OFFSET
1,2
COMMENTS
The sequence arose while solving puzzle 329 from Carlos Rivera's Prime Puzzles & Problems Connection site.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
Carlos Rivera, Puzzle 329. Odd abundant numbers not divided by 2 or 3, The Prime Puzzles & Problems Connection.
EXAMPLE
a(2) = 3 because the second prime being 3, then 3^2 * 5 * 7 * 11 = 3465 and sigma(3465) - 2*3465 = 558, a positive number (i.e., 3465 is abundant), but 3^2 * 5 * 7 = 315 and sigma(315) - 2*315 = -6, a nonpositive number (i.e., 315 is not abundant).
MATHEMATICA
abQ[n_] := DivisorSigma[1, n] > 2n; f[0] = 0; f[n_] := f[n] = Block[{k = f[n - 1]}, p = Fold[Times, Prime[n], Prime[ Range[n, n + k]]]; While[ !abQ[p], k++; p = p*Prime[n + k]]; k]; Table[ f[n], {n, 48}] (* Robert G. Wilson v, Sep 15 2005 *)
seq[len_] := Module[{s = {}, p = 2, k = 0, q = 2, c = 0, r}, r = 1 + 1/p + 1/p^2; While[c < len, While[r <= 2, p = NextPrime[p]; r *= (1 + 1/p); k++]; c++; AppendTo[s, k]; k--; r /= (1 + 1/q + 1/q^2); q = NextPrime[q]; r *= ((1 + 1/q + 1/q^2)/(1 + 1/q))]; s]; seq[48] (* Amiram Eldar, Feb 07 2026 *)
PROG
(PARI) forprime(p=2, 100, k=0; while(k++, if(sigma(n=p^2*prod(j=1, k, prime(j+primepi(p))))-n>n, print(k); break)))
(PARI) list(len) = {my(p = 2, k = 0, r = 1 + 1/p + 1/p^2, q = p, c = 0); while(c < len, while(r <= 2, p = nextprime(p+1); r *= (1 + 1/p); k++); c++; print1(k, ", "); k--; r /= (1 + 1/q + 1/q^2); q = nextprime(q+1); r *= ((1 + 1/q + 1/q^2)/(1 + 1/q))); } \\ Amiram Eldar, Feb 07 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Igor Schein, Sep 13 2005
EXTENSIONS
Edited and extended by Robert G. Wilson v, Sep 15 2005
STATUS
approved
