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Smallest number k of consecutive primes > p_n such that p_n^2 * p_(n+1) * p_(n+2) * ... * p_(n+k) is an abundant number.
3

%I #16 Dec 16 2016 16:33:46

%S 1,3,7,16,29,44,65,89,120,155,192,236,282,332,390,453,520,589,666,746,

%T 832,927,1026,1131,1239,1350,1467,1592,1725,1867,2017,2161,2313,2469,

%U 2634,2800,2975,3155,3339,3532,3729,3931,4143,4356,4579,4809,5051,5291

%N Smallest number k of consecutive primes > p_n such that p_n^2 * p_(n+1) * p_(n+2) * ... * p_(n+k) is an abundant number.

%C The sequence arose while solving puzzle 329 from Carlos Rivera's Prime Puzzles & Problems Connection site.

%H Carlos Rivera, <a href="http://primepuzzles.net/puzzles/puzz_329.htm">Puzzle 329. Odd abundant numbers not divided by 2 or 3.</a>

%e 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).

%t 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_ *)

%o (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)))

%Y Cf. A005101.

%Y Cf. A005231, A047802.

%K nonn

%O 1,2

%A _Igor Schein_, Sep 13 2005

%E Edited and extended by _Robert G. Wilson v_, Sep 15 2005