Luke Voyles - cp's OEIS frontend

User: Luke Voyles

Luke Voyles has authored 4 sequences.

A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

3, 4, 5, 1, 6, 8, 6, 8, 10, 2, 12, 16, 9, 12, 15, 3, 10, 18, 7, 14, 17, 12, 16, 20, 4, 17, 22, 3, 18, 24, 18, 19, 21, 11, 15, 27, 15, 20, 25, 4, 24, 32, 18, 24, 30, 6, 20, 36, 14, 28, 34, 2, 17, 40, 6, 32, 33, 21, 28, 35, 16, 23, 41, 5, 30, 40, 3, 36, 37, 27, 30, 37, 24, 32, 40, 8, 34, 44, 29, 34, 44, 6, 36, 48, 12, 19, 53, 27, 36, 45, 36, 38, 42

Offset: 1

Luke Voyles, Nov 05 2024

nonn

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.

The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.

			3^3+4^3+5^3=6^3
1^3+6^3+8^3=9^3
6^3+8^3+10^3=12^3
2^3+12^3+16^3=18^3
9^3+12^3+15^3=18^3

A. Russell and C. E. Gwyther, The Partition of Cubes, The Mathematical Gazette, Vol. 21, No. 242 (Feb., 1937), pp. 33-35 (3 pages).

The sum of each cubic number triple produce the sequence A023042. The comments produce another method to produce an infinite number of cubic number triples whose sum equals a cube that the method shown by Shiraishi according to A226903. The comments discuss qualities of Pythagorean triples A103606 and Pythagorean quadruples A096907. The title's structure drew inspiration from A291694.

If a^3+b^3+c^3=d^3, then any specific number k that has a zero as the last digit will make k(d^3) another cubic number through the formula k(a^3)+k(b^3)+k(c^3)=k(d^3)

A344990 Primes of the form m^k +- k, with m,k > 1.

Original entry on oeis.org

2, 5, 7, 11, 23, 37, 47, 61, 67, 79, 83, 167, 223, 227, 359, 439, 443, 503, 509, 521, 727, 839, 997, 1019, 1087, 1091, 1223, 1367, 1523, 1847, 2027, 2207, 2399, 2741, 3023, 3251, 3719, 3967, 4093, 4099, 4759, 5039, 5623, 5927, 6553, 6563, 6569, 7919, 8179

Offset: 1

Luke Voyles, Jun 05 2021

nonn

			    2 =  2^2 - 2,
    5 =  2^3 - 3,
    7 =  3^2 - 2,
   11 =  3^2 + 2,
   23 =  5^2 - 2,
   37 =  2^5 + 5,
   47 =  7^2 - 2,
   61 =  4^3 - 3,
   67 =  4^3 + 3,
   79 =  9^2 - 2,
   83 =  9^2 + 2,
  167 = 13^2 - 2,
  223 = 15^2 - 2,
  227 = 15^2 + 2,
  359 = 19^2 - 2,
  439 = 21^2 - 2,
  443 = 21^2 + 2,
  503 =  2^9 - 9,
  509 =  8^3 - 3,
  521 =  2^9 + 9,
  ...

Union of A099227 and A099228.

A344744 a(n) is the n-th power of the concatenation of the integers from 0 through n-1.

Original entry on oeis.org

0, 1, 1728, 228886641, 2861381721051424, 3539537889086624823140625, 437104634676747795452235896466702336, 5396563761318393964062660689603780554533710504641, 6662458388479360230805308787387369820914640828074410829911019008

Offset: 1

Luke Voyles, May 27 2021

			a(1) = 0^1 = 0;
a(2) = 01^2 = 1;
a(3) = 012^3 = 1728;
a(4) = 0123^4 = 228886641.

Cf. A007778, A007908, A061250.

a[n_] := FromDigits[Join @@ IntegerDigits @ Range[0, n - 1]]^n; Array[a, 9] (* Amiram Eldar, May 29 2021 *)

def a(n): return int("".join(str(i) for i in range(n)))**n
print([a(n) for n in range(1, 10)]) # Michael S. Branicky, May 29 2021

a(n) = A007908(n-1)^n.

