Christian Perfect - cp's OEIS frontend

A356739 a(n) is the smallest k such that k! has at least n consecutive zeros immediately after the leading digit in base 10.

7, 153, 197, 7399, 24434, 24434, 9242360, 238861211, 238861211

Offset: 1

Christian Perfect, Aug 25 2022

			a(1) = 7, because 7! = 5040 has one zero immediately after the leading digit, and there is no k<7 with that property.

Cf. A000142, A027869.

from itertools import count
t = 1
n = 1
for k in count(1):
    t *= k
    while str(t)[1:1+n] == '0'*n:
        print(n,k)
        n += 1

a(5)-a(6) from Amiram Eldar, Aug 25 2022

a(7)-a(9) from Jinyuan Wang, Aug 25 2022

A343934 Irregular triangle read by rows: row n gives the sequence of iterations of k - A006519(k), starting with k=n, until 0 is reached.

Original entry on oeis.org

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

Offset: 1

Christian Perfect, May 04 2021

Row n starts with n, then the highest power of 2 dividing n is subtracted to produce the next entry in the row.

n first appears at position A000788(n)+1.

			The triangle begins
1
2
3 2
4
5 4
6 4
7 6 4

Peter Kagey, Rows n = 1..1023, flattened

Cf. A000120 (row widths), A000788, A006519, A129760, A298011 (row sums).

Table[Most @ NestWhileList[# - 2^IntegerExponent[#, 2] &, n, # > 0 &], {n, 1, 30}] // Flatten (* Amiram Eldar, May 05 2021 *)

def gen_a():
    for n in range(1,100):
        k = n
        while k>0:
            yield k
            k = k & (k-1)
a = gen_a()

A338807 Numbers k such that the process starting at (k, 0) mapping (k, t) to (k+1, t+k) if t = 0 (mod k), and (k-1, t+k) otherwise, eventually reaches (1, T) for some T.

Original entry on oeis.org

1, 2, 8, 9, 11, 14, 18, 20, 23, 32, 35, 38, 40, 47, 49, 50, 53, 56, 57, 58, 59, 62, 67, 71, 73, 74, 77, 89, 91, 92, 95, 98, 101, 104, 106, 114, 116, 128, 134, 135, 137, 140, 148, 149, 152, 155, 156, 158, 159, 162, 164, 169, 172, 173, 185, 188, 191, 194, 197

Offset: 1

Christian Perfect, Nov 10 2020

At each step, let the state of the process be (i,t), i_max the greatest i seen so far and i_min > 1 the least i seen so far. Consider the triples (i,j,t%j) for 2 <= j <= i_max, where i_max is the largest i seen so far. If all of those triples have been seen in previous steps, then the next step will not produce a new triple either, so the process will never reach i=1.

			For k = 8, the process stops at T = 57: (8,0), (9,8), (8,17), (7,25), (6,32), (5,38), (4,43), (3,47), (2,50), (3,52), (2,55), (1,57).
For k = 4, the process never stops: (4,0), (5,4), (4,9), (3,13), (2,16), (3,18), (4,21), ...

def isok(n):
    t = 0
    seen = set()
    maxn = n
    steps = 0
    while n>1:
        maxn = max(maxn,n)
        tuples = set((n,m,t%m) for m in range(2,maxn+1))
        if tuples <= seen:
            break
        seen = seen.union(tuples)
        t += n
        if t%n==0:
            n += 1
        else:
            n -= 1
    return n==1

A326930 Height of the smallest stack in a block-stacking sequence.

Original entry on oeis.org

1, 3, 3, 3, 7, 6, 7, 8, 13, 10, 13, 12, 13, 15, 15, 15, 17, 21, 19, 21, 21, 21, 25, 24, 25, 27, 27, 27, 31, 30, 31, 32, 35, 34, 35, 39, 37, 39, 39, 39, 43, 42, 43, 44, 49, 46, 49, 48, 49, 51, 51, 51, 55, 54, 55, 56, 61, 58, 61, 60, 61, 63, 63, 63, 65, 69, 67, 69, 69, 69, 71, 75

Offset: 1

Christian Perfect, Oct 22 2019

This sequence describes a block-stacking process: at step 1, start with a single stack of height 1. At step n>1, if n is less than or equal to the height of the smallest stack, start a new stack of height n. Otherwise, add n to the height of the smallest stack.

seq(n)={my(L=List(), a=vector(n)); for(n=1, #a, if(#L && L[1]Andrew Howroyd, Oct 22 2019

def seq():
    towers = [0]
    for i in range(1, 100):
        towers = sorted(towers)
        if i <= towers[0]:
            towers = [i] + towers
        else:
            towers = [towers[0] + i] + towers[1:]
        yield min(towers)

A283714 a(n) is the first occurrence after a(n-1) of the n-th digit in the decimal expansion of Pi-3, beginning with a(0)=1.

