Gavin Lupo - cp's OEIS frontend

A384632 a(0)=0. For each digit d in the sequence, let a(n) equal the smallest unused integer which has at least d divisors.

0, 1, 2, 3, 4, 6, 12, 5, 7, 16, 24, 8, 18, 9, 10, 30, 11, 36, 48, 13, 14, 15, 17, 19, 20, 21, 28, 22, 40, 23, 25, 26, 27, 29, 32, 31, 42, 33, 60, 34, 35, 37, 38, 39, 54, 41, 43, 44, 45, 46, 49, 47, 50, 51, 52, 53, 56, 55, 72, 57, 58, 62, 59, 63, 61, 64, 65

Offset: 0

Gavin Lupo, Jun 05 2025

			a(0) = 0.
a(1) = Smallest unused integer with at least 0 divisors = 1.
a(2) = Smallest unused integer with at least 1 divisor = 2.
a(3) = Smallest unused integer with at least 2 divisors = 3.
a(4) = Smallest unused integer with at least 3 divisors = 4.
a(5) = Smallest unused integer with at least 4 divisors = 6.
a(6) = Smallest unused integer with at least 6 divisors = 12.
a(7) = Smallest unused integer with at least 1 divisor = 5.
a(8) = Smallest unused integer with at least 2 divisors = 7.

Cf. A000005, A362371, A362551.

from sympy import divisor_count
a = [0]
for n in range(100):
    if a[n] >= 10:
        split = [int(d) for d in str(a[n])]
    else:
        split = [a[n]]
    for s in split:
        num = 1
        while True:
            if divisor_count(num) >= s and num not in a:
                a.append(num)
                break
            num += 1
print(a)

A371620 Fractal sequence based on a heptagon: a(n) = number of unobscured heptagons remaining after completing step n.

Original entry on oeis.org

1, 1, 8, 48, 266

Offset: 1

Gavin Lupo, Mar 29 2024

Start with a single regular heptagon. The heptagon fractal is constructed by taking with the previous completed step, and overlapping it in its entirety to each vertex (excluding the top vertex as they overlap completely). Once constructed, any unobscured heptagons within are counted. "Unobscured" means no lines cross through it, and its shape is regular. See links for images.

			Start with a(1) = 1 unobscured heptagon. For the next step, take step 1's entire shape and place it at each vertex. Although the total number of heptagons is 7, only 1 unobscured heptagon remains, so a(2) = 1.

Gavin Lupo, Heptagon Construction, the initial steps
Gavin Lupo, Step 5

Cf. A371536.

A371536 a(n) is the number of unobscured pentagons remaining after completing step n.

Original entry on oeis.org

1, 1, 6, 41, 186

Offset: 1

Gavin Lupo, Mar 26 2024

Start with a single regular pentagon. The pentagon fractal is constructed by taking with the previous completed step, and overlapping it in its entirety to each vertex (excluding the top vertex as they overlap completely). Once constructed, any unobscured pentagons within are counted. "Unobscured" means no lines cross through it, and its shape is regular. See links for images.

			Start with a(1) = 1 unobscured pentagon. For the next step, take step 1's entire shape and place it at each vertex. Although the total number of pentagons is 5, only 1 unobscured pentagon remains, so a(2) = 1.

Gavin Lupo, Pentagon Construction, the initial steps
Gavin Lupo, Step 5

A363654 Lexicographically earliest sequence of positive integers such that the n-th pair of identical terms encloses exactly a(n) terms.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 1, 2, 4, 3, 4, 5, 3, 6, 7, 4, 3, 8, 6, 9, 7, 8, 4, 10, 3, 7, 4, 6, 8, 9, 11, 10, 12, 3, 9, 13, 7, 3, 11, 9, 14, 15, 13, 16, 17, 7, 18, 3, 19, 20, 11, 14, 9, 20, 17, 15, 19, 20, 18, 21, 11, 22, 3, 23, 24, 25, 26, 14, 17, 9, 27, 28, 29, 15, 26, 19

Offset: 1

Eric Angelini and Gavin Lupo, Jun 13 2023

nonn

Pairs are numbered according to the position of the second term.

			a(1) = 1. The 1st constructed pair encloses 1 term:  [1, 2, 1].
a(2) = 2. The 2nd constructed pair encloses 2 terms: [2, 1, 3, 2].
a(3) = 1. The 3rd constructed pair encloses 1 term:  [3, 2, 3].
a(4) = 3. The 4th constructed pair encloses 3 terms: [1, 3, 2, 3, 1].
a(5) = 2. The 5th constructed pair encloses 2 terms: [2, 3, 1, 2].
a(6) = 3. The 6th constructed pair encloses 3 terms: [3, 1, 2, 4, 3].
a(7) = 1. The 7th constructed pair encloses 1 term:  [4, 3, 4].
...

