonacci sequence A000045. (The program by Alonso del Arte below does involve a shifted version of this sequence.) - Petros Hadjicostas, May 11 2019

G. C. Greubel, Table of n, a(n) for n = 1..21
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, arXiv:1006.2117 [math.OC], 2010-2011.
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, Advances in Mathematics 226 (2011), 4667-4701.

Cf. A001622, A061021, A061292, A072878, A072879, A072880, A178768.

I:=[0,1,2]; [n le 3 select I[n] else Self(n-1)*Self(n-2) - Self(n-3): n in [1..15]];  // G. C. Greubel, Mar 01 2018

a[1] = 0; a[2] = 1; a[3] = 2; a[n_] := a[n] = a[n - 1] a[n - 2] - a[n - 3]; Table[a[n], {n, 1, 15}] (* Alonso del Arte, Jan 31 2011 *)
nxt[{a_,b_,c_}]:={b,c,c*b-a}; NestList[nxt,{0,1,2},15][[;;,1]] (* Harvey P. Dale, Mar 23 2025 *)

It appears that lim_{n->infinity} log(a(n))/phi^n = 0.07743008049000107520747623421744398272089261907514..., where phi = (1 + sqrt(5))/2 is the golden ratio A001622. - Petros Hadjicostas and Jon E. Schoenfield, May 11 2019

Name clarified by Michel Marcus, May 10 2019

A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 19, 119, 65339, 67258454, 959259994615659593, 171965197021698738644442682357, 12959040525296547835480490169418622922155526267774117749963303914461

Offset: 1

Benoit Cloitre, Jul 28 2002

A variation of a Somos-4 sequence with a(n-2)^4 in place of a(n-2)^2.

Seiichi Manyama, Table of n, a(n) for n = 1..17
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics 43(3) (2015), pp. 589-633.
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics 28 (2002), pp. 119-144.
David Gale, The strange and surprising saga of the Somos sequences, Math. Intelligencer 13(1) (1991), pp. 40-42.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), pp. L171-L177.

Cf. A006720, A022405, A061292, A065918, A072878, A072879, A072880, A074394, A178768.

L[0]:=0; L[1]:=0; L[2]:=0; L[3]:=0; for n from 0 to 4000 do L[n+4]:=evalf(ln(1+exp(L[n+3]+L[n+1]-4*L[n+2]))+4*L[n+2]-L[n]): od: for n from 3990 to 4000 do print(evalf(ln(L[n+4])/(n+4))): od: #Note: L[n] is log(a[n]) # Andrew Hone, Nov 15 2005

nxt[{a_,b_,c_,d_}]:={b,c,d,(d*b+c^4)/a}; NestList[nxt,{1,1,1,1},15][[All,1]] (* Harvey P. Dale, Jun 01 2022 *)

Lim_{n->infinity} (log(log(a(n))))/n = log(2+sqrt(3))/2 = A065918/2 or about 0.658. - Andrew Hone, Nov 15 2005; corrected by Michel Marcus, May 12 2019
From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*x^n + d0/x^n)) if n is even,
         ceiling(e^(c1*x^n + d1/x^n)) if n is odd,
where
  x  = sqrt(2 + sqrt(3)) = (sqrt(2) + sqrt(6))/2
  c0 =   0.024915247166055931001426396817534982995670642690...
  c1 =   0.029604794868229453467890216788323427656809346011...
  d0 = -10.535089427608481105514469573411011428431309483956...
  d1 =  -2.856773870202800001336732759121362374871088274450...
(End)

Definition corrected by Matthew C. Russell, Apr 24 2012

A061021 a(n) = a(n-1)*a(n-2) - a(n-3) with a(0) = a(1) = a(2) = 3.

Original entry on oeis.org

3, 3, 3, 6, 15, 87, 1299, 112998, 146784315, 16586334025071, 2434613678231239448367, 40381315689150066251526220641224742, 98312903521778500654864668915856114278134197773017871243

Offset: 0

Stephen G Penrice, May 23 2001

Any three consecutive terms are a solution to the Diophantine equation x^2 + y^2 + z^2 = xyz.

