Polydivisible number

Let n be a positive integer, and let k = \lfloor \log_{b}{n} \rfloor + 1 be the number of digits in n written in base b. The number n is a polydivisible number if for all 1 \leq i \leq k, : \left\lfloor\frac{n}{b^{k - i}}\right\rfloor \equiv 0 \pmod i. ; Example For example, 10801 is a seven-digit polydivisible number in base 4, as : \left\lfloor\frac{10801}{4^{7 - 1}}\right\rfloor = \left\lfloor\frac{10801}{4096}\right\rfloor = 2 \equiv 0 \pmod 1, : \left\lfloor\frac{10801}{4^{7 - 2}}\right\rfloor = \left\lfloor\frac{10801}{1024}\right\rfloor = 10 \equiv 0 \pmod 2, : \left\lfloor\frac{10801}{4^{7 - 3}}\right\rfloor = \left\lfloor\frac{10801}{256}\right\rfloor = 42 \equiv 0 \pmod 3, : \left\lfloor\frac{10801}{4^{7 - 4}}\right\rfloor = \left\lfloor\frac{10801}{64}\right\rfloor = 168 \equiv 0 \pmod 4, : \left\lfloor\frac{10801}{4^{7 - 5}}\right\rfloor = \left\lfloor\frac{10801}{16}\right\rfloor = 675 \equiv 0 \pmod 5, : \left\lfloor\frac{10801}{4^{7 - 6}}\right\rfloor = \left\lfloor\frac{10801}{4}\right\rfloor = 2700 \equiv 0 \pmod 6, : \left\lfloor\frac{10801}{4^{7 - 7}}\right\rfloor = \left\lfloor\frac{10801}{1}\right\rfloor = 10801 \equiv 0 \pmod 7. ==Enumeration==

For any given base b, there are only a finite number of polydivisible numbers. Maximum polydivisible number The following table lists maximum polydivisible numbers for some bases b, where represent digit values 10 to 35. Estimate for Fb(n) and Σ(b) Let n be the number of digits. The function F_b(n) determines the number of polydivisible numbers that has n digits in base b, and the function \Sigma(b) is the total number of polydivisible numbers in base b. If k is a polydivisible number in base b with n - 1 digits, then it can be extended to create a polydivisible number with n digits if there is a number between bk and b(k + 1) - 1 that is divisible by n. If n is less or equal to b, then it is always possible to extend an n - 1 digit polydivisible number to an n-digit polydivisible number in this way, and indeed there may be more than one possible extension. If n is greater than b, it is not always possible to extend a polydivisible number in this way, and as n becomes larger, the chances of being able to extend a given polydivisible number become smaller. On average, each polydivisible number with n - 1 digits can be extended to a polydivisible number with n digits in \frac{b}{n} different ways. This leads to the following estimate for F_{b}(n): :F_b(n) \approx (b - 1)\frac{b^{n-1}}{n!}. Summing over all values of n, this estimate suggests that the total number of polydivisible numbers will be approximately :\Sigma(b) \approx \frac{b - 1}{b}(e^{b}-1) ==Specific bases==

All numbers are represented in base b, using A−Z to represent digit values 10 to 35. Base 2 Base 3 Base 4 Base 5 The polydivisible numbers in base 5 are :1, 2, 3, 4, 11, 13, 20, 22, 24, 31, 33, 40, 42, 44, 110, 113, 132, 201, 204, 220, 223, 242, 311, 314, 330, 333, 402, 421, 424, 440, 443, 1102, 1133, 1322, 2011, 2042, 2200, 2204, 2231, 2420, 2424, 3113, 3140, 3144, 3302, 3333, 4022, 4211, 4242, 4400, 4404, 4431, 11020, 11330, 13220, 20110, 20420, 22000, 22040, 22310, 24200, 24240, 31130, 31400, 31440, 33020, 33330, 40220, 42110, 42420, 44000, 44040, 44310, 110204, 113300, 132204, 201102, 204204, 220000, 220402, 223102, 242000, 242402, 311300, 314000, 314402, 330204, 333300, 402204, 421102, 424204, 440000, 440402, 443102, 1133000, 1322043, 2011021, 2042040, 2204020, 2420003, 2424024, 3113002, 3140000, 3144021, 4022042, 4211020, 4431024, 11330000, 13220431, 20110211, 20420404, 24200031, 31400004, 31440211, 40220422, 42110202, 44310242, 132204314, 201102110, 242000311, 314000044, 402204220, 443102421, 1322043140, 2011021100, 3140000440, 4022042200 The smallest base 5 polydivisible numbers with n digits are :1, 11, 110, 1102, 11020, 110204, 1133000, 11330000, 132204314, 1322043140, none... The largest base 5 polydivisible numbers with n digits are :4, 44, 443, 4431, 44310, 443102, 4431024, 44310242, 443102421, 4022042200, none... The number of base 5 polydivisible numbers with n digits are :4, 10, 17, 21, 21, 21, 13, 10, 6, 4, 0, 0, 0... Base 10 The polydivisible numbers in base 10 are :1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 102, 105, 108, 120, 123, 126, 129, 141, 144, 147, 162, 165, 168, 180, 183, 186, 189, 201, 204, 207, 222, 225, 228, 243, 246, 249, 261, 264, 267, 282, 285, 288... The smallest base 10 polydivisible numbers with n digits are :1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800, ... The largest base 10 polydivisible numbers with n digits are :9, 98, 987, 9876, 98765, 987654, 9876545, 98765456, 987654564, 9876545640, 98765456405, 987606963096, 9876069630960, 98760696309604, 987606963096045, 9876062430364208, 98485872309636009, 984450645096105672, 9812523240364656789, 96685896604836004260, ... The number of base 10 polydivisible numbers with n digits are :9, 45, 150, 375, 750, 1200, 1713, 2227, 2492, 2492, 2225, 2041, 1575, 1132, 770, 571, 335, 180, 90, 44, 18, 12, 6, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... ==Programming example==

The example below searches for polydivisible numbers in Python. def find_polydivisible(base: int) -> list[int]: """Find polydivisible number.""" numbers = [] previous = [i for i in range(1, base)] new = [] digits = 2 while not previous == []: numbers.append(previous) for n in previous: for j in range(0, base): number = n * base + j if number % digits == 0: new.append(number) previous = new new = [] digits = digits + 1 return numbers ==Related problems==

Polydivisible numbers represent a generalization of the following well-known problem in recreational mathematics: : Arrange the digits 1 to 9 in order so that the first two digits form a multiple of 2, the first three digits form a multiple of 3, the first four digits form a multiple of 4 etc. and finally the entire number is a multiple of 9. The solution to the problem is a nine-digit polydivisible number with the additional condition that it contains the digits 1 to 9 exactly once each. There are 2,492 nine-digit polydivisible numbers, but the only one that satisfies the additional condition is :381 654 729 Other problems involving polydivisible numbers include: • Finding polydivisible numbers with additional restrictions on the digits - for example, the longest polydivisible number that only uses even digits is :48 000 688 208 466 084 040 • Finding palindromic polydivisible numbers - for example, the longest palindromic polydivisible number is :30 000 600 003 • A common, trivial extension of the aforementioned example is to arrange the digits 0 to 9 to make a 10 digit number in the same way, the result is 3816547290. This is a pandigital polydivisible number. ==References==