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.

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A309487 Positive integers represented by the quadratic form (the discriminant form) Δ = b^2 - 4ac, where a,b,c are consecutive palindromic primes.

Original entry on oeis.org

4437, 67088885, 608096563245, 6008043480300405, 60017281285205688005, 600012360124320087600005, 6000055320121974202106400005, 60000010840001925680009488000005, 600000005880000160040000148000000005, 6000000035120000052560000001460000000005
Offset: 1

Views

Author

Philip Mizzi, Sep 06 2019

Keywords

Comments

This is an interesting sequence because for most cases Δ<0. The cases where Δ>0 are sparse.
Based on a study of Δ for the case when a,b,c are consecutive primes I conjecture (but have no proof) that now Δ is always negative.
The conjecture in the previous comment is true. It says p(n)^2 <= 4*p(n-1)*p(n+1), and this follows from p(n)^2 <= 4*p(n-1)*p(n), i.e. p(n) <= 4*p(n-1), which is true (see A327447, also Mitrinovic, Sect. VII.18 (b)). - N. J. A. Sloane, Sep 10 2019
The corresponding least palindromic primes are: 11, 929, 98689, 9989899, 999727999, 99999199999, 9999987899999, 999999787999999, ...
Apart from the first term, it appears that the values of "a" and "b" are given by A028990 and A028989, respectively. - Daniel Suteu, Sep 08 2019

Examples

			Consecutive palindromic primes begin with 2,3,5. For a=2, b=3, c=5, Δ=b^2-4ac=-31. Since Δ<0 this is not a member of the sequence.
With consecutive palindromic primes 11,101,131 and a=11, b=101, c=131, Δ=b^2-4ac=4437, the first member of the sequence.
The corresponding values of a,b,c are given in the table bellow.
+----+---------------------+-----------------------+-----------------------+
|  n |          a          |           b           |            c          |
+----+---------------------+-----------------------+-----------------------+
|  1 |                  11 |                   101 |                   131 |
|  2 |                 929 |                 10301 |                 10501 |
|  3 |               98689 |               1003001 |               1008001 |
|  4 |             9989899 |             100030001 |             100050001 |
|  5 |           999727999 |           10000500001 |           10000900001 |
|  6 |         99999199999 |         1000008000001 |         1000017100001 |
|  7 |       9999987899999 |       100000323000001 |       100000353000001 |
|  8 |     999999787999999 |     10000000500000001 |     10000001910000001 |
|  9 |   99999999299999999 |   1000000008000000001 |   1000000032300000001 |
| 10 | 9999999992999999999 | 100000000212000000001 | 100000000252000000001 |
+----+---------------------+-----------------------+-----------------------+
		

References

  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer.

Crossrefs

See also A028990 and A028989.

Extensions

a(6)-a(8) from Daniel Suteu, Sep 08 2019
a(9) from Chai Wah Wu, Sep 09 2019
a(10) from Chai Wah Wu, Sep 12 2019

A324795 a(n) = 2*p(n)*p(n+2) - p(n+1)^2 where p(k) = k-th prime.

Original entry on oeis.org

11, 17, 61, 61, 205, 205, 421, 573, 585, 1185, 1173, 1501, 2005, 2349, 2737, 2985, 4185, 4173, 4741, 5889, 5877, 7173, 8181, 8569, 9781, 11005, 11005, 12301, 14917, 13477, 17637, 17649, 21505, 19777, 23985, 24577, 25869, 28509, 29857, 30585, 35617
Offset: 1

Views

Author

N. J. A. Sloane, Sep 10 2019

Keywords

Comments

Theorem: a(n) > 0. Proof: Use p(n+1) <= 2 p(n)^2 for n > 4. (See Sándor et al.) QED

References

  • József Sándor, Dragoslav S. Mitrinovic, and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Chapter VII, p. 247, section VII.18.b.

Crossrefs

Cf. A056221 (if leading coefficient 2 is changed to 1), A327447 or A309487 (if 2 is changed to 4).

Programs

  • Mathematica
    With[{p = Prime[Range[50]]}, 2 * p[[1;;-3]] * p[[3;;-1]] - p[[2;;-2]]^2] (* Amiram Eldar, Apr 25 2024 *)
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