A099807 -id:A099807 - cp's OEIS frontend

A098970 Numbers k such that (12*k)^2 can be expressed as the sum of the cubes of two distinct primes.

19, 67695, 411292, 1134035, 1184876, 2112836, 2455255, 4073384, 11293009, 16171470, 18589912, 34388501, 63609329, 63711615, 117446600, 166530856, 284034387, 449805631, 637548135, 685361103, 783484793, 888180400, 1121365940

Offset: 1

Hugo Pfoertner, Oct 24 2004

nonn

This sequence resulted from a discussion on the seqfan mailing list started by Ed Pegg Jr.
Dean Hickerson and Paul C. Leopardi have shown that if a and b are distinct primes with a^3 + b^3 = c^2, then c must be divisible by 12.
The numbers 12*k form a subsequence of A099426. - Hans Havermann, Oct 24 2004
All terms of this sequence are of the form M*N*(3*M^4+N^4)/2 for some pair M,N of relatively prime positive integers of opposite parity. For each n, A099806(n)^3 + A099807(n)^3 = (12*A098970(n))^2. - James R. Buddenhagen, Oct 26 2004

James Buddenhagen, Two Primes Cubed which Sum to a Square.

Cf. A099426.

Cf. A099806, A099807, A099808, A099809.

More terms from James R. Buddenhagen, Oct 26 2004

A099808 If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c.

Original entry on oeis.org

1, 15, 28, 35, 44, 44, 55, 56, 91, 90, 88, 119, 161, 165, 200, 184, 273, 319, 285, 357, 377, 400, 380, 434, 550, 517, 592, 615, 638, 667, 682, 666, 740, 697, 784, 688, 825, 682, 846, 770, 893, 814, 868, 925, 775, 899, 885, 1007, 1045, 1040, 1078, 1184, 1015

Offset: 0

James R. Buddenhagen, Oct 26 2004

For each n let a=A099806(n), b=A099807(n). Then sqrt((a+b)/48) is an integer and equals A099808(n). Note that a^3 + b^3 = c^2 factors as (a+b)*(a^2-a*b+b^2). The first factor (a+b) is 48*d^2, some d. This sequence tabulates the d values. Remember, a and b are prime numbers.

			From 11^3 + 37^3 = 228^2 we get sqrt((a+b)/48) = (11+37)/48 = 1, so 1 is in the sequence. [corrected by _Harvey P. Dale_, Apr 12 2011]

James Buddenhagen, Two Primes Cubed which Sum to a Square.

Cf. A099806, A099807, A098970, A099809.

A099806 If a,b are prime numbers satisfying the Diophantine equation a^3+b^3=c^2, then a is -1 mod 12 and b is 1 mod 12, or vice versa. Choose 'a' to be -1 mod 12. This is the sequence of 'a' values, sorted by the magnitude of c.

Original entry on oeis.org

11, 8663, 28703, 56999, 44111, 86291, 87959, 16931, 246011, 54083, 367823, 552011, 457511, 571019, 765983, 1586531, 1915163, 2437751, 16139, 2305883, 4074743, 3963299, 1296563, 5440991, 4683779, 2238023, 9682703, 8681639, 8142803

Offset: 0

James R. Buddenhagen, Oct 26 2004

nonn

All terms of this sequence are of the form 3*M^4-N^4+6*M^2*N^2 for some pair M,N of relatively prime positive integers of opposite parity.

			11 is in the sequence because 11 is -1 mod 12 and 11^3+37^3 = 228^2.

James Buddenhagen, Two Primes Cubed which Sum to a Square.

Cf. A099807, A098970, A099808, A099809.

A099809 Let a,b be prime numbers satisfying the Diophantine equation a^3+b^3=(a+b)(a^2-ab+b^2)=c^2. Then the second factor a^2-ab+b^2 is 3e^2 for some integer e. This sequence tabulates the 'e' values, sorted by magnitude of c.

Original entry on oeis.org

19, 4513, 14689, 32401, 26929, 48019, 44641, 72739, 124099, 179683, 211249, 288979, 395089, 386131, 587233, 905059, 1040419, 1410049, 2237011, 1919779, 2078209, 2220451, 2950963, 2767489, 4919971, 5582449, 5019889, 5255761

Offset: 0

James R. Buddenhagen, Oct 26 2004

nonn

For each n let a=A099806[n], b=A099807[n], c/12=A098970. Then a^3+b^3=c^2. The left side factors as (a+b)*(a^2-a*b+b^2). The second factor is 3*e^2 for some integer e. The sequence tabluates the 'e' values. These 'e' values all have the form 3*M^4+N^4, for some pair M,N of relatively prime integers of opposite parity. Remember, a and b are prime numbers.

			11^3+37^3=228^2, 11^2-11*37+37^2=3*e^2 with e=19, so 19 is in the sequence.

James Buddenhagen, Two Primes Cubed which Sum to a Square.

Cf. A099806, A099807, A098970, A099808.

cp's OEIS Frontend

A098970 Numbers k such that (12*k)^2 can be expressed as the sum of the cubes of two distinct primes.

Views

Author

Keywords

Comments

Links

Crossrefs

Extensions

A099808 If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A099806 If a,b are prime numbers satisfying the Diophantine equation a^3+b^3=c^2, then a is -1 mod 12 and b is 1 mod 12, or vice versa. Choose 'a' to be -1 mod 12. This is the sequence of 'a' values, sorted by the magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A099809 Let a,b be prime numbers satisfying the Diophantine equation a^3+b^3=(a+b)(a^2-ab+b^2)=c^2. Then the second factor a^2-ab+b^2 is 3e^2 for some integer e. This sequence tabulates the 'e' values, sorted by magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

cp's OEIS Frontend

A098970 Numbers k such that (12*k)^2 can be expressed as the sum of the cubes of two distinct primes.

Views

Author

Keywords

Comments

Links

Crossrefs

Extensions

A099808 If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A099806 If a,b are prime numbers satisfying the Diophantine equation a^3+b^3=c^2, then a is -1 mod 12 and b is 1 mod 12, or vice versa. Choose 'a' to be -1 mod 12. This is the sequence of 'a' values, sorted by the magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A099809 Let a,b be prime numbers satisfying the Diophantine equation a^3+b^3=(a+b)*(a^2-a*b+b^2)=c^2. Then the second factor a^2-a*b+b^2 is 3*e^2 for some integer e. This sequence tabulates the 'e' values, sorted by magnitude of c.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A099809 Let a,b be prime numbers satisfying the Diophantine equation a^3+b^3=(a+b)(a^2-ab+b^2)=c^2. Then the second factor a^2-ab+b^2 is 3e^2 for some integer e. This sequence tabulates the 'e' values, sorted by magnitude of c.