Walter Robinson - cp's OEIS frontend

A386936 Numbers that can be represented using their digits in the order of appearance, the operations +, -, *, /, ^, and any parentheses.

1, 2, 3, 4, 5, 6, 7, 8, 9, 343, 736, 1285, 2187, 2592, 2737, 3125, 3685, 3972, 4096, 6455, 11664, 14641, 15552, 15585, 15617, 15618, 15622, 15624, 15626, 15632, 15645, 15655, 15656, 15662, 15667, 15698, 16377, 16384, 17536, 19683, 23328, 24576, 27639

Offset: 1

Anuraag Pasula and Walter Robinson, Aug 09 2025

Each digit is its own operand (no concatenation of digits).
Real and imaginary intermediate values are allowed as long as the final value of the expression is an integer.
Unary minus is not allowed, otherwise we would have 127 = -1 + 2^7. - Sean A. Irvine, Aug 31 2025

			343 = (3+4)^3.
2737 = (2*7)^3-7.
46688 = (4 + 6^6/8)*8.

Walter Robinson, Expressions for Values up to 50000

Cf. A036057, A046469, A080035, A156954, A362769.

A386501 Numbers k divisible by A004719(k), excluding trivial cases.

Original entry on oeis.org

105, 108, 405, 1001, 1005, 1008, 1020, 2002, 2025, 2040, 2050, 3003, 3060, 4004, 4005, 4080, 5005, 6006, 6075, 7007, 7050, 8008, 9009, 10005, 10008, 10020, 10032, 10065, 10098, 10101, 10125, 10206, 10250, 16005, 19008, 20007, 20025, 20040, 20050

Offset: 1

Anuraag Pasula and Walter Robinson, Jul 23 2025

Trivial cases are identified as (1) values of k where there are already no 0s besides leading 0s, like 255 or 1296, such that A004719(k)=k, or (2) where k mod 10 = 0 and k/10 is already in the sequence or is itself a trivial case, like 10080 or 2550. In case (1), k/A004719(k) is equivalent to k/k (as in 255/255). In case (2), k/A004719(k) = 10 * (k/10)/A004719(k/10) when we already know that (k/10)/A004719(k/10) is already an integer (as in 1080/18).
Any number k of the form 1|(at least one 0)|5, such as 105 or 10000000005, will be included in this sequence because k will always be divisible by 3 and 5 due to divisibility rules, and thus will be divisible by A004719(k)=15.
Numbers of form 1|(at least one 0)|8, such as 108 or 10000008, or 4|(at least one 0)|5, such as 405 or 400005, will be included in this sequence for similar reasons.

			A004719(108)=18, 108/18=6.
A004719(9009)=99, 9009/99=91.
A004719(2040)=24, 2040/24=85, 2040 is nontrivial because 204/24=17/2.
50 is trivial because 50/10 = 5, and 5 is trivial because A004719(5)=5.

Subset of A090055.
Cf. A004719, A052382.
Cf. A096093, A090053.

def removeZeros(number):
    stringNum = str(number)
    stringNum = stringNum.replace("0", "")
    return int(stringNum)
for x in range(1, 100000):
    smallInt = removeZeros(x)
    if smallInt == x:
        continue
    if x % smallInt == 0:
        if x % 10 == 0:
            if (x//10) % removeZeros(x//10) == 0:
                continue
        print(x)

A377206 a(n) = ceiling(log(1/n)/log(1 - 1/n)).

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 16, 19, 22, 26, 29, 33, 36, 40, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 84, 88, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 142, 146, 151, 156, 160, 165, 170, 175, 180, 184, 189, 194, 199, 204, 209, 214, 219, 224, 229, 234

Offset: 2

Walter Robinson, Oct 19 2024

This sequence also describes the minimum number of n-player games, where each player has an equal chance of winning, that must be played for a given player to have an equal or greater chance of winning at least once than they have of losing a single game.

Cf. A031435, A094500.

Table[Ceiling[Log[1/n]/Log[1-1/n]], {n,2,58}] (* James C. McMahon, Nov 04 2024 *)

a(n) = A031435(n-1) + 1 for n >= 3.

A375972 a(n) = binomial(n,k(n)), where k(2) = 1, k(n) = k(n-1) + (a(n-1) mod 2).

Original entry on oeis.org

2, 3, 6, 10, 15, 35, 70, 126, 210, 330, 495, 1287, 3003, 6435, 12870, 24310, 43758, 75582, 125970, 203490, 319770, 490314, 735471, 2042975, 5311735, 13037895, 30421755, 67863915, 145422675, 300540195, 601080390, 1166803110, 2203961430, 4059928950, 7307872110, 12875774670, 22239974430

Offset: 2

Walter Robinson, Sep 04 2024

nonn

A way to travel down the Pascal's triangle starting in the center of the third row, where you turn left at every even number, and turn right at every odd number, while descending a row at every step.