Harry J. Smith, Table of n, a(n) for n = 0..17
Loren C. Larson, Solution to Problem Proposal 701, College Mathematics Journal 33 (2002), pp. 241-242.
Edward T. H. Wang, Problem Proposal 701, College Mathematics Journal 32 (2001), p. 211.

Cf. A022405, A061292, A072878, A072879, A072880, A074394, A178768.

a061021 n = a061021_list !! n
a061021_list = 3 : 3 : 3 : zipWith (-)
(tail $ zipWith (*) (tail a061021_list) a061021_list) a061021_list
-- Reinhard Zumkeller, Mar 25 2015

RecurrenceTable[{a[n] == a[n - 1] a[n - 2] - a[n - 3], a[0] == a[1] == a[2] == 3}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 21 2016 *)

for (n=0, 17, if (n>2, a=a1*a2 - a3; a3=a2; a2=a1; a1=a, if (n==0, a=a3=3, if (n==1, a=a2=3, a=a1=3))); write("b061021.txt", n, " ", a)) \\ Harry J. Smith, Jul 16 2009

From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*phi^n - c1/(-phi)^n))
where
  phi = (1 + sqrt(5))/2,
   c0 = 0.4004033011137849744572073756789830081726425559860...
   c1 = 0.2798639753144007577581523025628820390768226527315...
(End)

More terms from Erich Friedman, Jun 03 2001

Name clarified by Petros Hadjicostas, May 11 2019

A380462 a(n) is the number of positive solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxy*z such that x,y,z,w < e^n.

Original entry on oeis.org

1, 2, 2, 3, 4, 6, 6, 7, 10, 12, 16, 17, 18, 23, 25, 32, 35, 39, 43, 47, 55, 60, 64, 73, 76, 86, 94, 101, 111, 118, 132, 141, 150, 159, 168, 180, 192, 203, 220, 235, 247, 261, 275, 289, 302, 324, 340, 361, 376, 394, 413, 433, 454, 472, 498, 518, 536, 560, 584, 606, 632, 658, 684

Offset: 1

Bence Bernáth, Jun 22 2025

nonn

The trivial x=y=z=w=0 solution is not included. The solutions are generated by the Vieta jumping discussed in the Numberphile video.

			a(3)=2 because the solutions below e^3=20.085... are [2,2,2,2] and [2,2,2,6]. One quadruplet solution is counted only once.

Bence Bernáth, Table of n, a(n) for n = 1..650
Michael Magee and Brady Haran, A simple equation that behaves weirdly, Numberphile, Youtube video, 2025.

Cf. A061292.

import math
from collections import deque
def is_perfect_square(n):
    return (math.isqrt(n)) ** 2 == n
def generate_all_solutions(up_to_bound):
    solutions = set()
    visited = set()
    seed = (2, 2, 2, 2)
    queue = deque([seed])
    solutions.add(seed)
    while queue:
        quad = queue.popleft()
        for ii in range(4):
            rotated = list(quad[ii:] + quad[:ii])
            x, y, z, w = rotated
            product = y * z * w
            sumsq = y**2 + z**2 + w**2
            D = product**2 - 4 * sumsq
            if D < 0 or not is_perfect_square(D):
                continue
            sqrt_D = (math.isqrt(D))
            for sign in [+1, -1]:
                x_new = (product + sign * sqrt_D)
                if x_new % 2 != 0:
                    continue
                x_new //= 2
                if not (0 < x_new < up_to_bound):
                    continue
                new_quad = [x_new, y, z, w]
                if any(val >= up_to_bound for val in new_quad):
                    continue
                new_quad_sorted = tuple(sorted(new_quad))
                if new_quad_sorted not in visited:
                    solutions.add(new_quad_sorted)
                    queue.append(tuple(new_quad))
                    visited.add(new_quad_sorted)
    return sorted(solutions)
# Compute all solutions up to e^100
max_bound = math.exp(100)
all_solutions = generate_all_solutions(max_bound)
# Precompute max entry of each solution for fast thresholding
solution_maxes = [max(sol) for sol in all_solutions]
# Generate the sequence a(n) for n = 1 to 100
a_n = []
for n in range(1, 101):
    threshold = math.exp(n)
    count = sum(1 for m in solution_maxes if m < threshold)
    a_n.append(count)
print(a_n)

A385713 Numbers w + x + y + z where (w,x,y,z) are nonnegative solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxy*z.