Original entry on oeis.org

1, 3, 19, 37, 48, 55, 63, 69, 90, 91, 109, 113, 122, 139, 144, 170, 173, 194, 197, 201, 211, 221, 226, 227, 230, 231, 233, 237, 241, 242, 247, 252, 264, 275, 279, 305, 321, 324, 328, 343, 344, 347, 353, 358, 388, 391, 401, 405, 411, 417, 421, 444, 447, 456, 493, 496, 506, 511, 527, 528, 530, 534, 542

Offset: 0

Christian Perfect, Mar 15 2017

Skip the first decimal digit of Pi, and then look for the first occurrence of each of the digits in order, only moving forward.
This sequence arose as a result of the claim that the digits of Pi appear in order again later on, if you allow gaps between subsequent digits.
a(n) ~ 10n. - Robert G. Wilson v, Mar 15 2017

			a(1) = 3, because the first occurrence of the first decimal digit of Pi after the first decimal digit is at the 3rd decimal digit.
a(2) = 19, because the next occurrence of the second decimal digit of Pi after the 3rd decimal digit is at the 19th decimal digit.

Robert G. Wilson v, Table of n, a(n) for n = 0..10000 (corrected by Ray Chandler, Jan 19 2019)

Cf. A000796.

pid = RealDigits[Pi - 3, 10, 10000][[1]]; a[0] = 1; a[n_] := a[n] = SelectFirst[ Flatten[ Position[ pid, pid[[n]], 1, 200]], a[n -1] < # &]; Array[a, 100] (* Robert G. Wilson v, Mar 15 2017 *)

A269225 Smallest k such that k! > 2^n.

Original entry on oeis.org

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

Offset: 0

Christian Perfect, Jul 11 2016

			a(7) = 6 because 6! = 720 > 2^7 = 128, but 5! = 120 < 128.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A003070, A067850, A258782.

a[n_] := Block[{v=2^n, k=1}, While[++k! <= v]; k]; Array[a, 93, 0] (* Giovanni Resta, Jul 11 2016 *)
Module[{nn=30,f},f=Table[{k,k!},{k,nn}];Table[SelectFirst[f,#[[2]]>2^n&],{n,0,100}]][[;;,1]] (* Harvey P. Dale, Feb 19 2024 *)

a(n)=localprec(19); my(t=log(2)*n, x=ceil(solve(k=1, n/2+5, lngamma(k+1)-t))); while(x!<=2^n, x++); x \\ Charles R Greathouse IV, Jul 12 2016

def a269225():
   k = 1
   f = 1
   p = 1
   n = 0
   while True:
      while f<=p:
         k += 1
         f *= k
      yield k
      p *= 2
      n += 1

A268176 a(0) = a(1) = 1, and a(n) = a(n-1) + a( (a(n-1)-1) mod n ) for n>=2.

Original entry on oeis.org

1, 1, 2, 3, 5, 10, 13, 23, 36, 72, 73, 86, 87, 123, 196, 197, 202, 398, 399, 798, 1196, 1994, 2117, 2118, 2128, 2130, 4248, 4284, 8568, 8655, 8851, 9048, 11166, 11252, 20300, 40600, 44884, 44886, 44909, 45707, 49955, 50157, 50193, 50279, 59130, 118260, 163967

Offset: 0

Christian Perfect, Jan 28 2016

Alois P. Heinz, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Open Problems in the OEIS, Slides of Guest Lecture given in Doron Zeilberger's Experimental Mathematics Class, Rutgers University, May 2, 2016.

Cf. A125204.

a:= proc(n) option remember; `if`(n<2, 1,
       a(n-1)+a(irem((a(n-1)-1), n)))
    end:
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 08 2016

a[0] = a[1] = 1; a[n_] := a[n] = # + a@ Mod[# - 1, n] &@ a[n - 1]; Array[a, 47, 0] (* Michael De Vlieger, Apr 08 2016 *)

lista(nn) = {va = vector(nn); print1(va[1] = 1, ", "); print1(va[2] = 1, ", "); for (n=3, nn, va[n] = va[n-1] + va[((va[n-1]-1) % (n-1))+1]; print1(va[n], ", "););} \\ Michel Marcus, Apr 08 2016

from sympy.core.cache import cacheit
@cacheit
def a(n): return 1 if n<2 else a(n - 1) + a((a(n - 1) - 1)%n)
print([a(n) for n in range(51)]) # Indranil Ghosh, Aug 06 2017

A264688 The decimal digits of n appear n times in the decimal representation of n!.

Original entry on oeis.org

0, 1, 1170, 1528, 9877, 9886, 9897, 11535

Offset: 1

Christian Perfect, Jan 03 2016

There are no more terms up to n=21000, checked using the Python program below.

a(9) > 200000. - Dana Jacobsen, Jan 03 2016

			1170 belongs to this sequence because the digits 1, 7 and 0 appear 1170 times in total in the decimal representation of 1170!.
0! = 1, which contains no zeros, so 0 is a term.

use ntheory ":all"; sub is_a264688 { my $n = shift; $n == eval "factorial($n) =~ tr/[$n]//"; } # Dana Jacobsen, Jan 03 2016

from math import factorial
def in_a(n):
    f = str(factorial(n))
    s = set(str(n))
    return sum(f.count(d) for d in s)==n

A263458 Deal a pack of n cards into two piles and gather them up, n/2 times. All n such that this reverses the order of the deck.