Pascal's triangle is symmetric, thus switching the turn directions would not change the sequence.

			a(3) = 3, and k(3) = 1.
To derive a(4), we first find that k(4) = k(3) + (a(3) mod 2) = 1 + (3 mod 2) = 2.
Therefore a(4) = binomial(4,k(4)) = binomial(4,2) = 6.

Robert Israel, Table of n, a(n) for n = 2..3418

Cf. A007318.

Initial conditions of n=0, k(0)=0 would generate A000012.

k[2]:= 1:
for n from 2 to 50 do
  a[n]:= binomial(n, k[n]);
  k[n+1]:= k[n] + (a[n] mod 2);
od:
seq(a[n],n=2..50); # Robert Israel, Jan 27 2025

Module[{k}, FoldList[{Binomial[#2, k = #[[2]] + Mod[#[[1]], 2]], k} &, {2, 1}, Range[3, 50]][[All, 1]]] (* Paolo Xausa, Jan 28 2025 *)

lista(nn) = my(va=vector(nn), vk=vector(nn)); vk[2] = 1; va[2] = binomial(2, vk[2]); for (n=3, nn, vk[n] = vk[n-1] + va[n-1] % 2; va[n] = binomial(n, vk[n]);); vector(nn-1, k, va[k+1]); \\ Michel Marcus, Sep 27 2024

def genNextRow(arr):
    nextRow = []
    nextRow.append(arr[0])
    for i in range(1, len(arr)):
        nextRow.append(arr[i-1]+arr[i])
    nextRow.append(arr[len(arr)-1])
    return nextRow
pascal = [[1],[1,1]]
n = 0
index = 1
while n < 30:
    pascal.append(genNextRow(pascal[1]))
    pascal.pop(0)
    print(pascal[1][index])
    index = index + (pascal[1][index] % 2)
    n += 1

More terms from Michel Marcus, Sep 30 2024

A375864 Prime numbers that cannot be written as the sum of a prime number and a superior highly composite number.

Original entry on oeis.org

2, 3, 307, 911, 1201, 1259, 1693, 2179, 2381, 2927, 3191, 3499, 3557, 4201, 4441, 4721, 5573, 6121, 7207, 8219, 8273, 8537, 8627, 8999, 9137, 9203, 9811, 10133, 10357, 11597, 12211, 12343, 13217, 13421, 13921, 15053, 15401, 15551, 15959, 15991, 16411, 16561, 17117, 17207

Offset: 1

Walter Robinson, Aug 31 2024

nonn

			The prime number 37 can be written as the sum of prime number 31 and superior highly composite number 6 and thus is not in this sequence.

Cf. A000040, A002201, A375866.

from sympy import *
SHCN = [2, 6, 12, 60, 120, 360, 2520, 5040, 55440, 720720]
for x in range(3, 16000, 2):
    waysFound = 0
    if isprime(x):
        iterC = 0
        while iterC < len(SHCN) and SHCN[iterC] < x:
            if isprime(x - SHCN[iterC]):
                waysFound += 1
            iterC += 1
        if waysFound == 0:
            print(x)

A375866 Prime numbers that cannot be written as the sum of a prime number and a highly composite number.

Original entry on oeis.org

2, 1991443, 3637681, 18787511, 2644910729, 4113484691, 5071232129, 7424245487, 10770930019, 10979412541, 13313737777, 17341632041, 17413633157, 17687919407, 19380371939, 20357888099, 21939774683, 24643255853, 28106335199, 29007615421, 30228264713, 32284382539

Offset: 1

Walter Robinson, Aug 31 2024

nonn

			The prime number 5 can be written as the sum of prime number 3 and highly composite number 2 and thus is not in this sequence.

Cf. A000040, A002182.

a(5)-a(22) from Andrew Howroyd, Aug 31 2024

A375762 Maximum number of knights within an n X n chessboard, where each knight has a path to an edge.

Original entry on oeis.org

1, 4, 8, 14, 20, 30, 41, 55

Offset: 1

Walter Robinson, Aug 26 2024

Each knight must be either already on an edge square, or have a path of unoccupied squares which reach an unoccupied edge square (and without any other knights moving).

