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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A199767 Numbers m for which Sum_{i=1..k} (1+1/p_i) + Product_{i=1..k} (1+1/p_i) is an integer, where p_i are the k prime factors of m (with multiplicity).

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%I A199767 #29 May 30 2025 10:16:29
%S A199767 21,45,432,740,1728,3456,3888,5616,12096,23760,46656,52164,131328,
%T A199767 152064,186624,195656,233280,311472,606528,618192,746496,926208,
%U A199767 933120,979776,1273536,1403136,2985984,3221456,3732480,5178816,5412096,5971968,9704448,13651200
%N A199767 Numbers m for which Sum_{i=1..k} (1+1/p_i) + Product_{i=1..k} (1+1/p_i) is an integer, where p_i are the k prime factors of m (with multiplicity).
%C A199767 The numbers of the sequence are the solution of the differential equation m’=(a-k)*m-b, which can also be written as A003415(m)=(a-k)*m-A003958(m), where k is the number of prime factors of m, and a is the integer Sum_{i=1..k} (1+1/p_i) + Product_{1=1..k} (1+1/p_i).
%C A199767 The numbers of the sequence satisfy also Sum_{i=1..k} (1-1/p_i) - Product_{i=1..k} (1+1/p_i) = some integer.
%H A199767 J. M. Borwein and E. Wong, <a href="https://citeseerx.ist.psu.edu/document?doi=eefc789579b51c34a09658b91858447e89714d7c">A survey of results relating to Giuga’s conjecture on primality</a>, May 8, 1995.
%H A199767 Romeo Mestrovic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mestrovic/mes4.html">On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers</a>, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
%e A199767 740 has prime factors 2, 2, 5, 37. 1 + 1/2 + 1 + 1/2 + 1 + 1/5 + 1 + 1/37 = 967/185 is the sum over 1+1/p_i. (1+1/2) * (1+1/2) * (1+1/5) * (1+1/37) = 513/185 is the product over 1+1/p_i. 967/185 + 513/185 = 8 is an integer.
%p A199767 isA199767 := proc(n)
%p A199767     p := ifactors(n)[2] ;
%p A199767     add(op(2,d)+op(2,d)/op(1,d),d=p) + mul((1+1/op(1,d))^op(2,d),d=p) ;
%p A199767     type(%,'integer') ;
%p A199767 end proc:
%p A199767 for n from 20 do
%p A199767     if isA199767(n) then
%p A199767            printf("%d,\n",n);
%p A199767     end if;
%p A199767 end do: # _R. J. Mathar_, Nov 23 2011
%Y A199767 Cf. A003415, A003958, A007850, A198391.
%K A199767 nonn,easy
%O A199767 1,1
%A A199767 _Paolo P. Lava_, Nov 22 2011