A179239 -id:A179239 - cp's OEIS frontend

A306406 Numbers k such that the sum of the distinct prime divisors of the product of all legal permutations of the digits of k is equal to k-1.

1, 6, 102, 543, 37008

Offset: 1

Christopher Hohl, Feb 13 2019

It appears that these constitute all of the 'non-degenerate' cases for k less than 10^8.  That is, k is not allowed to have leading zeros, but all 'legal' permutations of k, where k has length m, must be of length m as well. Therefore leading zeros are allowed in the construction of the total permutation product.
From David A. Corneth, Feb 15 2019: (Start)
Let S(m) be the sum of the distinct prime divisors of the product of all legal permutations of the digits of m.
Let Z(m) be a number where a zero is inserted after the first digit of m (m > 0). For example, Z(1) = 10, Z(19) = 109.
All terms with at most k digits can be found by iterating only over terms in A179239 with at most k digits.
For example, 345 is in A179239. S(345) = S(543), namely 543. As 543 is a permutation of 345, s = 543 is in the sequence.
Similarily, 445 is in A179239 and S(445) = 341, 445 doesn't produce a term. As S(445) = S(454) = S(544), all these number don't produce a term and do not have to be checked.
We have S(Z(m)) >= S(m). Proof: The permutations of Z(m) give the same distinct prime factors as m does, and maybe more. Therefore, S(Z(m)) >= S(m).
This can be applied to eliminate candidates. For example, S(10378) = 1447642. The largest possible value a number with digits of Z(10378) = 100378 can have is 873100. But 1447642 > 873100. So 100378 can't produce a term and doesn't have to be checked.
To perhaps quickly eliminate a candidate without checking all permutations one might let the last digit d of a permutation be such that gcd(d, 10) = 1 to hopefully get large prime factors (if there is such d). For example, when checking if 1378 gives a candidate, start with the 12 permutations ending in 1 or 3.
To find S(m) where m has digits zero, one might use a known value of S(m') where m' has a digit 0 removed from m and proceed finding S(m) with permutations having leading nonzero digits only. (End)

			6 is a term because it is the product of its legal permutations. The distinct prime factors of 6 are 3 and 2, so 3+2=5 and 6-1=5.
102 is a term because the distinct prime factors of the product of its legal permutations, i.e., 102*120*210*201*21*12 = 130195900800, are 2,3,5,7,17, and 67. So, 2+3+5+7+17+67=101, and 102-1=101.

Cf. A179239.

A328517 Primitive sequence underlying A075053.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 3, 1, 1, 1, 3, 0, 1, 1, 2, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 3, 1, 4, 2, 1, 0, 0, 1, 0, 2, 0, 0, 1, 2, 1, 2, 1, 2, 0, 0, 2, 0, 0, 1, 2, 1, 3, 0, 0, 1, 0, 0, 0, 2, 1, 4, 2, 1, 2, 5, 0, 2, 3, 3, 9, 3, 3, 3, 6, 3, 5, 2, 5, 4, 4, 2, 6, 3, 3, 8, 6, 5, 7, 11

Offset: 0

David A. Corneth, Oct 18 2019

A179239 lists the least number that has its permutation of digits.

			a(34) = 4 as A075053(A179239(34)) = A075053(37) = 4. The four primes embedded in 37 according to A075053 are {3, 7, 37, 73}.

Cf. A075053.

a(n) = A075053(A179239(n)).

A357131 Numbers m such that A010888(m) = A031347(m) = A031286(m) = A031346(m); only the least of the anagrams are considered.

Original entry on oeis.org

0, 137, 11126, 111134, 111278, 1111223, 11111447, 111112247, 1111122227, 111111111137, 11111111111126, 111111111111134, 1111111111111223, 111111111111111111111111111111111111111111111111111111111111111111111111111111111111278

Offset: 1

Mohammed Yaseen, Sep 14 2022

The infinite sequence {(R_1)37, (R_91)37, (R_991)37, (R_9991)37, (R_99991)37, ...} is a subsequence, where R_k is the repunit of length k. Hence this sequence is infinite.
a(14) <= (R_84)278.
Some additional terms < 10^100: (R_84)278, (R_86)447, (R_86)2247, (R_86)22227, (R_91)37, (R_93)26, (R_94)34, (R_93)278, (R_94)223, (R_95)447, (R_95)2247, (R_95)22227.
If a term k > 0 then k cannot contain a digit 0 as if it does A031347(k) = 0 while A031346(k) = 1, contradicting equality. - David A. Corneth, Sep 15 2022
a(14) > 10^50. - Michael S. Branicky, Sep 16 2022
From Michael S. Branicky, Sep 17 2022: (Start)
(R_{10^k})37 and (R_{2*10^k - 10})37 also form infinite subsequences for k >= 0.
Indeed, terms of the form (R_k)e form infinite subsequences for each e in 26, 34, 37, 223, 278, 447, 2247, 22227 for k such that A007953(k + A007953(e)) = 2.
a(2)-a(14) and all terms of the forms above have 2 = A010888(m) = A031347(m) = A031286(m) = A031346(m).
(R_{t})277 where t+2+7+7 = 4 followed by 55555 9's is a term with 4 = A010888(m) = A031347(m) = A031286(m) = A031346(m).
Likewise, there exists a term of the form (R_{t})5579 with 5 = A010888(m) = A031347(m) = A031286(m) = A031346(m), where t+26 is part of the additive persistence chain ending ..., 5999999, 59, 14, 5. Likewise for 888899 and 6, and so on.
However, there are no terms with A010888(m) = A031347(m) = A031286(m) = A031346(m) = 1 or 3. (End)

