A073859 a(1)=1 and a(n>1)=min{primes up to n-1 | a(1)+a(2)+...+a(n-1)+a(n) is a cube}.

1, 7, 19, 37, 61

Offset

1,2

Comments

The sequence is complete. Proof: We must find a prime p>61 and an integer m such that p = m^3 - (1+7+19+37+61) = m^3 - 125 = (m - 5)*(m^2 + 5*m + 25). Since p is prime, we must have p=1*p, therefore set m=6 to make p = (6-5)(6^2 + 5*6 + 25) = 1*91. However, 91=7*13 (not prime). Since m=6 is the only answer for which m-5=1, we can conclude that there is no prime p satisfying the equation. - Francois Jooste (pin(AT)myway.com), Mar 09 2003

Links

Table of n, a(n) for n=1..5.

Maple

a := proc(r, M) local tmp, f, i; description "returns the sequence (first term is 1, all others are prime, up to the (M-1)-th prime) such that the partial sum of all the previous terms in the sequence form a cube (that is to say, 'a(1)=1' and 'a(n>1)=min{primes up to n-1 | a(1)+a(2)+...+a(n-1)+a(n) is a cube}')."; for i from 1 to M do tmp := `if`(i>1, [seq(f(j), j=1..i-1), ithprime(i-1)], [1]); f(i) := `if`(proot(`+`(op(tmp)), r)<>_NOROOT, tmp[i], 0); od; return seq(`if`(f(j)=0, NULL, f(j)), j=1..M); end proc;

Crossrefs

Sequence in context: A241984 A301717 A152540 * A038847 A130056 A136057

Adjacent sequences: A073856 A073857 A073858 * A073860 A073861 A073862

Keyword

nonn,fini,full

Author

Amarnath Murthy, Aug 15 2002

Status

approved