Aidan Clarke - cp's OEIS frontend

A358431 a(0) = 1; a(n+1) = 1 if a(n) > n, otherwise a(n+1) = a(n) + a(a(n)).

1, 1, 2, 4, 1, 2, 4, 5, 7, 12, 1, 2, 4, 5, 7, 12, 16, 32, 1, 2, 4, 5, 7, 12, 16, 32, 1, 2, 4, 5, 7, 12, 16, 32, 48, 1, 2, 4, 5, 7, 12, 16, 32, 48, 1, 2, 4, 5, 7, 12, 16, 32, 48, 55, 1, 2, 4, 5, 7, 12, 16, 32, 48, 55, 57, 62, 110, 1, 2, 4, 5, 7, 12, 16, 32, 48, 55, 57, 62, 110

Offset: 0

Aidan Clarke, Nov 15 2022

nonn

Cf. A062039.

a[0] = 1; a[n_] := a[n] = If[a[n - 1] > n - 1, 1, a[n - 1] + a[a[n - 1]]]; Array[a, 80, 0] (* Amiram Eldar, Dec 06 2022 *)

from functools import cache
@cache
def a(n): return 1 if n==0 else (1 if a(n-1) > n-1 else a(n-1) + a(a(n-1)))
print([a(n) for n in range(76)]) # Michael S. Branicky, Nov 15 2022

A356675 Lexicographically earliest infinite sequence satisfying a(1) > -1 and a(n-1) = A075826(a(n)).

Original entry on oeis.org

1, 5, 9, 16, 27, 38, 48, 58, 66, 76, 87, 98, 117, 136, 155, 177, 198, 215, 235, 254, 275, 295, 310, 333, 350, 372, 394, 411, 433, 452, 474, 495, 514, 535, 555, 576, 598, 615, 635, 650, 669, 689, 705, 728, 749, 773, 795, 810, 833, 850, 872, 894, 913, 934, 950, 973, 994, 1013, 1034, 1050, 1071, 1093

Offset: 1

Aidan Clarke, Aug 22 2022

There is at least one infinite sequence that satisfies the parameters of the sequence, because it is always true that A075826(k) < k for any k. (For the statement to be false, the number of letters must be less than or equal to 0.)

All known values in this sequence are guaranteed to be in any infinite sequence which can be produced, since iterating A075826(k) where the starting value of k is any number greater than 11126 converges at 6890 or larger. This proves any infinite sequence must contain 6890 and all values returned by iterating A075826 down to 0.

			A075826 yields the number of letters in US English spelling of each number subtracted from the number's value. For example, A075826(5) yields 1, because 5 - 4 (F-I-V-E) is 1. Our formula reveals that if 5 is in our sequence, it must come immediately after 1, which it does. However, 4, which must come immediately after 0, is not in our sequence, which we can prove because there exists no number k such that A075826(k) = 4. Each number less than the largest known value in this sequence produces a finite sequence unless it is in this sequence, because some number eventually is a dead end like 4.

Michel Marcus, Table of n, a(n) for n = 1..11623

Cf. A005589, A075826.

f(k) = k - A005589(k);
listd(nn) = {nn *= 2; my(vs = vector(nn)); my(list = List()); my(m=1, lbound = 0); listput(list, m); while (m < nn, if (vs[m] == 0, vs[m] = Vec(select(x->(x==m), vector(100, k, f(k+m-1)), 1)); if (#vs[m], vs[m] = apply(x->(x+m-1), vs[m]))); my(ok = 1, vc = vs[m]); if (! #vc, ok = 0, vc = select(x->(x>lbound), vc); if (! #vc, ok = 0);); if (!ok, lbound = list[#list]; listpop(list); if (! #list, return()); m = list[#list];, lbound = 0; m = vc[1]; listput(list, m););); Vec(select(x->(x<=nn/2), list));} \\ Michel Marcus, Aug 31 2022

a(n-1) = A075826(a(n)).

A338174 The largest value that integer n is connected to in a web where every integer n has n connections, each integer is connected to the lowest values possible, and new values are added in phases.

