Benjamin Knight - cp's OEIS frontend

0, 15, 8, 10, 11, 12, 7, 5, 6, 0, 7, 6, 6, 5, 9, 5, 6, 4, 5, 15, 8, 6, 4, 5, 8, 5, 4, 5, 7, 8, 7, 7, 6, 8, 7, 6, 6, 5, 6, 10, 7, 6, 6, 5, 6, 6, 5, 7, 6, 11, 6, 6, 6, 4, 5, 7, 4, 5, 4, 12, 4, 6, 7, 7, 8, 4, 4, 5, 3, 7, 4, 5, 6, 7, 6, 4, 5, 6, 4, 5, 6, 5, 4, 4, 6, 5, 5, 4, 7, 6, 4, 5, 4, 6, 4, 4, 4, 6

Offset: 1

Benjamin Knight, Oct 15 2018

7 is the only fixed point less than 10.

			For n=1, a(1) = 0, because all powers are identical to 1, and it is not possible to get any other digits.
For n=3, a(3) = 8 because the following are needed to use all nonzero digits: 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^8 = 6561.
For n=10, a(10) = 0, because one can only get digits 0 and 1.

Cf. A090493.

a[n_] := If[IntegerQ[Log10[n]], 0, Module[{s={0}, m=1}, While[Length[s]<10, s=DeleteDuplicates@Catenate[{s,IntegerDigits[n^m]}]; m++]; m-1]]; Array[a,100] (* Amiram Eldar, Nov 14 2018 *)

a(n) = {if (10^valuation(n, 10) == n, return(0)); v = []; kn = n; for (m=1, oo, v = Set(concat(v, digits(n))); v = select(x->(x>0), v); if (#v == 9, return (m)); n *= kn;);} \\ Michel Marcus, Oct 17 2018

a(n) = {if(vecsum(digits(n))==1, return(0)); my(v = vector(9), todo = 9, t = n); for(i=1, oo, d=digits(t); for(j = 1, #d, if(d[j] > 0 && v[d[j]] == 0, todo--; v[d[j]] = 1; if(todo <= 0, return(i)))); t*=n)} \\ David A. Corneth, Oct 17 2018

def A320572(n):
    n = int(str(n).strip('0'))
    if n != 1:
        s = set(range(1, 10))
        a = 0
        m = 1
        while s:
            a += 1
            m *= n
            s.difference_update(int(z) for z in str(m))
        return a
    else:
        return 0

a(n) = a(10*n).

a(n) <= A090493(n). - Rémy Sigrist, Oct 16 2018

A304207 a(1)=17; for n>1, a(n) = (a(n-1)^2 - 1)/2 if n is even, a(n-1) + 1 if n is odd.

Original entry on oeis.org

17, 144, 145, 10512, 10513, 55261584, 55261585, 1526921388356112, 1526921388356113, 1165744463109679828308252234384, 1165744463109679828308252234385, 679480076635437837059150531810555804350472205781672488164112, 679480076635437837059150531810555804350472205781672488164113

Offset: 1

Benjamin Knight, May 08 2018

nonn

{17, 144, 145}, {145, 10512, 10513}, {10513, 55261584, 55261585}, ... are sides {a < b < c} of the right triangles, with hypotenuse c = b + 1.

Eric Chen, Table of n, a(n) for n = 1..21

Same basic form as A076601, A076602, A076603, and A076604.

nxt[{n_,a_}]:={n+1,If[OddQ[n],(a^2-1)/2,a+1]}; NestList[nxt,{1,17},20][[All,2]] (* Harvey P. Dale, Mar 27 2021 *)

a(n) = if(n==1,17,if(n%2,a(n-1)+1,(a(n-1)^2 - 1)/2)) \\ Eric Chen, Jun 09 2018

A304120 a(n) is the least k such that there is a permutation of the positive integers from 1 through k for which every pair of adjacent terms sums to a perfect power with exponent n.

Original entry on oeis.org

2, 15, 305, 6479

Offset: 1

Benjamin Knight, May 06 2018

From Jordan D Fredette, May 28 2019: (Start)
I was helped immensely by my friend Christian Burns who is an expert programmer.
I also worked with my friend Josiah Findley. Without him I likely would have never figured this out.
Essentially, I discovered that when arranging the numbers 1..k so that any two next to each other add up to a perfect power, we need not check every single permutation.
In fact, the best method is to list all the different ways to add up two numbers to get each of the powers which are greater than 1 but less than 2k.
If any number shows up only once in this list, then it must be an endpoint.
Once the endpoints are determined, any number that only shows up twice must be connected to the two other numbers with which it is paired in the list.
This will result in strings of numbers which must be a part of the permutation.
If the same number is at the end of two different strings then those strings must be joined.
Thus, the number of permutations to be checked is further decreased.
Basically, Christian was able to implement this as a program and found that 6479 is the smallest k for which the numbers 1..k can be arranged so that any two next to each other add up to a perfect fourth power. (End)

			a(1) = 2 as the permutation (1, 2) of the integers has the property that the adjacent terms sum to a power of 1.
a(2) = 15 as the permutation of the positive integers 1 through 15 [8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9] has the property that every pair of adjacent terms sums to a power with exponent n = 2 (a square).

Moritz Firsching, Arranging numbers from 1 to n such that the sum of every two adjacent numbers is a perfect power, MathOverflow.
Carlos Rivera, Puzzle 311: Sum to a cube

For n=2, see A090461.

Sage
```
See MathOverflow link.
```

a(4) from Jordan D Fredette, May 28 2019

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User: Benjamin Knight

A327152 r values of Triphosian primes.

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A307694 b values of Triphosian primes.

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A307692 g values of Triphosian primes.

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A320572 The smallest integer m such that each nonzero digit appears in the decimal representation of the sequence n^1, n^2, ..., n^x, where 1 <= x <= m, or 0 if no such m exists.

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A304207 a(1)=17; for n>1, a(n) = (a(n-1)^2 - 1)/2 if n is even, a(n-1) + 1 if n is odd.

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A304120 a(n) is the least k such that there is a permutation of the positive integers from 1 through k for which every pair of adjacent terms sums to a perfect power with exponent n.

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