A342443 -id:A342443 - cp's OEIS frontend

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

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

Offset: 1

Bernard Schott, Mar 12 2021

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.

			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)

Cf. A067377, A342439, A342440, A342443, A342453, A342454.

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

Bernard Schott, Mar 13 2021

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

Cf. A342440, A342441, A342443, A342444, A342453.

a(5)-a(14) from Jon E. Schoenfield, Mar 14 2021

a(15) from Max Alekseyev, Dec 11 2024

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

Bernard Schott, Mar 12 2021

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

			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.

Dreamshire, Project Euler 50 Solution.
Archive.today, trizen / experimental-projects.
Project Euler, Problem 50: Consecutive prime sum.

Cf. A034962, A034965, A082246, A082251, A127340, A127341, A215991, A067377.

Cf. A342440, A342443, A342444, A342453, A342454.

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

A342440 The longest length of consecutive primes which sums to prime = A342439(n) < 10^n.

Original entry on oeis.org

2, 6, 21, 65, 183, 543, 1587, 4685, 13935, 41708, 125479, 379317, 1150971, 3503790, 10695879, 32729271, 100361001, 308313167, 948694965

Offset: 1

Bernard Schott, Mar 12 2021

Inspired by the 50th problem of Project Euler (see link).
The corresponding largest primes obtained are in A342439.
Solutions and Python program are proposed in Dreamshire and archive.today links. - Daniel Suteu, Mar 12 2021

			A342439(1) = 5 = 2+3, hence a(1) = 2 since there are 2 terms in this longest sum.
A342439(2) = 41 = 2 + 3 + 5 + 7 + 11 + 13 hence a(2) = 6 since there are 6 terms in this longest sum.

Archive.today, trizen / experimental-projects.
Dreamshire, Project Euler 50 Solution.
Project Euler, Problem 50: Consecutive prime sum.

Cf. A342439, A342443, A342444.

a(4)-a(17) from Daniel Suteu, Mar 12 2021

a(18)-a(19) from Martin Ehrenstein, Mar 13 2021

A342453 When A342439(n) is the largest prime < 10^n obtained with the longest sum of the A342440(n) consecutive primes, then a(n) is the first prime of these A342440(n) consecutive primes.

Original entry on oeis.org

2, 2, 7, 3, 3, 7, 7, 7, 11, 2, 19, 5, 5, 2, 13, 5, 5, 7, 11

Offset: 1

Bernard Schott, Mar 14 2021

Inspired by the 50th problem of Project Euler (see link).
There must be at least two consecutive primes in the sum.
The terms a(4)-a(17) come from the Perl program and the results proposed by Daniel Suteu in the link Archive.today.

			A342439(2) = 41 = 2 + 3 + 5 + 7 + 11 + 13 hence a(2) = 2.

Archive.today, trizen / experimental-projects.
Dreamshire, Project Euler 50 Solution.
Project Euler, Problem 50: Consecutive prime sum.

Cf. A342439, A342440, A342443, A342444, A342454

a(4)-a(17) from Daniel Suteu, Mar 14 2021
a(18)-a(19) from Martin Ehrenstein, Mar 14 2021
a(7) and a(15) corrected by Martin Ehrenstein, Mar 14 2021

cp's OEIS Frontend

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

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

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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).

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A342440 The longest length of consecutive primes which sums to prime = A342439(n) < 10^n.

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A342453 When A342439(n) is the largest prime < 10^n obtained with the longest sum of the A342440(n) consecutive primes, then a(n) is the first prime of these A342440(n) consecutive primes.

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