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A336826
Bogotá numbers: numbers k such that k = m*p(m) where p(m) is the digital product of m.
17
0, 1, 4, 9, 11, 16, 24, 25, 36, 39, 42, 49, 56, 64, 75, 81, 88, 93, 96, 111, 119, 138, 144, 164, 171, 192, 224, 242, 250, 255, 297, 312, 336, 339, 366, 378, 393, 408, 422, 448, 456, 488, 497, 516, 520, 522, 525, 564, 575, 648, 696, 704, 738, 744, 755, 777, 792
OFFSET
1,3
COMMENTS
Named Bogotá numbers by Tomás Uribe and Juan Pablo Fernández based on similarity of the construction to the Colombian numbers (A003052).
Some questions about these numbers:
(i) Some Bogotá numbers occur in pairs (such as 24 and 25). Are there infinitely many such pairs?
(ii) More generally, can arbitrarily long sets of consecutive numbers be found all of which are Bogotá numbers?
(iii) Can the gap between two consecutive Bogotá numbers be arbitrarily large? Answer: Yes.
From David A. Corneth, Aug 06 2020: (Start)
The only primes in this sequence are A004022.
To see if a number is a Bogotá number, we only have to look at its 7-smooth divisors. Proof: If a number k is a Bogotá number then k = m*p(m) where p(m) is 7-smooth as it's a product of digits. Furthermore, if k = m*p(m) then p(m) | k. Q.e.d. Below is an example using this idea.
To find Bogotá numbers k up to N we can make a list of 7-smooth numbers up to sqrt(N) and list the factorizations into single-digit numbers of each of these 7-smooth numbers that when concatenated give m such that m * p(m) = k where p(m) is that 7-smooth number.
For example, 10 is a 7-smooth number. Its factorizations into single-digit numbers are 2*5, 5*2, 1*2*5 and so on. This tells us that 10*25 = 250, 10*52 = 520, 10*125 = 1250 all are Bogotá numbers.
Similarily we can find odd Bogotá numbers to then find consecutive Bogotá numbers (See A336864). (End)
LINKS
Sean A. Irvine, Java program (github)
Math Stackexchange, Gaps between Bogotá numbers, 2020.
Puzzling Stackexchange, Pairs of Bogotá numbers, 2020.
EXAMPLE
From David A. Corneth, Aug 06 2020: (Start)
520 is a term because 52 * p(52) = 52 * 10 = 520.
Example using we only have to look at 7-smooth divisors:
520 is a term as its 7-smooth divisors d are 1, 2, 4, 5, 8, 10, 20, 40. values 520/d are 520, 260, 130, 104, 65, 52, 26, 13 of which 52 * (5*2) = 520 where (5*2) are the products of 52. (End)
PROG
(PARI) f(n) = vecprod(digits(n))*n; \\ A098736
isok(n) = my(k=0); for (k=0, n, if (f(k) == n, return(1))); \\ Michel Marcus, Aug 06 2020
(PARI) is(n) = { my(f = factor(n), s7 = 1, d, sl = sqrtint(n)); for(i = 1, #f~, if(f[i, 1] > 7, break ); s7 *= f[i, 1]^f[i, 2]; ); d = divisors(s7); for(i = 1, #d, if(d[i] > sl, return(0)); if(n/d[i] * vecprod(digits(n/d[i])) == n, return(1); ) ); 0 } \\ David A. Corneth, Aug 06 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Sean A. Irvine, Aug 05 2020
STATUS
approved