			For n=3, the following board, with X for each knight, is the unique solution a(3) = 8 and which cannot be 9 since the central square has no move to anywhere within the board.
  XXX
  X-X
  XXX
For n=4, the following is a solution for a(4) = 14, with each of the 4 central knights able to make a single move to one of the unoccupied corner squares.
  -XX-
  XXXX
  XXXX
  XXXX
For n = 8, one 55 knight solution is:
  XXXXXXXX
  XXXXXXXX
  XX-X-XXX
  XX-X-XXX
  -XX---XX
  XXXX-XXX
  XXXXXXXX
  XXXXXXXX

Cf. A335445 (rooks), A337746 (bishops), A337722 (knights moving off the board).

A365583 Numbers k with property that k can be represented by the digits present in k using the operations specified in the comment, and requiring fewer digits than the number of digits in k.

Original entry on oeis.org

1024, 1253, 1287, 1296, 1331, 2048, 2163, 2187, 2435, 2500, 2564, 2568, 2916, 3025, 3125, 3216, 3375, 3437, 3645, 3729, 4088, 4096, 4256, 4375, 4625, 5129, 5243, 6250, 6254, 7128, 7293, 7343, 7776, 8256, 9025, 9216, 9375, 9512, 10003, 10004

Offset: 1

Valentin Miakinen and Walter Robinson, Sep 20 2023

The only operations allowed are addition, subtraction, multiplication, division, exponentiation, parenthesizing, and concatenation.

Real and imaginary intermediate values are allowed as long as the final value of the expression is an integer. - Walter Robinson, Aug 22 2025

			For k = 3125, k can be represented as 5^5, using only 2 digits, which is less than the length of k, 4.
For k = 3437, k can be represented as (7^3)||7, using only 3 digits which is less than the length of k, 4.
For k = 10003, k can be represented as ((1||0)^3)||3, using only 4 digits, which is less than the length of k, 5.

Walter Robinson, Python Program

Cf. A043537, A362769.

Python
```
# See Robinson link.
```

A362769 Minimum number of digits required to represent n only using digits present in n.

Original entry on oeis.org

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

Offset: 1

Oskar Macholl, Valentin Miakinen, and Walter Robinson, May 02 2023

The only operations allowed are addition, subtraction, multiplication, division, exponentiation, and parenthesizing. Concatenation is not allowed.

Real and imaginary intermediate values are allowed as long as the final value of the expression is an integer. - Walter Robinson, Aug 22 2025

			For n = 10, a solution for a(10)=6 is (1+1+1)^(1+1)+1.
For n = 27, a solution for a(27)=4 is 2+(2-7)^2.

Valentin Miakinen, Python program

Cf. A043537.

Python
```
# See Miakinen link.
```

Edited by N. J. A. Sloane, Oct 16 2023.

A359013 Numbers k that can be written as the sum of a perfect square and a factorial in exactly 3 distinct ways.

Original entry on oeis.org

145, 46249, 63121, 42916624, 18700677890064, 28112213204100, 41654823930457982576640000, 445860623276908458083942400, 666474080134036599385635225600

Offset: 1

Walter Robinson, Dec 11 2022

This does not count x^2 and (-x)^2 as distinct, nor does it count 0! and 1! as distinct.

a(10) > 10^30 if it exists. - David A. Corneth, Dec 11 2022

			145 = 5^2 + 5! = 11^2 + 4! = 12^2 + 1!.

Cf. A000142, A000290.

Subset of A358071.

import math
for x in range(1, 120000000):
    total = 0
    prod = 1
    factInc = 2
    while prod <= x:
        sq = math.sqrt(x - prod)
        if sq % 1 == 0:
            total = total + 1
        prod = prod * factInc
        factInc = factInc + 1
    if total == 3:
        print(x)

a(5)-a(9) from David A. Corneth, Dec 11 2022

cp's OEIS Frontend

User: Walter Robinson

A386936 Numbers that can be represented using their digits in the order of appearance, the operations +, -, *, /, ^, and any parentheses.

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A386501 Numbers k divisible by A004719(k), excluding trivial cases.

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Python

A377206 a(n) = ceiling(log(1/n)/log(1 - 1/n)).

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Mathematica

Formula

A375972 a(n) = binomial(n,k(n)), where k(2) = 1, k(n) = k(n-1) + (a(n-1) mod 2).

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A375864 Prime numbers that cannot be written as the sum of a prime number and a superior highly composite number.

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Python

A375866 Prime numbers that cannot be written as the sum of a prime number and a highly composite number.

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Author

Keywords

Examples

Crossrefs

Extensions

A375762 Maximum number of knights within an n X n chessboard, where each knight has a path to an edge.

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Keywords

Comments

Examples

Crossrefs

A365583 Numbers k with property that k can be represented by the digits present in k using the operations specified in the comment, and requiring fewer digits than the number of digits in k.

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Crossrefs

Programs

Python

A362769 Minimum number of digits required to represent n only using digits present in n.

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