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A350744
Numbers m such that A061078(m)/A061077(m) = 4/5.
0
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 51, 52, 53, 54, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 101, 102, 103, 104, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 151, 152, 153, 154, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 201, 202, 203, 204, 205, 210, 215, 220, 225
OFFSET
1,1
COMMENTS
All positive multiples of 5 are terms of the sequence.
REFERENCES
Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.
LINKS
FORMULA
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.
Limit_{n->oo} A061078(n)/A061077(n) = 4/5.
EXAMPLE
30 is a term, in fact A061078(30)=320, A061077(30)=400 and a(n) = 320/400 = is 4/5.
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.
MATHEMATICA
Flatten[Position[(Accumulate[Times @@@ IntegerDigits[Range[2, 10000, 2]]]/
Accumulate[Times @@@ IntegerDigits[Range[1, 9999, 2]]]), 4/5]]
PROG
(PARI) pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
isok(k) = sum(i=1, k, pd(2*i))/sum(i=1, k, pd(2*i-1)) == 4/5; \\ Michel Marcus, Mar 21 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Luca Onnis, Mar 20 2022
STATUS
approved