Original entry on oeis.org

2, 3, 5, 7, 10, 11, 13, 16, 20, 25, 21, 22, 23, 24, 25, 26, 28, 31, 35, 40, 46, 54, 64, 76, 90, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 80, 82, 84, 86, 88, 90, 92, 96, 102, 110, 120, 132, 146

Offset: 1

Aidan Clarke, Nov 18 2020

nonn

The concept of this sequence is based on a web where every possible connection in the web is made with the numbers written so far, and then new numbers are added for every connection that needs to be made.

			Start with a map of nothing but 1.
   1 has 0 connections.
So we add 2 and connect it.
   1 has 1 connection (2)
   2 has 1 connection (1)
Now we need one more number to connect with 2, so we add 3.
   2 has 2 connections (1,3)
   3 has 1 connection (2)
Now because 3 needs two more numbers, we add 4 and 5 at once, and make as many connections as possible.
   3 has 3 connections (2,4,5)
   4 has 2 connections (3,5)
   5 has 2 connections (3,4)
We need five more numbers because 4 lacks two connections, and 5 lacks three connections, so we add 6,7,8,9,10.
   4 has 4 connections (3,5,6,7)
   5 has 5 connections (3,4,8,9,10)
   6 has 5 connections (4,7,8,9,10)
   7 has 5 connections (4,6,8,9,10)
   8 has 5 connections (5,6,7,9,10)
   9 has 5 connections (5,6,7,8,10)
   10 has 5 connections (5,6,7,8,9)
This continues indefinitely.

A337804 Lexicographically earliest triangle of nonnegative integers read by rows such that for each pair (x,y) != (0,0), there is at most one pair (n,k) such that T(n,k) = T(n+x,k+y).

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 0, 3, 4, 0, 3, 5, 2, 6, 3, 2, 7, 8, 5, 1, 9, 1, 0, 9, 10, 11, 7, 2, 6, 4, 12, 13, 14, 15, 0, 8, 9, 11, 16, 17, 18, 19, 20, 6, 5, 5, 15, 21, 22, 23, 24, 25, 21, 3, 10, 8, 1, 3, 26, 27, 28, 29, 7, 16, 1, 4, 2, 19, 30, 31, 32, 33, 34, 35, 30, 2, 12, 11

Offset: 1

Aidan Clarke, Sep 22 2020

Each value is determined by placing the least possible nonnegative integer that will abide by the rules of the sequence.

			Triangle begins:
0;
0, 0;
1, 2, 1;
0, 3, 4, 0;
3, 5, 2, 6, 3;
2, 7, 8, 5, 1, 9;
...

Rémy Sigrist, Table of n, a(n) for n = 1..10011 (rows for n = 1..141, flattened)
Rémy Sigrist, Colored representation of the first 500 rows (where the hue is function of T(n,k))
Rémy Sigrist, Colored scatterplot of (x, y) such that T(n, k) = T(n+x, k+y) and max(n, n+x) <= 500 and (x, y) <> (0, 0) (where the hue is function of T(n, k))
Rémy Sigrist, PARI program for A337804

Cf. A337226 (linear version).

T(n)={my(v=vector(n), S=Set(), L=List());
  for(n=1, #v, v[n]=vector(n); for(k=1, n, my(i=1);
    while(i<=#L, my(P=Set([[n-p[1], k-p[2]] | p<-L[i]])); if(!#setintersect(P,S), S = setunion(S,P); break); i++);
    if(i>#L, listput(L, []));
    L[i] = concat(L[i], [[n,k]]);
    v[n][k] = i-1 )); v
}
concat(T(12)) \\ Andrew Howroyd, Sep 24 2020

PARI
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See Links section.
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Terms a(46) and beyond from Andrew Howroyd, Sep 24 2020

A334146 Numbers with at least two prime factors greater than 3 counted with multiplicity.

Original entry on oeis.org

25, 35, 49, 50, 55, 65, 70, 75, 77, 85, 91, 95, 98, 100, 105, 110, 115, 119, 121, 125, 130, 133, 140, 143, 145, 147, 150, 154, 155, 161, 165, 169, 170, 175, 182, 185, 187, 190, 195, 196, 200, 203, 205, 209, 210, 215, 217, 220, 221, 225, 230, 231, 235, 238, 242

Offset: 1

Aidan Clarke, Apr 16 2020

			25 = 5*5
35 = 5*7
49 = 7*7
50 = (2)*5*5
55 = 5*11
65 = 5*13
70 = (2)*5*7
75 = (3)*5*5

Cf. A093641.

