cp's OEIS Frontend

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.

Showing 1-5 of 5 results.

A115652 Brilliant numbers (A078972) which are the sum of distinct double factorials (A006882).

Original entry on oeis.org

4, 6, 9, 10, 14, 15, 21, 25, 49, 121, 169, 403, 407, 437, 451, 493, 517, 551, 949, 961, 1003, 1007, 1067, 1073, 1079, 1121, 1333, 1343, 1349, 1357, 1387, 1403, 1457, 1501, 3869, 3901, 3953, 4331, 4891, 5183, 5293, 10403, 10807, 11413, 11449
Offset: 1

Views

Author

Giovanni Resta, Jan 28 2006

Keywords

Comments

Double factorials 0!! and 1!! are not considered distinct. Note that double factorial (n!!) is different from (n!)!.

Examples

			949 = 13*73 = 9!! + 3!! + 1!!.
		

Crossrefs

A115653 Prime numbers which are the sum of distinct double factorials (A006882).

Original entry on oeis.org

2, 3, 5, 11, 13, 17, 19, 23, 29, 53, 59, 61, 67, 71, 73, 107, 109, 113, 131, 157, 163, 167, 173, 179, 181, 389, 397, 401, 409, 433, 443, 449, 457, 461, 491, 499, 503, 509, 541, 547, 557, 563, 947, 953, 971, 997, 1009, 1013, 1019, 1021, 1051, 1061, 1063
Offset: 1

Views

Author

Giovanni Resta, Jan 28 2006

Keywords

Comments

Double factorials 0!! and 1!! are not considered distinct. Note that double factorial (n!!) is different from (n!)!.

Examples

			947 is prime and 947 = 9!! + 2!!.
		

Crossrefs

A115654 Semiprimes (A001358) which are the sum of distinct double factorials (A006882).

Original entry on oeis.org

4, 6, 9, 10, 14, 15, 21, 25, 26, 49, 51, 57, 58, 62, 65, 69, 74, 77, 106, 111, 115, 118, 119, 121, 122, 123, 129, 133, 134, 155, 158, 159, 161, 166, 169, 177, 178, 386, 393, 394, 395, 398, 403, 407, 411, 413, 437, 445, 446, 447, 451, 453, 458, 489, 493, 497
Offset: 1

Views

Author

Giovanni Resta, Jan 28 2006

Keywords

Comments

Double factorials 0!! and 1!! are not considered distinct. Note that double factorial (n!!) is different from (n!)!.

Examples

			384 = 2*19 = 8!!+2!!.
		

Crossrefs

Programs

  • Mathematica
    Union[Select[Total/@Subsets[Range[10]!!,10],PrimeOmega[#]==2&]] (* Harvey P. Dale, Aug 24 2012 *)

A115648 Square numbers which are the sum of distinct double factorials (A006882).

Original entry on oeis.org

1, 4, 9, 16, 25, 49, 64, 121, 169, 400, 441, 961, 1444, 3844, 3969, 4225, 4356, 4900, 5184, 10404, 11449, 11881, 14400, 15625, 47089, 47524, 56644, 57600, 139129, 145924, 149769, 182329, 192721, 695556, 705600, 792100, 837225, 2073600
Offset: 1

Views

Author

Giovanni Resta, Jan 28 2006

Keywords

Comments

Double factorials 0!! and 1!! are not considered distinct. Note that double factorial (n!!) is different from (n!)!.

Examples

			10404 = 102^2 = 1!! + 4!! + 11!!.
		

Crossrefs

A173518 Solutions z of the Diophantine equation x^3 + y^3 = 6z^3.

Original entry on oeis.org

21, 960540, 16418498901144294337512360, 436066841882071117095002459324085167366543342937477344818646196279385305441506861017701946929489111120
Offset: 1

Views

Author

Michel Lagneau, Feb 20 2010

Keywords

Comments

A. Nitaj proved Erdős's conjecture (1975) and claimed that there exist infinitely many triples of 3-powerful numbers a,b,c with (a,b) = 1, such that a+b=c, because the equation x^3 + y^3 = 6z^3 admits an infinite number of solutions, and given by the recurrence equations (see formula). It is proved that a=x(k)^3, b=y(k)^3, and c=6c^3, and are 3-powerful numbers for each k >= 1.

Examples

			37^3 + 17^3 = 6*21^3.
		

References

  • J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 348.
  • Mordell, L. J. (1969). Diophantine equations. Academic Press. ISBN 0-12-506250-8

Crossrefs

Programs

  • Maple
    x0:=37:y0:=17:z0:=21: for p from 1 to 5 do: x1:=x0*(x0^3+ 2*y0^3):y1:=-y0*(2*x0^3+ y0^3):z1:=z0*(x0^3- y0^3): print(z1) : x0 :=x1 :y0 :=y1 :z0 :=z1 :od :

Formula

We generate the solutions (x(k),y(k),z(k)) from the initial solution x(0) = 37, y(0)=17, z(0)=21 x(k+1) = x(k)*(x(k)^3 + 2*y(k)^3) y(k+1) = -y(k)*(2*x(k)^3 + y(k)^3) z(k+1) = z(k)*(x(k)^3 - y(k)^3).
Showing 1-5 of 5 results.