A050793 -id:A050793 - cp's OEIS frontend

A050794 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Arrange solutions by increasing values of z (see A050791). Sequence gives values of x^3 + y^3 = z^3 + 1. For corresponding values of x, y, z see A050792, A050793, A050791 respectively.

1729, 1092728, 3375001, 15438250, 121287376, 401947273, 3680797185, 6352182209, 7856862273, 12422690497, 73244501505, 145697644729, 179406144001, 648787169394, 938601300672, 985966166178, 1594232306569

Offset: 1

Patrick De Geest, Sep 15 1999

nonn

Note that a(1)=1729 is the Hardy-Ramanujan number. - Omar E. Pol, Jan 28 2009

			577^3 + 2304^3 = 2316^3 + 1 = 12422690497.

Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.
David Wells, "Curious and Interesting Numbers", Revised Ed. 1997, Penguin Books, On number "1729", p. 153.

Uwe Hollerbach and David Rabahy, Table of n, a(n) for n = 1..368, a(n) for n = 75..368 by David Rabahy, Oct 13 2015.
Shyam Sunder Gupta, On Some Special Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 22, 527-565.
Eric Weisstein's World of Mathematics, Diophantine Equation - 3rd Powers.

Cf. A050791, A050792, A050793, A259753.

Extended through 1594232306569 by Jud McCranie, Dec 25 2000

A050791 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Sequence gives values of z in monotonic increasing order.

Original entry on oeis.org

12, 103, 150, 249, 495, 738, 1544, 1852, 1988, 2316, 4184, 5262, 5640, 8657, 9791, 9953, 11682, 14258, 21279, 21630, 31615, 36620, 36888, 38599, 38823, 40362, 41485, 47584, 57978, 59076, 63086, 73967, 79273, 83711, 83802, 86166, 90030

Offset: 1

Patrick De Geest, Sep 15 1999

Numbers n such that n^3+1 is expressible as the sum of two nonzero cubes (both greater than 1).
Values of z associated with A050794.
Sequence is infinite. One subsequence is (from x = 1 + 9 m^3, y = 9 m^4, z = 3*m*(3*m^3 + 1), x^3 + y^3 = z^3 + 1): z(m) = 3*m*(3*m^3 + 1) = {12, 150, 738, 2316, 5640, 11682, 21630, 36888, 59076, 90030, ...} = a (1, 3, 6, 10, 13, 17, 20, 23, 30, 37, ...). - Zak Seidov, Sep 16 2013
Numbers n such that n^3+1 is a member of A001235. - Altug Alkan, May 09 2016

			12 is a term because 10^3 + 9^3 = 12^3 + 1 (= 1729).
2316 is in the sequence because 577^3 + 2304^3 = 2316^3 + 1.

Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.

Lewis Mammel, Table of n, a(n) for n = 1..368
Noam Elkies, Rational points near curves and small nonzero |x^3-y^2| via lattice reduction, arXiv:math/0005139 [math.NT], 2000.
S. Ramanujan, Question 681, J. Ind. Math. Soc.
Eric Weisstein's World of Mathematics, Diophantine Equation - 3rd Powers

Cf. A050792, A050793, A050794, A050787, A229383.

r[z_] := Reduce[ 1 < x < y < z && x^3 + y^3 == z^3 + 1, {x, y}, Integers]; z = 4; A050791 = {}; While[z < 10^4, If[r[z] =!= False, Print[z]; AppendTo[A050791, z]]; z++]; A050791 (* Jean-François Alcover, Dec 27 2011 *)

is(n)=if(n<2,return(0));my(c3=n^3);for(a=2,sqrtnint(c3-5,3),if(ispower(c3-1-a^3,3),return(1)));0 \\ Charles R Greathouse IV, Oct 26 2014

T=thueinit('x^3+1); is(n)=n>8&&#select(v->min(v[1], v[2])>1, thue(T, n^3+1))>0 \\ Charles R Greathouse IV, Oct 26 2014

More terms from Michel ten Voorde
Extended through 47584 by Jud McCranie, Dec 25 2000
More terms from Don Reble, Nov 29 2001
Edited by N. J. A. Sloane, May 08 2007

A050792 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Arrange solutions by increasing values of z (see A050791). Sequence gives values of x.

Original entry on oeis.org

9, 64, 73, 135, 334, 244, 368, 1033, 1010, 577, 3097, 3753, 1126, 4083, 5856, 3987, 1945, 11161, 13294, 3088, 10876, 16617, 4609, 27238, 5700, 27784, 11767, 26914, 38305, 6562, 49193, 27835, 35131, 7364, 65601, 50313, 9001, 11980, 39892, 20848

Offset: 1

Patrick De Geest, Sep 15 1999

nonn

"One of the simplest cubic Diophantine equations is known to have an infinite number of solutions (Lehmer, 1956; Payne and Vaserstein, 1991). Any number of solutions to the equation x^3 + y^3 + z^3 = 1 can be produced through the use of the algebraic identity (9t^3+1)^3 + (9t^4)^3 + (-9t^4-3t)^3 = 1 by substituting in values of t. ...
"Although these are certainly solutions, the identity generates only one family of solutions. Other solutions such as (94, 64, -103), (235, 135, -249), (438, 334, -495), ... can be found. What is not known is if it is possible to parameterize all solutions for this equation. Put another way, are there an infinite number of families of solutions? Probable yes, but that too remains to be shown." [Herkommer]
Values of x associated with A050794.

			577^3 + 2304^3 = 2316^3 + 1.

Mark A. Herkommer, Number Theory, A Programmer's Guide, McGraw-Hill, NY, 1999, page 370.
Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.