Original entry on oeis.org

4, 6, 12, 22, 28, 30, 36, 46, 52, 60, 70, 78, 100, 102, 108, 126, 148, 150, 156, 166, 172, 180, 190, 196, 198, 222, 228, 238, 262, 268, 270, 276, 292, 310, 316, 348, 358, 366, 372, 382, 388, 396, 420, 430, 438, 460, 462, 478, 486, 502, 508, 540, 556, 598

Offset: 1

Christian Perfect, Oct 19 2015

nonn

This seems to be A003628(n)-1; that is, each element of this sequence is one less than a prime congruent to 5 or 7 modulo 8.

			Take a deck of 52 playing cards. Deal it into two piles, then pick up the first pile and put it on top of the other. Do this 26 times. The order of the deck is reversed, so 52 belongs to this sequence.
6 is in the sequence because the 3 shuffles are [1, 2, 3, 4, 5, 6] -> [5, 3, 1, 6, 4, 2] -> [4, 1, 5, 2, 6, 3] -> [6, 5, 4, 3, 2, 1], original reversed. 8 is not in the sequence because the 4 shuffles are [1, 2, 3, 4, 5, 6, 7, 8] -> [7, 5, 3, 1, 8, 6, 4, 2] -> [4, 8, 3, 7, 2, 6, 1, 5] -> [1, 2, 3, 4, 5, 6, 7, 8] -> [7, 5, 3, 1, 8, 6, 4, 2], not the original reversed. - _R. J. Mathar_, Aug 02 2024

Cf. A003628, A373416.

isA263458 := proc(n)
    local L,itr ;
    L := [seq(i,i=1..n)] ;
    for itr from 1 to n/2 do
        L := pileShuf(L) ; # function code in A323712
    end do:
    for i from 1 to nops(L) do
        if op(-i,L) <> i then
            return false ;
        end if;
    end do:
    true ;
end proc:
n := 1;
for k from 2 do
    if isA263458(k) then
        printf("%d %d\n",n,k) ;
        n := n+1 ;
    end if;
end do: # R. J. Mathar, Aug 02 2024

from itertools import cycle
def into_piles(r,deck):
    packs = [[] for i in range(r)]
    for card, pack in zip(range(1,deck+1),cycle(range(r))):
        packs[pack].insert(0,card)
    out = sum(packs,[])
    return Permutation(out)
def has_reversing_property(deck):
    p = power(into_piles(2,deck), deck/2)
    return p==into_piles(1,deck)
[i for i in range(2,400,2) if has_reversing_property(i)]

A262719 a(n) is the smallest nonnegative k such that there is no 3 X 3 matrix with entries in {1,...,n} whose determinant is k.

Original entry on oeis.org

1, 6, 21, 55, 110, 203, 357, 544, 808, 1177, 1670, 2215, 2865, 3599, 4558, 5621, 6637, 8041, 9769, 11413, 13394, 15593, 17683, 20317, 23249, 26063, 29506, 33287, 37461, 41692, 46306, 50707, 55667, 61723, 67547, 73939, 80767, 87941, 94913, 101613, 111422

Offset: 1

Christian Perfect, Sep 28 2015

nonn

			For n=1, the only matrix is the matrix of all 1s, which has determinant 0. Hence, a(1)=1.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..50

from itertools import product, groupby, count
def det(m):
    a, b, c, d, e, f, g, h, i = m
    return abs(a*(e*i-f*h)-b*(d*i-f*g)+c*(d*h-e*g))
def a262719(n):
    s = list(product(range(1, n+1), repeat=9))
    i = 0
    for k, ms in groupby(sorted(s, key=det), key=det):
        if k!=i:
            return i
        i += 1
    return i

a(7)-a(41) from Hiroaki Yamanouchi, Oct 17 2015

cp's OEIS Frontend

User: Christian Perfect

A356739 a(n) is the smallest k such that k! has at least n consecutive zeros immediately after the leading digit in base 10.

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A343934 Irregular triangle read by rows: row n gives the sequence of iterations of k - A006519(k), starting with k=n, until 0 is reached.

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Mathematica

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A338807 Numbers k such that the process starting at (k, 0) mapping (k, t) to (k+1, t+k) if t = 0 (mod k), and (k-1, t+k) otherwise, eventually reaches (1, T) for some T.

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A326930 Height of the smallest stack in a block-stacking sequence.

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PARI

Python

A283714 a(n) is the first occurrence after a(n-1) of the n-th digit in the decimal expansion of Pi-3, beginning with a(0)=1.

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Mathematica

A269225 Smallest k such that k! > 2^n.

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Mathematica

PARI

Python

A268176 a(0) = a(1) = 1, and a(n) = a(n-1) + a( (a(n-1)-1) mod n ) for n>=2.

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Maple

Mathematica

PARI

Python

A264688 The decimal digits of n appear n times in the decimal representation of n!.

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Perl

Python

A263458 Deal a pack of n cards into two piles and gather them up, n/2 times. All n such that this reverses the order of the deck.