A350744 - OEIS

A350744

Numbers m such thatA061078(m)/A061077(m) = 4/5.

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

(list; graph; refs; listen; history; text; internal format)

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

Table of n, a(n) for n=1..61.

Luca Onnis,On the general Smarandache's sigma product of digits,arXiv:2203.07227 [math.GM], 2022.

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 factA061078(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

Cf.A061076,A061077,A061078.

Sequence in context:A063284 A257222 A092454*A248359 A008587 A008706

Adjacent sequences:A350741 A350742 A350743*A350745 A350746 A350747

KEYWORD

nonn

AUTHOR

Luca Onnis,Mar 20 2022

STATUS

approved