Select[Range[250], CompositeQ[# / 2^IntegerExponent[#, 2] / 3^IntegerExponent[#, 3]] &] (* Amiram Eldar, May 01 2020 *)

is(n)=n>>=valuation(n,2);n/=3^valuation(n,3);n>1 && !isprime(n) \\ Charles R Greathouse IV, Apr 16 2020

a(n) ~ n. - Charles R Greathouse IV, Apr 16 2020

A333614 Lexicographically first sequence of nonnegative integers such that no two terms share all but one digit in common, nor does interchanging any two adjacent digits in any term change it into another term of the sequence.

Original entry on oeis.org

0, 11, 22, 33, 44, 55, 66, 77, 88, 99, 102, 110, 121, 134, 145, 153, 167, 178, 186, 201, 212, 220, 235, 243, 254, 268, 276, 287, 304, 313, 326, 330, 341, 352, 365, 379, 398, 403, 414, 425, 432, 440, 451, 469, 497, 506, 515, 523, 531, 542, 550, 564, 589, 605, 616, 624, 637, 648, 659

Offset: 1

Aidan Clarke, Mar 28 2020

The rules of this sequence are designed to decrease the likelihood of confusing any two numbers in the sequence. This would be ideal, for example, in the case of generating account numbers.

			a(1) = 0 because it is the smallest nonnegative integer.
a(2) = 11 because all integers between 1 and 10 only differ from 0 by one digit.
a(3) = 22 because all integers between 12 and 21 only differ from 11 by one digit.
a(11) = 102 because 101 becomes 011 when interchanging the first two adjacent digits.

Rémy Sigrist, PARI program for A333614

Cf. A008585 (binary analog).

PARI
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See Links section.
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A322523 a(n) is the least nonnegative integer k for which there does not exist i < j with i+j=n and a(i)=a(j)=k.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 2, 2, 0, 2, 1, 0, 1, 2, 0, 3, 2, 0, 3, 1, 0, 1, 3, 0, 3, 3, 0, 3, 1, 0, 1, 3, 0, 2, 2, 0, 2, 1, 0, 1, 2, 0, 4, 3, 0, 4, 1, 0, 1, 4, 0, 4, 3, 0, 4, 1, 0, 1, 4, 0, 2, 2, 0, 2, 1, 0, 1, 2, 0, 4, 4, 0, 4, 1, 0, 1, 4, 0, 4, 4, 0, 4, 1, 0, 1, 4, 0, 2, 2, 0, 2, 1, 0, 1, 2, 0, 3, 4, 0

Offset: 1

Aidan Clarke, Aug 28 2019

If x is an integer that we are checking whether it is an option for a(n), at position n = 3(3^(x+1)-1)/2 there appears to begin a repeating sequence (containing 3^(x+1) terms) of whether it can or cannot be used for a(n) that continues infinitely.

The variant where we drop the condition "i < j" corresponds to A007814. - Rémy Sigrist, Sep 06 2019

			a(1) = 0.
a(2) = 0.
a(3) = 1 (because a(1) and a(2) both equal 0).
a(5) = 1 (because a(1) and a(4) both equal 0).
a(8) = 2 (because a(1) and a(7) equal 0, and a(3) and a(5) equal 1).

Jianing Song, Table of n, a(n) for n = 1..10000 (first 995 terms from Aidan Clarke)

Cf. A007814, A033627, A218579.

for n from 1 to 100 do
  forbid:= {seq(A[i],i= select(i -> A[i]=A[n-i],[$1..(n-1)/2]))};
  if forbid = {} then A[n]:= 0 else A[n]:= min({$0..max(forbid)+1} minus forbid) fi;
od:
seq(A[i],i=1..100); # Robert Israel, Sep 06 2019

least(v, n) = {my(found = []); for (i=1, n, if (i >= n-i, break, if (v[i] == v[n-i], found = Set(concat(found, v[i]))));); if (#found == 0, return(0)); my(m = vecmax(found)); for (i=0, m, if (!vecsearch(found, i), return (i))); return (m+1);}
lista(nn) = {my(v = vector(nn)); for (n=1, nn, v[n] = least(v, n);); v;} \\ Michel Marcus, Sep 07 2019