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Eric Angelini, A seq with a strange behavior (or not)?
Neal Gersh Tolunsky, Scatterplot of first 100000 terms

Cf. A026272, A363708, A363757.

from itertools import count, islice
def agen(): # generator of terms
    a, indexes = [1], {1: 0}
    yield a[-1]
    for i in count(0):
        num = 1
        while True:
            if num in indexes:
                if (len(a) - indexes[num]) == (a[i]+1):
                    an = num; indexes[an] = len(a); a.append(an); yield an
                    break
                else:
                    num += 1
            else:
                an = max(a)+1; indexes[an] = len(a); a.append(an); yield an
                num = 1
print(list(islice(agen(), 100))) # Gavin Lupo and Michael S. Branicky, Jun 13 2023

A363247 a(0) = 0. a(n) = the smallest nonnegative integer, excluding a(n-1), not occurring in all occurrences of a(n-1)'s eight neighbors thus far in a square spiral.

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 5, 2, 4, 5, 3, 6, 0, 7, 1, 8, 0, 9, 2, 8, 6, 4, 7, 8, 10, 0, 11, 1, 10, 2, 11, 4, 12, 2, 13, 3, 9, 5, 12, 3, 14, 0, 15, 1, 14, 5, 13, 4, 15, 2, 16, 1, 17, 0, 18, 1, 19, 0, 20, 1, 21, 0, 22, 1, 23, 0, 24, 1, 25, 0, 26, 1, 27, 0, 28, 2, 17, 3

Offset: 0

Gavin Lupo, May 24 2023

nonn

			The spiral begins:
  0---8---1---7---0
  |               |
  9   4---3---2   6
  |   |       |   |
  2   1   0---1   3   .
  |   |           |   .
  8   5---2---4---5   .
  |                   |
  6---4---7---8---10--0
a(0) = 0
a(1) = 1. Smallest integer that's not a neighbor to every 0 = 1.
a(2) = 2. Smallest integer that's not a neighbor to every 1 = 2.
a(3) = 3. Smallest integer that's not a neighbor to every 2 = 3.
a(4) = 4. Smallest integer that's not a neighbor to every 3 = 4.
a(5) = 1. Smallest integer that's not a neighbor to every 4 = 1.

Scott R. Shannon, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 10000 terms on the square spiral. The value colors are scaled across the spectrum from red to violet to show their relative size. Zoom in to see the numbers.

Cf. A275609, A358048.

A363083 a(0)=a(1)=1. For n>1, if the number of occurrences of a(n-1) is less than abs(a(n-1)), then a(n)=a(n-1)-a(n-2). Otherwise, a(n)=a(n-1)+a(n-2).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 3, -2, -5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 2, 7, 5, -2, 3, 1, 4, 3, 7, 4, -3, -7, -4, 3, -1, 2, 1, 3, 4, 1, 5, 4, 9, 5, 14, 9, -5, -14, -9, 5, -4, -9, -5, 4, -1, 3, 2, 5, 7, 2, 9, 7, -2, 5, 3, 8, 5, 13, 8, -5, -13, -8, 5, -3, 2, -1, 1, 0

Offset: 0

Gavin Lupo, May 18 2023

			a(0) = 1
a(1) = 1
a(2) = 2. Two 1's in the list so far.    2 > abs(1).   1 + 1 = 2.
a(3) = 1. One 2 in the list so far.      1 < abs(2).   2 - 1 = 1.
a(4) = 3. Three 1's in the list so far.  3 > abs(1).   1 + 2 = 3.

Gavin Lupo, Table of n, a(n) for n = 0..50000
Gavin Lupo, Image of the first 100000 terms.
Gavin Lupo, Image of the first 1000000 terms.
Gavin Lupo, Image of the first 10000000 terms.
Gavin Lupo, Image of the first 25000000 terms.