Original entry on oeis.org

0, 8, 12, 32, 108, 292, 392, 1452, 3392, 3712, 5408, 11808, 20172, 34872, 40332, 50572, 75272, 162212, 280908, 480512, 518252, 595052, 700352, 1048352, 1639308, 3912492, 4599172, 4911172, 5725732, 6049292, 6508932, 7132608, 8279808, 9739812

Offset: 1

Bence Bernáth, Jul 07 2025

nonn

The trivial sum with x=y=z=w=0 is included. The solutions are generated by the Vieta jumping discussed in the Numberphile video. The algorithm finds all the quadruplet solutions within the radius of a 4D sphere. Then the quadruplets are sorted based on their sum in increasing order.

			8 is a term because (w,x,y,z) = (2,2,2,2) is the first nontrivial solution.
12 is a term because (2,2,2,6) is a solution.

Bence Bernáth, Table of n, a(n) for n = 1..10000
Michael Magee and Brady Haran, A simple equation that behaves weirdly, Numberphile, Youtube video, 2025.

Cf. A061292, A380462.

import math
def is_perfect_square(n):
    return (math.isqrt(n)) ** 2 == n
def generate_all_solutions(radius):
    solutions = set()
    visited = set()
    # Initial known solution
    seed = (2, 2, 2, 2)
    queue = [seed]
    solutions.add(seed)
    while queue:
        quad = queue.pop(0)
        for ii in range(4):
            # Cyclically rotate variables: solve for the ii-th one
            rotated = list(quad[ii:] + quad[:ii])
            x, y, z, w = rotated
            # Use Vieta's formula: x^2 - (yzw)x + (y² + z² + w²) = 0
            product = y * z * w
            sumsq = y**2 + z**2 + w**2
            D = product**2 - 4 * sumsq
            if D < 0 or not is_perfect_square(D):
                continue
            sqrt_D = (math.isqrt(D))
            for sign in [+1, -1]:
                x_new = (product + sign * sqrt_D)
                if x_new % 2 != 0:
                    continue
                x_new //= 2
                if not (0 < x_new < radius):
                    continue
                new_quad = [x_new, y, z, w]
                if any(val >= radius for val in new_quad):
                    continue
                new_quad_sorted = tuple(sorted(new_quad))
                if new_quad_sorted not in visited:
                    solutions.add(new_quad_sorted)
                    queue.append(tuple(new_quad))
                    visited.add(new_quad_sorted)
    # Return solutions sorted by sum of elements
    return sorted([list(sol) for sol in solutions], key=lambda tup: sum(tup))
print([sum(sol) for sol in generate_all_solutions(10000000)])

A146974 Numbers k such that there is no nonzero integer solution for the Diophantine equation x_1^2 + x_2^2 + ... + x_k^2 = x_1x_2...*x_k.

Original entry on oeis.org

2, 6, 9, 11, 12, 15, 16, 18, 20, 21, 24, 29, 32, 33, 36, 41, 42, 45, 48, 50, 51, 56, 57, 60, 66, 72, 76, 77, 81, 82, 84, 90, 96, 99, 101, 102, 105, 106, 108, 113, 114, 120, 122, 123, 126, 132, 136, 137, 140, 141, 144, 146, 156, 162, 164, 168, 171, 176, 177, 180

Offset: 1

Zhao Hui Du, Nov 04 2008

nonn

In the link, a C++ program calling the GMP library is provided to solve such kinds of equations.

If the equation has nonzero solutions and k > 2, then there is a positive integer solution (x_1, x_2, ..., x_k) such that 3 <= x_1*x_2*...*x_(k-2) <= n and x_(k-1) <= sqrt((x_1^2 + x_2^2 + ... + x_(k-2)^2)/(x_1*x_2*...*x_(k-2) - 2)).

