A386311 a(1) = 2, a(n+1) is the largest prime p such that b(n+1) = b(n)*(p + a(n))/(p - a(n)) is a positive integer, where b(1) = 1.

2, 3, 13, 29, 71, 73, 3673, 3677, 1970327, 8879341, 30578677

Offset: 1

Author

Thomas Ordowski, Jul 18 2025

Keywords

• nonn
• fini
• full

Comments

This sequence is finite and full.

Note that a(n) < a(n+1) <= 2*b(n) + a(n).

b(1) = 1, b(n+1) is the smallest k such that a(n+1) = a(n)*(k + b(n))/(k - b(n)) is a prime, where a(1) = 2.

b(n) = 1, 5, 8, 21, 50, 3600, 3746, 6883275, 6909014, 10849668, and 19729009.

Conjecture: a'(n) = prime(n) for "the smallest prime p" and b'(n) = A352743(n-1) for "the largest k".

If a(n+1) <= 2*b(n) + a(n), then a(11) = 30578677 is the last term. - M. F. Hasler, Jul 19 2025

Crossrefs

Cf. A000040, A352743 (see author's conjecture).

Programs

PARI

{a=List(2); b=List(1); for(n=1,oo, print1(a[n]", "); my(an=a[n], bn=b[n], p=precprime(2*bn+an)); iferr(while(bn*(p+an)%(p-an), p=precprime(p-1)), E, break); listput(a, p); listput(b, bn*(p+an)\(p-an))); print("that's all."); a=Vec(a)} \\ M. F. Hasler, Jul 19 2025

Formula

Product_{k=1..n} (a(k+1) + a(k))/(a(k+1) - a(k)) = b(n+1).

a(n+1)/a(n) = (b(n+1) + b(n))/(b(n+1) - b(n)).

b(n+1)/b(n) = (a(n+1) + a(n))/(a(n+1) - a(n)).

Extensions

a(7)-a(10) from M. F. Hasler, Jul 18 2025

a(11) = 2*b(10)+a(10) from Thomas Ordowski, Jul 19 2025