Cf. A329985, A362746, A362890, A363086.

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    anprev, an, c = 1, 1, Counter([1])
    yield 1
    while True:
        yield an
        c[an] += 1
        anprev, an = an, an-anprev if c[an] < abs(an) else an+anprev
print(list(islice(agen(), 80))) # Michael S. Branicky, May 18 2023

A363086 a(0)=a(1)=1. For n>1, let c=count of all occurrences of a(n-1) in the list so far. If c < abs(a(n-1)), then a(n)=c-a(n-1). Otherwise, a(n)=c.

Original entry on oeis.org

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

Offset: 0

Gavin Lupo, May 18 2023

This is a variant of A363083.

			a(0) =  1
a(1) =  1
a(2) =  2. Two 1's in the list so far.  2 > abs(1).      c =  2.
a(3) = -1. One 2 in the list so far.    1 < abs(2).  1 - 2 = -1.
a(4) =  1. One -1 in the list so far.   1 = abs(-1).     c =  1.

Gavin Lupo, Table of n, a(n) for n = 0..50000
Gavin Lupo, Image of the first 1000000 terms

Cf. A363083.

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an, c = 1, Counter([1, 1])
    yield from [1, 1]
    while True:
        an = c[an]-an if c[an] < abs(an) else c[an]
        c[an] += 1
        yield an
print(list(islice(agen(), 80))) # Michael S. Branicky, May 19 2023

A362896 a(0)=2. For n>0, let d = n-th digit in the sequence thus far. a(n) = a(n-1) + d if d is even. Otherwise, a(n) = a(n-1) - d.

Original entry on oeis.org

2, 4, 8, 16, 15, 21, 20, 15, 17, 16, 18, 18, 17, 12, 11, 4, 3, 9, 8, 16, 15, 23, 22, 15, 14, 16, 15, 14, 18, 15, 6, 14, 13, 19, 18, 13, 15, 12, 14, 16, 15, 10, 9, 13, 12, 18, 17, 12, 11, 15, 14, 22, 21, 16, 22, 21, 25, 24, 21, 20, 11, 10, 18, 17, 14, 13, 8, 7

Offset: 0

Gavin Lupo, May 09 2023

"-" signs on negative values are ignored when determining the n-th digit.

			a(0) =  2.
a(1) =  4. 1st digit is 2, which is even. 2 + 2 = 4.
a(2) =  8. 2nd digit is 4, which is even. 4 + 4 = 8.
a(3) = 16. 3rd digit is 8, which is even. 8 + 8 = 16.
a(4) = 15. 4th digit is 1, which is odd.  16 - 1 = 15.
a(5) = 21. 5th digit is 6, which is even. 15 + 6 = 21.

Gavin Lupo, Table of n, a(n) for n = 0..10000

Cf. A362371, A362551.

a = [2]
split = []
for i in range(100):
    split += [int(j) for j in str(abs(a[i]))]
    if split[i] % 2 == 0:
        a.append((a[i] + split[i]))
    else:
        a.append((a[i] - split[i]))
print(a)

A362746 a(1)=a(2)=1; a(n)=The count of all occurrences in the list so far where integer a(n-1) appears adjacent to integer a(n-2).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 3, 1, 1, 4, 1, 2, 3, 2, 3, 3, 2, 4, 1, 3, 2, 5, 1, 1, 6, 1, 2, 4, 2, 3, 6, 1, 3, 3, 4, 1, 4, 4, 2, 4, 4, 4, 5, 1, 2, 5, 2, 3, 7, 1, 1, 8, 1, 2, 6, 1, 4, 5, 2, 4, 5, 3, 1, 4, 6, 1, 5, 3, 2, 8, 1, 3, 5, 3, 4, 2, 6, 2, 3, 9, 1, 1, 10, 1, 2, 7

Offset: 1

Gavin Lupo, May 01 2023

			a(1) = 1.
a(2) = 1.
a(3) = 2. How many 1's so far are adjacent to a 1? = 2.
a(4) = 1. How many 2's so far are adjacent to a 1? = 1.
a(5) = 2. How many 1's so far are adjacent to a 2? = 2.
a(6) = 2. How many 2's so far are adjacent to a 1? = 2.