			For k=3, there are nonzero integer solutions 3^2 + 3^2 + 3^2 = 3*3*3; 3^2 + 6^2 + 15^2 = 3*6*15.
For k=4, there are nonzero integer solutions 2^2 + 2^2 + 2^2 + 2^2 = 2*2*2*2; 2^2 + 6^2 + 22^2 + 262^2 = 2*6*22*262.
However, for k=2, there is no nonzero integer solution for the equation a^2 + b^2 = a*b.

Link for the problem to be solved [broken link?]
A Chinese webpage where the problem is raised

Cf. A002559, A061292.

is(w, k) = my(p, s); for(x=w[k], sqrtint((s=sum(i=1, k, w[i]^2))\p=vecprod(w)-2), if(issquare((p^2+4*p)*x^2-4*s), return(1)))
lista(nn) = my(b, t, v=List([])); for(n=2, nn, b=1; for(i=1, #v, if(n%vecprod(v[i])==0&&v[i][1]<=t=n\vecprod(v[i]), listput(v, concat(t, v[i])))); listput(v, [n]); for(m=2, #v, if(is(concat(vector(n-2-#v[m], i, 1), v[m]), n-2), b=0; break)); if(b, print1(n, ", "))) \\ Jinyuan Wang, Oct 04 2021

Edited by Jon E. Schoenfield, Aug 09 2015

More terms from Jinyuan Wang, Oct 04 2021

A385121 a(n+1) = 12*a(n) - a(n-1), a(0) = a(1) = 2, a(n) = a(1-n).

Original entry on oeis.org

2, 2, 22, 262, 3122, 37202, 443302, 5282422, 62945762, 750066722, 8937854902, 106504192102, 1269112450322, 15122845211762, 180205030090822, 2147337515878102, 25587845160446402, 304906804409478722, 3633293807753298262, 43294618888630100422

Offset: 0

Michael Somos, Jun 18 2025

If x = 2, y = 6, z = a(n), w = a(n+1), then x^2+y^2+z^2+w^2 = x*y*z*w.

			G.f. = 2 + 2*x + 22*x^2 + 262*x^3 + 3122*x^4 + 37202*x^5 + ...

Index entries for linear recurrences with constant coefficients, signature (12,-1).

Cf. A061292, A077417.

a[ n_] := Which[n<1, a[1-n], n==1, 2, True, 12*a[n-1] - a[n-2]];

{a(n) = if(n<1, a(1-n), n==1, 2, 12*a(n-1) - a(n-2))};

G.f.: (2 - 22*x)/(1 - 12*x + x^2).
0 = 40 + a(n)^2 - 12*a(n)*a(n+1) + a(n+1)^2 for all n in Z.
a(n) = 2 * A077417(n-1).
E.g.f.: 2*exp(6*x)*(7*cosh(sqrt(35)*x) - sqrt(35)*sinh(sqrt(35)*x))/7. - Stefano Spezia, Aug 29 2025

cp's OEIS Frontend

A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

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A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

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A061021 a(n) = a(n-1)*a(n-2) - a(n-3) with a(0) = a(1) = a(2) = 3.

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A380462 a(n) is the number of positive solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = w*x*y*z such that x,y,z,w < e^n.

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A385713 Numbers w + x + y + z where (w,x,y,z) are nonnegative solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = w*x*y*z.

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A146974 Numbers k such that there is no nonzero integer solution for the Diophantine equation x_1^2 + x_2^2 + ... + x_k^2 = x_1*x_2*...*x_k.

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A385121 a(n+1) = 12*a(n) - a(n-1), a(0) = a(1) = 2, a(n) = a(1-n).

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A380462 a(n) is the number of positive solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxy*z such that x,y,z,w < e^n.

A385713 Numbers w + x + y + z where (w,x,y,z) are nonnegative solutions to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxy*z.

A146974 Numbers k such that there is no nonzero integer solution for the Diophantine equation x_1^2 + x_2^2 + ... + x_k^2 = x_1x_2...*x_k.