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#28 by Bruno Berselli at Fri Feb 01 11:56:52 EST 2019
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#27 by Michel Marcus at Thu Jan 31 06:18:00 EST 2019
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#26 by Daniel Suteu at Mon Jan 28 08:34:28 EST 2019
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#25 by Daniel Suteu at Mon Jan 28 08:33:35 EST 2019
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(PARI) a(n) = my(r=0, t=0, b=0); forsum(k=1, n, my(b=bigomega(k); )); t += +=b; r += k*b-t); r; );
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proposed
editing
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#24 by Michael De Vlieger at Sat Jan 19 13:16:57 EST 2019
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#23 by Michael De Vlieger at Sat Jan 19 13:16:54 EST 2019
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Array[Sum[PrimeOmega@ Binomial[#, k], {k, 0, #}] &, 57] (* Michael De Vlieger, Jan 19 2019 *)
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proposed
editing
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#22 by David A. Corneth at Wed Jan 16 11:29:20 EST 2019
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Discussion
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| Wed Jan 16
| 11:30
| David A. Corneth: (instead of n + 1 terms)
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#21 by David A. Corneth at Wed Jan 16 11:26:37 EST 2019
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(PARI) first(n) = my(res = List([0]), r=0, t=0, b=0); for(k=1, n, b=bigomega(k); t += b; r += k*b-t; listput(res, r)); res \\ adapted from Daniel Suteu \\ David A. Corneth, Jan 16 2019
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proposed
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Discussion
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| Wed Jan 16
| 11:29
| David A. Corneth: Your prog suits perfectly for finding the first terms as it uses a recurrence on a(n) to find a(n+1). One might change "for(i = 1, n, " to "for(i = 1, n-1, " to actually get n terms.
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#20 by Daniel Suteu at Tue Jan 15 10:56:02 EST 2019
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#19 by Daniel Suteu at Tue Jan 15 10:55:58 EST 2019
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(PARI) a(n) = bigomega(prodsum(k=0, n, bigomega(binomial(n, k)));
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proposed
editing
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