Lewis Mammel, Table of n, a(n) for n = 1..368
Eric Weisstein's World of Mathematics, Diophantine Equation - 3rd Powers

Cf. A050791, A050793, A050794.

More terms from Michel ten Voorde.
Extended through 26914 by Jud McCranie, Dec 25 2000
More terms from Don Reble, Nov 29 2001
Edited by N. J. A. Sloane, May 08 2007

A141326 Subsequence of 'Fermat near misses' which is generated by a simple formula based on the cubic binomial expansion along with formulas for the corresponding terms in the expression, x^3 + y^3 = z^3 + 1.

Original entry on oeis.org

12, 150, 738, 2316, 5640, 11682, 21630, 36888, 59076, 90030, 131802, 186660, 257088, 345786, 455670, 589872, 751740, 944838, 1172946, 1440060, 1750392, 2108370, 2518638, 2986056, 3515700, 4112862, 4783050, 5531988, 6365616, 7290090, 8311782, 9437280, 10673388

Offset: 1

Lewis Mammel (l_mammel(AT)att.net), Aug 03 2008

From Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008: (Start)
In Ramanujan's parametric equation: (ax+y)^3 + (b+x^2y)^3 = (bx+y)^3 + (a+x^2y)^3
where a^2 + ab + b^2 = 3xy^2.
This sequence is obtained by setting a=0, y=1 and finding the solution to b^2=3x:
b=3n, x=3n^2. (End)

			For a(1)=12: 1 + 12^3 = 9^3 + 10^3 = 1729.

Colin Barker, Table of n, a(n) for n = 1..1000
Tito Piezas III and Eric Weisstein, Diophantine Equation--3rd Powers [Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008]
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A050791, A050792, A050793, A050794.

Vec(6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5 + O(x^40)) \\ Colin Barker, Oct 26 2019

a(n) = 9*n^4 + 3*n, with b(n) = 9*n^4 and c(n) = 9*n^3 + 1 we have 1 + a(n)^3 = b(n)^3 + c(n)^3.
From Colin Barker, Oct 25 2019: (Start)
G.f.: 6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
a(n) = 3*(n + 3*n^4).
(End)

Edited by Joerg Arndt, Oct 26 2019

A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.

Original entry on oeis.org

1, 30, 71, 100, 104, 485, 1169, 705, 887, 1727, 421, 775, 4499, 4260, 3180, 5748, 9719, 307, 6092, 18521, 20304, 18825, 32255, 6174, 33082, 7601, 29400, 17607, 13457, 52487, 1727, 44794, 41772, 76328, 1801, 29707, 80999, 119789, 111226, 132105, 122730, 171071, 123117, 237275

Offset: 1

David Rabahy, Jul 21 2015

nonn

It seems to me the sequence can never drop all the way to 1 again.
From Robert Israel, Oct 13 2015: (Start)
There are only finitely many n with a(n) = 1.
Such n correspond to solutions of the Diophantine equation x (2 x^2 + 3 x + 3) = z^3.
Since gcd(x, 2 x^2 + 3 x + 3) = 1 or 3, we get two cases:
  if x is not divisible by 3, x = s^3, z = s^3 t^3 where 2 s^6 + 3 s^3 + 3 = t^3,
  otherwise x = 9  s^3, z = 3 s t, where 54 s^6 + 9 s^3 + 1 = t^3.
The algebraic curves 2 s^6 + 3 s^3 + 3 = t^3 and 54 s^6 + 9 s^3 + 1 = t^3 both have genus 4, so by Faltings's theorem they have only finitely many rational solutions.
(End)

			10 - 9 = 1 is the first number in the sequence because 10^3 + 9^3 = 12^3 + 1^3 and no other lower z produces a result.

Robert G. Wilson v, Table of n, a(n) for n = 1..368

Cf. A001235.

Cf. x = A050792, y = A050793, z = A050791 , x^3+y^3 = A050794.

Cubes:= {seq(x^3, x=2..10^4)}:
count:= 0:
for z from 1 to 10^4 do
  s:= z^3+1;
  M:= map(t -> s-t, select(`<`,Cubes,floor(s/2))) intersect Cubes;
  for m in M do
    count:= count+1;
    y:= simplify(m^(1/3));
    x:= simplify((s-m)^(1/3));
    A[count]:= y-x;
  od
od:
seq(A[i],i=1..count); # Robert Israel, Oct 13 2015

y = 3; lst = {}; While[y < 100001, x = 2; While[x < y, z = (y^3 + x^3 - 1)^(1/3); If[IntegerQ[z], AppendTo[lst, {z, y, x, y - x}]; Print[{z, y, x, y - x}]]; x++]; y++]; Last@ Transpose@ Sort@ lst (* Robert G. Wilson v, Jul 21 2015 and modified Oct 14 2015 *)

a(n) = A050793(n) - A050792(n). - Robert G. Wilson v, Jul 21 2015

cp's OEIS Frontend

A050794 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Arrange solutions by increasing values of z (see A050791). Sequence gives values of x^3 + y^3 = z^3 + 1. For corresponding values of x, y, z see A050792, A050793, A050791 respectively.

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A050791 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Sequence gives values of z in monotonic increasing order.

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A050792 Consider the Diophantine equation x^3 + y^3 = z^3 + 1 (1 < x < y < z) or 'Fermat near misses'. Arrange solutions by increasing values of z (see A050791). Sequence gives values of x.

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Extensions

A141326 Subsequence of 'Fermat near misses' which is generated by a simple formula based on the cubic binomial expansion along with formulas for the corresponding terms in the expression, x^3 + y^3 = z^3 + 1.

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Extensions

A259753 For increasing z > 0, integers, y - x, where x^3 + y^3 = z^3 + 1, with y > x > 1.

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