			137 is in the sequence as A010888(137) = 137 mod 9 = 2, A031347(137) = 2 via 1*3*7 = 21 -> 2*1 = 2 < 10, A031286(137) = 2 via 1+3+7 = 11 -> 1+1 = 2 so two steps, A031346(137) = 2 via 1*3*7 = 21 -> 2*1 = 2 so two steps. As all these outcomes are 2, 137 is a term. - _David A. Corneth_, Sep 15 2022

Eric Weisstein's World of Mathematics, Additive Persistence
Eric Weisstein's World of Mathematics, Digital Root
Eric Weisstein's World of Mathematics, Multiplicative Digital Root
Eric Weisstein's World of Mathematics, Multiplicative Persistence

Subsequence of A179239.
Cf. A010888, A031347, A031286, A031346.
Cf. A064702, A239427.

a(8)-a(13) from Pontus von Brömssen, Sep 14 2022

a(14) confirmed by Michael S. Branicky, Sep 17 2022

A374349 Integers >=0 whose decimal digits are topologically distinct from those of any smaller number.

Original entry on oeis.org

0, 1, 8, 10, 11, 18, 40, 48, 88, 100, 101, 108, 111, 118, 188, 400, 408, 488, 888, 1000, 1001, 1008, 1011, 1018, 1088, 1111, 1118, 1188, 1888, 4000, 4008, 4088, 4888, 8888, 10000, 10001, 10008, 10011, 10018, 10088, 10111, 10118, 10188, 10888, 11111, 11118

Offset: 1

Charles L. Hohn, Jul 05 2024

Assumes 0 without a slash or a center dot, closed 4, 6, and 9, and no overlapping of multiple digits. Digits homologous to a (flattened) sphere: 1, 2, 3, 5, 7; to a torus: 0, 4, 6, 9; to a double torus: 8. Sequence is a run of the terms in ascending numeric order.

All topologically distinct terms can be represented by nondecreasing sequences of strings of 0s, 1s, and 8s. However, terms cannot begin with 0. Therefore, if a string has 0s, then (i) if there are any 1s, one of them moves to the front, (ii) else, the first 0 is replaced with 4. Sequence is the resulting strings sorted as base-10 numbers. - Michael S. Branicky, Jul 11 2024

			0 is homologous to 1 torus, so a(1)=0.
1 is homologous to 1 sphere, so a(2)=1.
2 is homologous to 1 sphere, same as 1, so it is not in the sequence.
4 is homologous to 1 torus, same as 0, so it is not in the sequence.
8 is homologous to 1 double torus, so a(3)=8.
10 is homologous to 1 sphere and 1 torus, so a(4)=10.
11 is homologous to 2 spheres, so a(5)=11.
14 is homologous to 1 sphere and 1 torus, same as 10, so it is not in the sequence.
41 is homologous to 1 sphere and 1 torus, same as 10, so it is not in the sequence.

Cf. A179239, A064692, A249572.

df(d, c)=(10^c-1)/9*d
n=0; a=0; at=1; while(true, a++; at+=a+1; ac=0; for(b=0, a, for(c=0, b, n++; print(n, " ", if(n<=2, n-1, ac+b-c+1

from itertools import count, islice, combinations_with_replacement as cwr
def agen(): # generator of terms
    after = {"1":"018", "4":"08", "8":"8"}
    yield from (0, 1, 8)
    for digits in count(2):
        for first in "148":
            for rest in cwr(after[first], digits-1):
                yield int(first + "".join(rest))
print(list(islice(agen(), 50))) # Michael S. Branicky, Jul 07 2024

A381509 Numbers whose nonzero digits are in nondecreasing order and any zeros appear at the end.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 40, 44, 45, 46, 47, 48, 49, 50, 55, 56, 57, 58, 59, 60, 66, 67, 68, 69, 70, 77, 78, 79, 80, 88, 89, 90, 99, 100, 110, 111, 112, 113, 114, 115, 116, 117

Offset: 1

Keenin D. Krehbiel, Feb 25 2025

This sequence includes all non-negative integers where non-zero digits (1-9) are in non-decreasing order and zeros are at the end.

Each term is a unique multiset of digits in canonical form.

			112 is in the sequence because 1 <= 1 <= 2.
120 is in the sequence because 1 <= 2, then 0.
21 is not in the sequence because 2 > 1.
102 is not in the sequence because the zero is not at the end.

A variant of A179239.

from itertools import combinations_with_replacement as cwr, count, islice
def agen(): # generator of terms
    yield 0
    for d in count(1):
        yield from sorted(int(f+"".join(mc)) for f in "123456789" for mc in cwr([str(i) for i in range(int(f), 10)]+["0"], d-1))
print(list(islice(agen(), 1000))) # Michael S. Branicky, Apr 11 2025

cp's OEIS Frontend

A306406 Numbers k such that the sum of the distinct prime divisors of the product of all legal permutations of the digits of k is equal to k-1.

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A328517 Primitive sequence underlying A075053.

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A357131 Numbers m such that A010888(m) = A031347(m) = A031286(m) = A031346(m); only the least of the anagrams are considered.

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A374349 Integers >=0 whose decimal digits are topologically distinct from those of any smaller number.

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A381509 Numbers whose nonzero digits are in nondecreasing order and any zeros appear at the end.

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Python