Robin Powell - cp's OEIS frontend

A362557 Start with first term 0, then add paired terms counting every preceding term up to the largest term so far and loop back to 0 after every pair has been counted.

0, 1, 0, 1, 1, 2, 0, 3, 1, 1, 2, 1, 3, 3, 0, 6, 1, 2, 2, 3, 3, 1, 6, 4, 0, 8, 1, 4, 2, 5, 3, 2, 4, 1, 5, 2, 6, 1, 8, 5, 0, 11, 1, 7, 2, 6, 3, 3, 4, 3, 5, 4, 6, 1, 7, 2, 8, 1, 11, 6, 0, 14, 1, 9, 2, 9, 3, 5, 4, 5, 5, 6, 6, 2, 7, 3, 8, 2, 9, 2, 11, 1, 14, 7, 0

Offset: 1

Robin Powell, Apr 24 2023

Same as A055186, except previous pairs from the same row are included in the count.

			Write "0". There is now "1 0". Now there is "1 1". We can't find any terms greater than 1, so we recheck the sequence for 0s and find "2 0(s)". Listing these terms in the order read out loud yields the sequence "0, 1, 0, 1, 1, 2, 0, ...".

Cf. A217760, A055186, A342585.

seq(n)={my(L=List([0]), m=0, k=0); while(#Lt==k, L)); if(c, listput(L,c); listput(L,k); m=max(m,c));  k=if(k==m, 0, k+1)); Vec(L)} \\ Andrew Howroyd, May 02 2023

A361614 Set a(1)=0 and a(2)=1. For n > 1, if a(n) has already appeared in the sequence, then a(n+1) = number of steps since its first appearance. If a(n) has not appeared before, search instead for a(n)-1, then a(n)-2, etc., until you find a number that has appeared before.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 1, 5, 2, 4, 4, 1, 10, 5, 6, 7, 1, 15, 5, 11, 7, 5, 14, 3, 18, 7, 10, 14, 5, 21, 5, 23, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2

Offset: 1

Robin Powell, Mar 17 2023

The first 32 terms are distributed chaotically, after which the sequence alternates between 2 and n-6 indefinitely.

			We start with a(1) = 0 and a(2) = 1. 1 has not appeared before, so we search for the greatest valid integer less than 1, which in this case is 0. 0 first occurred at a(1), which is 1 term before a(2) so a(3) = 1.
1 first occurred 1 term before, so a(4) = 1.
1 appeared at term a(1), which is 2 terms before a(4), so a(5) = 2.
2 has not appeared before, so we search for 1, which occurred 3 terms before at a(1). a(6) = 3.
And so on.

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A328294, A181391.

A333725 Number of primes between pairs of consecutive highly composite numbers (A002182).

Original entry on oeis.org

0, 1, 1, 2, 4, 2, 4, 2, 13, 11, 11, 20, 56, 18, 59, 58, 105, 307, 284, 278, 528, 515, 501, 241, 1684, 466, 456, 2491, 2403, 4676, 4561, 4459, 4396, 12839, 4202, 8317, 4111, 26274, 25673, 50073, 48866, 47998, 47441, 139491, 45881, 90692, 134351, 220465, 173831, 257677

Offset: 1

Robin Powell, Apr 03 2020

nonn

			There are no primes between HCN(1) and HCN(2), so a(1) = 0. The next term a(2) is equal to 1 as 3 is the only prime between HCN(2) and HCN(3); the prime 2 is not greater than HCN(2) and so is omitted here. The first gap to contain more than one prime occurs at a(4) = 2, which alludes to 7 and 11 being the only primes contained within HCN(4) and HCN(5).

Charles R Greathouse IV, Table of n, a(n) for n = 1..200

Cf. A002182, A000720, A262501.

Join[{0},PrimePi[#[[2]]]-PrimePi[#[[1]]]&/@Partition[DeleteDuplicates[Table[ {n,DivisorSigma[ 0,n]},{n,2,22*10^5}],GreaterEqual[#1[[2]],#2[[2]]]&][[All,1]],2,1]] (* Harvey P. Dale, Jan 09 2023 *)

a(n) = A000720(A002182(n+1)) - A000720(A002182(n)) for n > 1. - Amiram Eldar, Apr 26 2020

More terms from Giovanni Resta, Apr 04 2020

A331703 Index of first occurrence of n in A328294, or 0 if n never appears.

