A338134 Initial terms are the first 11 odd prime numbers. After that each new term in the sequence is the sum of whichever subset of the previous 11 terms gives the smallest prime number that is larger than the largest term so far.

3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 67, 71, 83, 97, 109, 127, 131, 149, 167, 179, 197, 251, 263, 307, 337, 367, 443, 587, 653, 683, 743, 797, 821, 907, 1087, 1291, 1493, 2003, 2237, 2243, 2333, 2447, 2791, 3019, 3691, 4583, 4787, 6287

Offset

1,1

Comments

The sequence is finite with the final term a(1022) having 94 decimal digits.

Instead of 11, other choices k of the number of initial odd primes and number of previous terms to consider when picking the next prime give different sequence lengths L and final terms m. Values are shown in the following table.

=======================================================

k L log(m) m

-----+-------------------------------------------------

4 | 13 6.42 53

5 | 18 7.77 2377

6 | 26 8.42 4547

7 | 66 19.32 246069541

8 | 176 46.61 174401554254658978301

9 | 313 74.93 348416283338512248746770398000349

10 | 657 155.19 25092330625366012736545948...

11 | 1022 214.78 18914773701982111040121422...

12 | 2575 596.03 71213007235221948920803479...

13 | 5142 1172.46 15648776898352974934674946...

Links

Jinyuan Wang, Table of n, a(n) for n = 1..1022

Prog

(PARI)
F(u, m, test)={my(recurse(k, s, b)=if(s<b, if(s>m&&test(s), b=s); while(k<#u, k++; b=self()(k, s+u[k], b))); b); recurse(0, 0, oo)}
lista(n, k=11)={my(v=vector(k)); for(r=1, n, my(t=if(r<=k, prime(r+1), F(v, v[#v], ispseudoprime))); print1(t, ", "); if(t==oo, break); v=concat(v[2..k], t))}
{ lista(60) } \\ Andrew Howroyd, Oct 13 2020

Crossrefs

Cf. A065091.

Sequence in context: A215697 A322184 A245072 * A138980 A191378 A191376

Adjacent sequences: A338131 A338132 A338133 * A338135 A338136 A338137

Keyword

nonn,fini,full

Author

Anthony Winkelspecht, Oct 11 2020

Status

approved