Gavin Lupo, Table of n, a(n) for n = 1..10000
Gavin Lupo, Image of the first 100000 terms.

Cf. A342585, A355271, A362890.

K = {1, 1}; While[Length@K < 87, A = Position[K, Last@K]; c = 0; For[a = 1, a <= Length@A, a++, If[K[[A[[a]] - 1]] == {K[[Length@K - 1]]} || K[[A[[a]] + 1]] == {K[[Length@K - 1]]}, c++]]; AppendTo[K, c]]; Print[K] (* Samuel Harkness, May 08 2023 *)

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    aprev, an, anext, c = 0, 1, 1, Counter({(1, 1)})
    while True:
        aprev, an, anext = an, anext, c[an, anext]
        c[an, anext] += 1
        if aprev != anext: c[anext, an] +=  1
        yield an
print(list(islice(agen(), 100))) # Michael S. Branicky, May 02 2023

A362190 Triangle read by rows: T(n,k) is the smallest integer not already in the same row or column and also not diagonally adjacent to an equal integer.

Original entry on oeis.org

0, 1, 2, 3, 0, 1, 2, 4, 3, 0, 5, 1, 2, 4, 3, 4, 3, 0, 1, 2, 5, 6, 5, 4, 3, 0, 1, 2, 7, 8, 6, 2, 4, 3, 0, 1, 9, 10, 5, 7, 1, 2, 4, 3, 0, 8, 6, 9, 10, 5, 0, 1, 2, 4, 3, 10, 7, 8, 6, 9, 4, 3, 0, 1, 2, 5, 11, 9, 10, 5, 7, 6, 8, 4, 3, 0, 1, 2

Offset: 1

Gavin Lupo, Apr 10 2023

This sequence is A025581 without diagonal adjacent restriction.

			Triangle begins:
  n\k|  1  2  3  4  5  6  7  8 ...
  ---+----------------------------
   1 |  0
   2 |  1  2
   3 |  3  0  1
   4 |  2  4  3  0
   5 |  5  1  2  4  3
   6 |  4  3  0  1  2  5
   7 |  6  5  4  3  0  1  2
   8 |  7  8  6  2  4  3  0  1

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Cf. A025581.

\\ here f(S) gives smallest not in set S.
f(S)={for(k=0, oo, if(!setsearch(S,k), return(k)))}
T(n)={my(M=matrix(n,n)); for(i=2, n, for(j=1, i, my(S=Set(concat([M[j..i-1,j]~, M[i,1..j-1], M[i-1,max(1,j-1)], M[i-1,min(j+1,i-1)]]))); M[i,j]=f(S))); M}
{ my(A=T(10)); for(i=1, #A, print(A[i, 1..i])) } \\ Andrew Howroyd, Apr 10 2023

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User: Gavin Lupo

A384632 a(0)=0. For each digit d in the sequence, let a(n) equal the smallest unused integer which has at least d divisors.

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A371620 Fractal sequence based on a heptagon: a(n) = number of unobscured heptagons remaining after completing step n.

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A371536 a(n) is the number of unobscured pentagons remaining after completing step n.

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A363654 Lexicographically earliest sequence of positive integers such that the n-th pair of identical terms encloses exactly a(n) terms.

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Python

A363247 a(0) = 0. a(n) = the smallest nonnegative integer, excluding a(n-1), not occurring in all occurrences of a(n-1)'s eight neighbors thus far in a square spiral.

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A363083 a(0)=a(1)=1. For n>1, if the number of occurrences of a(n-1) is less than abs(a(n-1)), then a(n)=a(n-1)-a(n-2). Otherwise, a(n)=a(n-1)+a(n-2).

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Python

A363086 a(0)=a(1)=1. For n>1, let c=count of all occurrences of a(n-1) in the list so far. If c < abs(a(n-1)), then a(n)=c-a(n-1). Otherwise, a(n)=c.

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A362896 a(0)=2. For n>0, let d = n-th digit in the sequence thus far. a(n) = a(n-1) + d if d is even. Otherwise, a(n) = a(n-1) - d.

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A362746 a(1)=a(2)=1; a(n)=The count of all occurrences in the list so far where integer a(n-1) appears adjacent to integer a(n-2).

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