A344555 Numbers k such that the infinite sequence of digits consisting of the final digit of k^m for m = 2, 3, 4, ... is the same as the sequence of digits obtained by concatenating infinitely many copies of k.

Original entry on oeis.org

0, 1, 5, 6, 11, 19, 55, 64, 66, 111, 555, 666, 1111, 1919, 4268, 4862, 5555, 6464, 6666, 9317, 9713, 11111, 55555, 66666, 111111, 191919, 555555, 646464, 666666, 1111111, 5555555, 6666666, 11111111, 19191919, 42684268, 48624862, 55555555, 64646464, 66666666, 93179317

Offset: 1

Luke Voyles, May 22 2021

The numbers k of this sequence repeat from k^2 onward.  For example, if number ends in 8, the last digit of the square of k will always be 4, the last digit of the cube of k will always be 2, the last digit of the fourth power of k will always be 6, and the fifth power of k will always be 8.  Base numbers that end in 0, 1, 5, and 6 will always result in numbers with the same digit when they have positive integers as exponents.  For k that have a 2 in the ones place, then the square of k will have 4 in the ones place, the cube of k will have 8 in the ones place, the fourth power of k will have 6 in the ones place, and the fifth power of k will have 2 in the ones place.
For any integer k, the ones digit of each higher power, i.e., k^2, k^3, k^4, etc., depends only on the ones digit of k as follows:
.
Ones digits in k and larger powers of k
--+------------------------------------  Resulting string of
k | k^2 k^3 k^4 k^5 k^6 k^7 k^8 k^9 ...  concatenated digits
--+------------------------------------  -------------------
0 |  0   0   0   0   0   0   0   0  ...  0000000000000000...
1 |  1   1   1   1   1   1   1   1  ...  1111111111111111...
2 |  4   8   6   2   4   8   6   2  ...  4862486248624862...
3 |  9   7   1   3   9   7   1   3  ...  9713971397139713...
4 |  6   4   6   4   6   4   6   4  ...  6464646464646464...
5 |  5   5   5   5   5   5   5   5  ...  5555555555555555...
6 |  6   6   6   6   6   6   6   6  ...  6666666666666666...
7 |  9   3   1   7   9   3   1   7  ...  9317931793179317...
8 |  4   2   6   8   4   2   6   8  ...  4268426842684268...
9 |  1   9   1   9   1   9   1   9  ...  1919191919191919...
.
This sequence consists of each of the nonnegative integers that, when repeated infinitely, yields one of the digit strings in the column at the right.

			The patterns that I have noticed and seen confirmed demonstrate that the infinite patterns that result with the end digits of exponents when n has a particular numerical value from k^2, k^3, k^4, and k^5 before they repeat are as follows: k with final digit 0 (0000); k with final digit 1 (1111); k with final digit 2 (4862); k with final digit 3 (9713); k with final digit 4 (6464); k with final digit 5 (5555); k with final digit 6 (6666); k with final digit 7 (9317); k with final digit 8 (4268); and k with final digit 9 (1919).
Therefore, the number 64 infinitely repeats because 64^2 equals 4096 (which ends in 6), 64^3 equals 262144 (which ends in 4), 64^4 equals 16777216 (which ends in 6), and 1073741824 (which ends in 4). 64 repeated twice in the previous demonstration, but all numbers infinitely repeat in the same way.
Additionally, 4862^2=23639044 (ends in 4), 4862^3=114933031928 (ends in 8), 4862^4=558804401233936 (ends in 6), and 4862^5=2716906998799396832 (ends in 2). The 4862 sequence among the final digits of the power for 4862 then continues infinitely as 4862^6 ends in 4, 4862^7 ends in 8, 4862^8 ends in 6, 4862 ends in 2, and so on.
One interesting fact about this sequence is that only the last digit of an odd-numbered power of k is necessary to determine the last digit of k itself.

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User: Luke Voyles

A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

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A344990 Primes of the form m^k +- k, with m,k > 1.

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A344744 a(n) is the n-th power of the concatenation of the integers from 0 through n-1.

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A344555 Numbers k such that the infinite sequence of digits consisting of the final digit of k^m for m = 2, 3, 4, ... is the same as the sequence of digits obtained by concatenating infinitely many copies of k.

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