Original entry on oeis.org

1, 2, 7, 14, 15, 13, 21, 28, 19, 40, 159, 79, 56, 47, 27, 36, 33, 58, 64, 81, 68, 90, 60, 85, 83, 132, 170, 107, 127, 80, 114, 55, 51, 109, 191, 152, 134, 171, 202, 192, 164, 138, 513, 96, 97, 166, 129, 126, 377, 651, 369, 95, 351, 232, 530, 244, 137, 110

Offset: 1

Robin Powell, Jan 25 2020

nonn

Cf. A328294, A171862.

A328294 Set a(1)=1 and a(2)=2. For n > 2, if a(n) had already appeared in the sequence, then a(n+1) = number of steps since its most recent appearance, as in Van Eck's sequence A181391. If a(n) had not appeared before, search instead for a(n)-1, then a(n)-2, etc., until you find a number that has appeared before.

Original entry on oeis.org

1, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 1, 6, 4, 5, 1, 4, 3, 9, 6, 7, 1, 6, 3, 6, 2, 15, 8, 7, 8, 2, 5, 17, 6, 9, 16, 9, 2, 7, 10, 3, 17, 9, 6, 10, 5, 14, 2, 10, 4, 33, 9, 9, 1, 32, 13, 7, 18, 16, 23, 2, 13, 6, 19, 6, 2, 5, 21, 4, 19, 6, 6, 1, 19, 4, 6, 4, 2, 12, 30

Offset: 1

Robin Powell, Oct 11 2019

nonn

In other words, let a(1)=1, a(2)=2, and for any n >= 2, let v be the greatest value <= a(n) among the first n-1 terms; a(n+1) is the least d > 0 such that a(n-d) = v.

			We start with a(1) = 1 and a(2) = 2. 2 has not appeared before, so we search for the greatest valid integer less than 2, which in this case is 1. 1 last occurred at a(1), which is 1 term before, so a(3) = 1.
1 occurred 2 terms before, so a(4) = 2.
2 appeared at term a(2), which is 2 terms before, so a(5) = 2.
2 appeared most recently at term a(5), which is 1 term earlier, so a(6) = 1.
And so on.

Rémy Sigrist, Table of n, a(n) for n = 1..25000

Cf. A181391, A171911.

seq(n)={my(a=vector(n)); a[1]=1; a[2]=2; for(n=2, n-1, my(m=1); for(i=2, n-1, if(a[i] <= a[n] && a[i] >= a[m], m=i)); a[n+1]=n-m); a} \\ Andrew Howroyd, Oct 25 2019

A306904 The geometric mean of the first n integers, rounded to the nearest integer.

Original entry on oeis.org

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

Offset: 1

Robin Powell, Mar 15 2019

nonn

a(n) is the least k such that (k-1/2)^n < n!. - Robert Israel, May 05 2019

			a(5) is the 5th root of the product of the first 5 integers (approx. 2.605171) rounded up to 3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000142, A061375 (floor instead of round), A214046.

Res:= 1: last:= 1: v:= 1:
for n from 2 to 100 do
  v:= n*v; t:= 2^n*v;
  for k from last while (2*k-1)^n < t do od:
  last:= k-1;
  Res:= Res, last;
od:
Res; # Robert Israel, May 05 2019

Array[Round[#!^(1/#)] &, 68] (* Michael De Vlieger, Mar 31 2019 *)
Table[Round[GeometricMean[Range[n]]],{n,70}] (* Harvey P. Dale, Mar 19 2023 *)

a(n) = round(n!^(1/n)); \\ Michel Marcus, May 05 2019

from sympy import integer_nthroot, factorial
def A306904(n): return (m:=int(integer_nthroot(f:=int(factorial(n)),n)[0]))+int((f<=((m<<1)+1)**n) # Chai Wah Wu, Jun 06 2025

a(n) = round(n!^(1/n)).

a(n) ~ n/e + log(n)/(2*e). - Robert Israel, May 05 2019

Corrected by Robert Israel, May 05 2019

A306560 Smallest number of 1's required to build n using +, *, ^ and tetration.

