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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Michel Mizony

Michel Mizony's wiki page.

Michel Mizony has authored 3 sequences.

A213687 Numbers which are the values of the quadratic polynomial 3+4*k+7*t+8*k*t on nonnegative integers.

Original entry on oeis.org

3, 7, 10, 11, 15, 17, 19, 22, 23, 24, 27, 31, 34, 35, 37, 38, 39, 43, 45, 46, 47, 51, 52, 55, 57, 58, 59, 63, 66, 67, 70, 71, 73, 75, 77, 79, 80, 82, 83, 87, 91, 94, 95, 97, 99, 101, 103, 106, 107, 108, 111, 112, 115, 117, 118, 119, 122, 123, 126, 127, 129
Offset: 1

Views

Author

Michel Mizony, Jun 18 2012

Keywords

Comments

For all these numbers a(n) we have the following Erdos-Straus decomposition: 4/p = 4/(3+4*k+7*t+8*k*t) = 1/(2*(3+4*k+7*t+8*k*t)*(1+k)) + 1/((1+k)*(2*t+1)) + 1/(2*(1+k)*(2*t+1)*(3+4*k+7*t+8*k*t));
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2t(t+1),2t+1, 2t(t+1)+1].
In 1948 Erdos and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

Examples

			31 is a term because the solutions to 3+4*k+7*t+8*k*t = 31 are {k = 0, t = 4}, {k = 7, t = 0}.

References

  • I. Gueye and M. Mizony, Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
  • M. Mizony and I. Gueye, Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 141-150.

Links

Crossrefs

Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A073101 (number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z).

Programs

  • Maple
    H:=(k, t) -> 4/(3+4*k+7*t+8*k*t) = [1/2*1/((3+4*k+7*t+8*k*t)*(1+k)), 1/((1+k)*(2*t+1)), 1/2*1/((1+k)*(2*t+1)*(3+4*k+7*t+8*k*t))]:
cousin:=proc(p)
local n,k;
for n from 0 to (p-3)/7 do
if (p-3-7*n) mod (4+8*n)=0  then k:=(p-3-7*n)/(4+8*n):
return([p,n,H(k,n)]) fi;od;
end:
L:=NULL:for p from 2 to 500 do L:=L,cousin(p): od:{L}[1..10];map(u->op(1,u),{L});map(u->op(2,u),{L});

A213686 Numbers which are the values of the quadratic polynomial 13+20*t+24*k+32*k*t at nonnegative integers.

Original entry on oeis.org

13, 33, 37, 53, 61, 73, 85, 89, 93, 109, 113, 133, 141, 145, 153, 157, 173, 181, 193, 201, 205, 213, 229, 233, 245, 253, 257, 273, 277, 293, 297, 301, 313, 317, 325, 333, 349, 353, 369, 373, 393, 397, 401, 405, 413, 421, 425, 433, 445, 453, 469, 473, 481
Offset: 1

Views

Author

Michel Mizony, Jun 17 2012

Keywords

Comments

For all these numbers a(n) we have the following Erdos-Straus decomposition: 4/p=4/(13+24*k+20*t+32*k*t) = 1/(6*k+8*k*t+4+6*t) + 1/((13+24*k+20*t+32*k*t)*(5+8*k)*(3*k+4*k*t+2+3*t)) + 1/(2*(5+8*k)*(3*k+4*k*t+2+3*t)).
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2n(n+1),2n+1,2n(n+1)+1], where n=2+4k.
In 1948 Erdos and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

Examples

			For n=12 the a(12)=133  solutions are {k = 0, t = 6},{k = 5, t = 0}.

References

  • I. Gueye and M. Mizony : Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
  • M. Mizony and I. Gueye : Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2,p pp 141-150.

Links

Crossrefs

Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A073101 (number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z).

Programs

  • Maple
    G:=(n,p)->4/p = [2*(2*n+1)/(n*p+p+1), 4/p/(n*p+p+1), 2/(n*p+p+1)]:
cousin:=proc(p)
local n;
for n from 0 to 300 do
if n*p+p+1 mod 4*(2*n+1)=0 then return([p,n,G(n,p)]);fi:
od:
end:
L:=NULL:for m to 400 do L:=L,cousin(4*m+1): od:{L}[1..4];map(u->op(1,u),{L});

A213340 Numbers which are the values of the quadratic polynomial 5+8t+12k+16kt on nonnegative integers.

Original entry on oeis.org

5, 13, 17, 21, 29, 37, 41, 45, 53, 61, 65, 69, 77, 85, 89, 93, 97, 101, 109, 113, 117, 125, 133, 137, 141, 149, 153, 157, 161, 165, 173, 181, 185, 189, 197, 205, 209, 213, 221, 229, 233, 237, 241, 245, 253, 257
Offset: 1

Views

Author

Michel Mizony, Jun 09 2012

Keywords

Comments

For all these numbers a(n) we have the following Erdős-Straus decomposition: 4/p = 4/(5+8*t+12*k+16*k*t) = 1/(2*(2*k+1)*(2+3*t+3*k+4*k*t)) + 1/(2+3*t+3*k+4*k*t) + 1/(2*(5+8*t+12*k+16*k*t)*(2*k+1)*(2+3*t+3*k+4*k*t)).
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2n(n+1), 2n+1,2n(n+1)+1], where n=2+4t+6k+8kt.
In 1948 Erdős and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

Examples

			For n=5 the a(5)=29 solutions are {k=0, t=3}, {k=2, t=0}.

References

  • I. Gueye and M. Mizony, Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
  • I. Gueye and M. Mizony, Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 141-150.

Links

Crossrefs

Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A195770 (positive integers a for which there is a 1-Pythagorean triple (a,b,c) satisfying a<=b).
A073101 number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Programs

  • Maple
    G:=(p,d)->4/p = [4*d/(p+d)/(p+1), 4/(p+d), 4*d/(p+d)/(p+1)/p]:
cousin:=proc(p)
local d;
for d from 3 by 4 to 100 do
if ((p+1)/2) mod d=0 and (p+d)*(p+1) mod d=0 then
return([p,G(p,d)]) fi;od;
end:
for k to 20 do cousin(4*k+1) od;

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