a(n) = my(v=valuation(n,3)); n=n/3^v; if(n==2 || n%3==1, v, A215879((n-5)/3)+1+v) \\ Jianing Song, Aug 23 2022; see A215879 for its program

def A322523(n):
    c, m = 0, n
    while not (a:=divmod(m,3))[1]:
        c += 1
        m = a[0]
    if m==2 or m%3==1: return c
    m = (m+1)//3-2
    while (a:=divmod(m,3))[1]:
        c += 1
        m = a[0]
    return c+1 # Chai Wah Wu, Oct 15 2022

a(n) = 0 iff n belongs to A033627. - Rémy Sigrist, Sep 06 2019
From Jianing Song, Aug 23 2022: (Start)
Properties of this sequence:
(1) a(n) = a(n/3) + 1 for n == 0 (mod 3).
Proof is based on induction on n. Obviously a(n) != 0. For 1 <= e <= a(n/3), there exists i+j = n/3, i < j such that a(i) = a(j) = e-1, by induction hypothesis we have a(3*i) = a(3*j) = e, so a(n) != e. a(n) = a(n/3) + 1 is OK. Suppose otherwise that n = i+j, i < j and a(i) = a(j) = a(n/3) + 1 > 0, then i,j == 0 (mod 3), by induction hypothesis we have a(i/3) = a(j/3) = a(n/3), a contradiction.
(2) a(n) = A215879((n-5)/3) + 1 for n == 2 (mod 3) and n > 2.
If A215879((n-5)/3) = t, then n = 3*(Sum_{0<=r<=t-1} d_r*3^r + O(3^(t+1))) + 5, d_r = 1 or 2. Suppose that a(m) = A215879((m-5)/3) + 1 for m == 2 (mod 3) and 2 < n < m.
Obviously a(n) != 0. From (1) we know that a(3^e) = a(2*3^e) = t. For 1 <= e <= t, n - d_e*3^e = 3*(Sum_{0<=r<=t-1, r!=e-1} d_r*3^r + O(3^(t+1))) + 5 = 3*(Sum_{0<=r<=e-2} d_r*3^r + O(3^e)) + 5, so A215879((n-d_e*3^e-5)/3) = e-1. Since 5 <= n - d_e*3^e == 2 (mod 3), by induction hypothesis we have a(n-d_e*3^e) = e = a(d_e*3^e), so a(n) != e.
a(n) = t+1 is OK. Suppose otherwise that n = i+j, i != j and a(i) = a(j) = t+1 > 0, then {i,j} == {0,2} (mod 3) and i,j > 2. Suppose that i == 2 (mod 3), by induction hypothesis A215879((i-5)/3) = t. Write i = 3*(Sum_{0<=r<=t-1} d'_r*3^r + O(3^(t+1))) + 5, 1 <= d'_r <= d_r. If d'_r = d_r for all r, then j = n-i is divisible by 3^(t+2), so a(j) >= t+2, a contradiction. If d'_r != d_r for some r, then j = 3^(r_0+1) + O(3^(r_0+2)) where r_0 is the smallest index such that d'_r != d_r, so a(j) = r_0+1 + a(1+O(3^1)) = r_0+1 <= t, also a contradiction.
Recursive formulas:
a(n) = 0 for n = 2 or n == 1 (mod 3);
a(n) = 1 for n == 5 (mod 9);
a(n) = a(n/3) + 1 for n == 0 (mod 3);
a(n) = a((n+4)/3) + 1 for n == 2 (mod 9) and n > 2;
a(n) = a((n+7)/3) + 1 for n == 8 (mod 9).
Let A_1 = {3}, B_1 = {5}, A_{t+1} = {3*n: n in A_t, B_t}, B_{t+1} = {3*n-7, 3*n-4: n in B_t}, then for t >= 1, {n: a(n) = t} = (Union_{k>=0} {n+k*3^(t+1): n in A_t, B_t}) U {2*3^t}.
General formula: write n = s*3^t, gcd(s,3) = 1, then a(n) = t if s = 2 or s == 1 (mod 3), A215879((s-5)/3) + 1 + t otherwise. (End)

A299790 Positive integers that do not equal any number minus its number of factorizations.

