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%I #44 Mar 22 2022 18:49:06
%S 5,10,15,20,25,30,35,40,45,50,51,52,53,54,55,60,65,70,75,80,85,90,95,
%T 100,101,102,103,104,105,110,115,120,125,130,135,140,145,150,151,152,
%U 153,154,155,160,165,170,175,180,185,190,195,200,201,202,203,204,205,210,215,220,225
%N Numbers m such that A061078(m)/A061077(m) = 4/5.
%C All positive multiples of 5 are terms of the sequence.
%D Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.
%H Luca Onnis, <a href="https://arxiv.org/abs/2203.07227">On the general Smarandache's sigma product of digits</a>, arXiv:2203.07227 [math.GM], 2022.
%F Let k be a positive integer not divisible by 5 and j >= 0; then 5*k*10^j, 5*k*10^j+1, ..., 5*k*10^j+(5/9)*(10^j-1) are all terms of the sequence.
%F Limit_{n->oo} A061078(n)/A061077(n) = 4/5.
%e 30 is a term, in fact A061078(30)=320, A061077(30)=400 and a(n) = 320/400 = is 4/5.
%e 500, 501, 502, ..., 554, 555 are all terms. In fact 500=5*10^2 and for the formula above also 501, ..., 500+(5/9)*(10^2-1) = 555 are all terms of the sequence.
%t Flatten[Position[(Accumulate[Times @@@ IntegerDigits[Range[2, 10000, 2]]]/
%t Accumulate[Times @@@ IntegerDigits[Range[1, 9999, 2]]]), 4/5]]
%o (PARI) pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
%o isok(k) = sum(i=1, k, pd(2*i))/sum(i=1, k, pd(2*i-1)) == 4/5; \\ _Michel Marcus_, Mar 21 2022
%Y Cf. A061076, A061077, A061078.
%K nonn
%O 1,1
%A _Luca Onnis_, Mar 20 2022