Original entry on oeis.org

1, 2, 3, 4, 5, 5, 6, 5, 5, 6, 7, 7, 8, 8, 8, 5, 6, 7, 8, 8, 9, 9, 10, 8, 7, 8, 5, 6, 7, 8, 9, 7, 8, 8, 9, 7, 8, 9, 10, 10, 11, 11, 10, 11, 10, 11, 12, 8, 8, 9, 9, 10, 11, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 7, 8, 9, 10, 10, 11, 11, 12, 9, 10, 10, 10, 11, 12, 11, 12, 10, 7, 8, 9, 9, 10, 11, 10

Offset: 1

Robin Powell, Feb 23 2019

nonn

			a(16) = 5 because 16 = (1+1)^^(1+1+1). (Note that 16 is also the smallest index at which this sequence differs from A025280.)
a(34) = 8 because 34 = ((1+1)^^(1+1+1)+1)*(1+1). - _Jens Ahlström_, Jan 11 2023

Index to sequences related to the complexity of n

Cf. A005245, A025280, A323727.

from functools import lru_cache
from sympy import factorint, divisors
tetration = {2**2**2: 5, 2**2**2**2: 6, 3**3: 5, 4**4: 6, 5**5: 7}
@lru_cache(maxsize=None)
def a(n):
    res = n
    if n < 6:
        return res
    if n in tetration:
        return tetration[n]
    for i in range(1, n):
        res = min(res, a(i) + a(n-i))
    for d in [i for i in divisors(n) if i not in {1, n}]:
        res = min(res, a(d) + a(n//d))
    factors = factorint(n)
    exponents = set(factors.values())
    if len(exponents) == 1:
        e = exponents.pop()
        if e > 1:
            res = min(res, a(sum(factors.keys())) + a(e))
    return res
# Jens Ahlström, Jan 11 2023

a(34) onward corrected by Jens Ahlström, Jan 11 2023

A323727 Number of 1's required to build n using +, -, *, ^ and tetration.

Original entry on oeis.org

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

Offset: 1

Robin Powell, Jan 25 2019

nonn

			a(14) = 7 because 14 = (1+1)^^(1+1+1)-1-1. (Note that 14 is also the smallest index at which this sequence differs from A091334.)

Wikipedia, Tetration
Index to sequences related to the complexity of n

Cf. A091333, A091334, A025280.

A272615 Numbers with digits in descending numerical order in base 2, 3 and 4 expansions.

Original entry on oeis.org

0, 1, 2, 3, 4, 8, 12, 63, 240

Offset: 1

Robin Powell, May 03 2016

a(10), if it exists, has more than 1000 decimal digits. Conjecture: there are no more terms in this sequence. - Charles R Greathouse IV, May 03 2016

			12 is 1100 in base 2, 110 in base 3 and 30 in base 4; in each representation every digit is smaller than or equal to the one proceeding so 12 is a term.
Similarly, 63 is 111111 in base 2, 2100 in base 3 and 333 in base 4 so it is also a term.

Intersection of A023758, A023759, and A023760.

dec(n,b)=my(v=digits(n,b)); v==vecsort(v,,4)
is(n)=dec(n,2) && dec(n,3) && dec(n,4) \\ Charles R Greathouse IV, May 03 2016

dec(n,b)=my(v=digits(n,b)); v==vecsort(v,,4)
list(lim)=my(v=List([0]),t); for(i=1,logint(lim\1+1,4), t=4^i-1; while(t<=lim, if(dec(t,3), listput(v,t)); t*=4); t=2*4^i-2; while(t<=lim, if(dec(t,3), listput(v,t)); t*=4)); Set(v) \\ Charles R Greathouse IV, May 03 2016

A267218 a(n) is the a(n-1)-st a(n-2)-dimensional orthoplex number, starting with the terms 1, 2.

Original entry on oeis.org

1, 2, 2, 4, 16, 22016

Offset: 1

Robin Powell, Jan 18 2016

nonn

			16 is the 4th 2-orthoplex number = A000290(4).
22016 is the 16th 4-orthoplex number = A014820(16).
The next term will be the 22016th 16-orthoplex number.

OEIS Wiki, Orthoplex numbers

Cf. A000290, A014820, A260161, A051285.

cp's OEIS Frontend

User: Robin Powell

A362557 Start with first term 0, then add paired terms counting every preceding term up to the largest term so far and loop back to 0 after every pair has been counted.

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A361614 Set a(1)=0 and a(2)=1. For n > 1, if a(n) has already appeared in the sequence, then a(n+1) = number of steps since its first appearance. If a(n) has not appeared before, search instead for a(n)-1, then a(n)-2, etc., until you find a number that has appeared before.

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A333725 Number of primes between pairs of consecutive highly composite numbers (A002182).

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Mathematica

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A331703 Index of first occurrence of n in A328294, or 0 if n never appears.

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PARI

A306904 The geometric mean of the first n integers, rounded to the nearest integer.

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A306560 Smallest number of 1's required to build n using +, *, ^ and tetration.

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Python

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A323727 Number of 1's required to build n using +, -, *, ^ and tetration.

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A272615 Numbers with digits in descending numerical order in base 2, 3 and 4 expansions.

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