Original entry on oeis.org

3, 9, 15, 21, 26, 29, 34, 35, 38, 39, 43, 45, 50, 51, 54, 57, 62, 69, 74, 86, 87, 90, 98, 101, 103, 107, 110, 111, 114, 118, 123, 124, 134, 135, 137, 142, 146, 151, 152, 155, 158, 161, 163, 170, 173, 174, 179, 186, 188, 189, 191, 194, 197, 206, 214, 218, 220, 221, 227, 229, 230, 231, 237, 239, 242, 244, 246, 248

Offset: 1

Aidan Clarke, Jan 21 2019

nonn

Cf. A001055 (number of factorizations).

fcnt(n, m) = {local(s); s=0; if(n == 1, s=1, fordiv(n, d, if(d > 1 & d <= m, s=s+fcnt(n/d, d)))); s};
f(n) = n - fcnt(n, n);
is(n) = j=1;x=2^(floor(log(n)/log(2))+2);while(jJinyuan Wang, Feb 14 2019

More terms from Jinyuan Wang, Feb 14 2019

A317888 The minimum value in the central cell of an n-dimensional cube, three cells wide in each dimension, such that each cell is the product of its adjacent cells, each cell is a whole number, and no two cells have the same value.

Original entry on oeis.org

6, 120, 362880, 355687428096000, 8683317618811886495518194401280000000, 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000

Offset: 1

Aidan Clarke, Aug 10 2018

We can prove that the corner cells are filled with values from 2 to 2^n+1, because in order to prove that none of the corner cells have to skip a value so as not to repeat a value within any of the other cells, we must prove that the non-corner cells all have values greater than 2^n+1 in at least one arrangement of corner cell values. 2(2^n-n+2) can describe the lowest value of cells in between two corner cells in every dimension from one to infinity, excluding the third dimension, because in the ideal arrangement of the three-dimensional grid, 12 must appear between 3 and 4, while the formula returns the value 14. Regardless, 12 is still greater than 2^n+1, which is 9 in three dimensions. Every value for 2(2^n-n+2) is greater than 2^n+1 as can clearly be observed from a graph. The formula 2(2^n-n+2) was derived from the fact that the smallest non-corner values appear next to 2 in every dimension except the third, and next to 3 in the first three dimensions. So with the exclusion of the third dimension due to its limited number of corner cells, these values must be equal to the product of 2 and the lowest number that is paired with 2. The lowest number that is paired with 2 in an ideal arrangement of values is always equal to one less than the number of dimensions, n-1, subtracted from the highest corner value, 2^n+1, thus resulting in 2^n+1-(n-1), or in its simplified form, 2^n-n+2. When multiplying this by two, we receive the lowest non-corner value possible in every dimension except the third.

			One arrangement for n=2 is:
   2   10   5
   8  120  15
   4   12   3
a(2) = 120 because this is the minimum possible value for the central cell.

Aidan Clarke, Table of n, a(n) for n = 1..8 (shortened by _N. J. A. Sloane_, Jan 17 2019)

Cf. A000051, A000142.

a(n) = (2^n+1)!.

a(n) = A000142(A000051(n)). - Michel Marcus, Aug 11 2018

cp's OEIS Frontend

User: Aidan Clarke

A358431 a(0) = 1; a(n+1) = 1 if a(n) > n, otherwise a(n+1) = a(n) + a(a(n)).

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A356675 Lexicographically earliest infinite sequence satisfying a(1) > -1 and a(n-1) = A075826(a(n)).

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A338174 The largest value that integer n is connected to in a web where every integer n has n connections, each integer is connected to the lowest values possible, and new values are added in phases.

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A337804 Lexicographically earliest triangle of nonnegative integers read by rows such that for each pair (x,y) != (0,0), there is at most one pair (n,k) such that T(n,k) = T(n+x,k+y).

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PARI

PARI

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A334146 Numbers with at least two prime factors greater than 3 counted with multiplicity.

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Mathematica

PARI

Formula

A333614 Lexicographically first sequence of nonnegative integers such that no two terms share all but one digit in common, nor does interchanging any two adjacent digits in any term change it into another term of the sequence.

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PARI

A322523 a(n) is the least nonnegative integer k for which there does not exist i < j with i+j=n and a(i)=a(j)=k.

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Maple

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A299790 Positive integers that do not equal any number minus its number of factorizations.

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PARI