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 1-4 of 4 results.

A342439 Let S(n,k) denote the set of primes < 10^n which are the sum of k consecutive primes, and let K = maximum k >= 2 such that S(n,k) is nonempty; then a(n) = max S(n,K).

Original entry on oeis.org

5, 41, 953, 9521, 92951, 997651, 9964597, 99819619, 999715711, 9999419621, 99987684473, 999973156643, 9999946325147, 99999863884699, 999999149973119, 9999994503821977, 99999999469565483, 999999988375776737, 9999999776402081701
Offset: 1

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Author

Bernard Schott, Mar 12 2021

Keywords

Comments

Inspired by the 50th problem of Project Euler (see link).
There must be at least two consecutive primes in the sum.
The corresponding number K of consecutive primes to get this largest prime is A342440(n) and the first prime of these A342440(n) consecutive primes is A342453(n).
It can happen that the sums of K = A342440(n) consecutive primes give two (or more) distinct n-digit primes. In that case, a(n) is the greatest of these primes. Martin Ehrenstein proved that there are only two such cases when 1 <= n <= 19, for n = 7 and n = 15 (see corresponding examples).
Solutions and Python program are proposed in Dreamshire and Archive.today links. - Daniel Suteu, Mar 12 2021

Examples

			a(1) = 5 = 2+3.
a(2) = 41 = 2 + 3 + 5 + 7 + 11 + 13; note that 97 = 29 + 31 + 37 is prime, sum of 3 consecutive primes, but 41 is obtained by adding 6 consecutive primes, so, 97 is not a term.
A342440(7) = 1587, and there exist two 7-digit primes that are sum of 1587 consecutive primes; as 9951191 = 5+...+13399 < 9964597 = 7+...+13411 hence a(7) = 9964597.
A342440(15) = 10695879 , and there exist two 15-digit primes that are sum of 10695879 consecutive primes; as 999998764608469 = 7+...+192682309 < 999999149973119 = 13+...+192682337, hence a(15) = 999999149973119.
		

Crossrefs

Extensions

Name improved by N. J. A. Sloane, Mar 12 2021
a(4)-a(17) from Daniel Suteu, Mar 12 2021
a(18)-a(19) from Martin Ehrenstein, Mar 13 2021
a(7) and a(15) corrected by Martin Ehrenstein, Mar 15 2021

A342443 a(n) is the largest prime < 10^n that is the sum of at least two consecutive primes.

Original entry on oeis.org

5, 97, 991, 9949, 99971, 999983, 9999991, 99999989, 999999937, 9999999943, 99999999977, 999999999989, 9999999999763, 99999999999959, 999999999999989
Offset: 1

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Author

Bernard Schott, Mar 12 2021

Keywords

Comments

The minimum corresponding number of consecutive primes to get this largest prime a(n) is A342444(n) and the first prime of these A342444(n) consecutive primes is A342454(n).
Differs from A342439 where the corresponding primes result of the longest sum < 10^n of consecutive primes.
a(n) is the largest n-digit prime A003618(n) for n = 2, 6, 7, 8, 9, 11, 12, ...
a(13) >= k = 10^13 - 237. If a(13) > k then it is the sum of at least 30000 primes. k can be written as the sum of 6449 consecutive primes. - David A. Corneth, Mar 13 2021
No sum of 30000 or more consecutive primes is in the interval [10^13 - 237, 10^13 - 1], so a(13) = 10^13 - 237. - Jon E. Schoenfield, Mar 14 2021

Examples

			a(1) = 5 = 2 + 3, since it is not possible to obtain the greatest 1-digit prime 7 when adding consecutive primes.
a(2) = 29 + 31 + 37 = 97, since (29, 31, 37) are consecutive primes and 97 is the largest 2-digit prime.
		

Crossrefs

Formula

A342439(n) <= a(n) <= A003618(n).

Extensions

a(9) from Jinyuan Wang, Mar 13 2021
a(10) from David A. Corneth, Mar 13 2021
a(11)-a(12) from Jinyuan Wang, Mar 13 2021
a(13)-a(14) from Jon E. Schoenfield, Mar 13 2021
a(15) from Max Alekseyev, Dec 11 2024

A342444 a(n) is the smallest number of consecutive primes that are necessary to add to obtain the largest prime = A342443(n) < 10^n.

Original entry on oeis.org

2, 3, 5, 9, 5, 29, 281, 1575, 599, 7, 17, 3, 6449, 2725361, 163315
Offset: 1

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Author

Bernard Schott, Mar 12 2021

Keywords

Comments

There are at least two consecutive primes in each sum.
The corresponding largest primes obtained are in A342443, and the first primes of these a(n) consecutive primes are in A342454.

Examples

			A342443(1) = 5 = 2 + 3, hence a(1) = 2.
A342443(2) = 97 = 29 + 31 + 37, hence a(2) = 3.
From _Jon E. Schoenfield_, Mar 14 2021: (Start)
                                                    a(n) =
              sum of consecutive primes           number of
      -----------------------------------------  consecutive
   n   A342454(n)   +   ...    =    A342443(n)      primes
  --  -----------------------------------------  -----------
   1             2  +  3       =              5          2
   2            29  + 31 + 37  =             97          3
   3           191  +   ...    =            991          5
   4          1087  +   ...    =           9949          9
   5         19979  +   ...    =          99971          5
   6         34337  +   ...    =         999983         29
   7         34129  +   ...    =        9999991        281
   8         54829  +   ...    =       99999989       1575
   9       1665437  +   ...    =      999999937        599
  10    1428571363  +   ...    =     9999999943          7
  11    5882352691  +   ...    =    99999999977         17
  12  333333333299  +   ...    =   999999999989          3
  13    1550560001  +   ...    =  9999999999763       6449
  14      13384757  +   ...    = 99999999999959    2725361
(End)
		

Crossrefs

Extensions

a(6)-a(9) from Jinyuan Wang, Mar 13 2021
a(10) from David A. Corneth, Mar 13 2021
a(11)-a(14) from Jon E. Schoenfield, Mar 14 2021
a(15) from Max Alekseyev, Dec 11 2024

A342454 a(n) = first prime of the A342444(n) consecutive primes summing to A342443(n).

Original entry on oeis.org

2, 29, 191, 1087, 19979, 34337, 34129, 54829, 1665437, 1428571363, 5882352691, 333333333299, 1550560001, 13384757, 6121296037
Offset: 1

Views

Author

Bernard Schott, Mar 13 2021

Keywords

Examples

			A342439(1) = 2 + 3 = 5 hence a(1) = 2.
A342439(2) = 29 + 31 + 37 = 97 hence a(2) = 29.
		

Crossrefs

Extensions

a(5)-a(14) from Jon E. Schoenfield, Mar 14 2021
a(15) from Max Alekseyev, Dec 11 2024
Showing 1-4 of 4 results.