Martin Y. Champel - cp's OEIS frontend

A264117 Largest integer which cannot be partitioned using only parts from the set {perfect powers excluding the n smallest}.

23, 55, 87, 94, 119, 178, 271, 312, 335, 403, 501, 551, 598, 717, 861, 861, 903, 1022, 1119, 1248, 1463, 1535, 1688, 2031, 2067, 2416, 2535, 2976, 3064, 3164, 3407, 3552, 3552, 4023, 4143, 4416, 4633, 4663, 5424, 5424, 5688, 6000, 6455

Offset: 1

Martin Y. Champel, Nov 03 2015

nonn

It appears, but has not been proved, that for n>28, a(n) < a(n-1) + A001597(n).

			a(1) = 23 as 23 cannot be obtained by any combination of {4, 8, 9, 16} but the 4 following integers can:
24 (6*4) a combination of {4, 8, 9, 16},
25 (1*25) a combination of {4, 8, 9, 16, 25},
26 (1*8+2*9) a combination of {4, 8, 9, 16, 25},
27 (1*27) a combination of {4, 8, 9, 16, 25, 27} so all following integers can.
a(2) = 55 as 55 cannot be obtained by any combination of {8, 9, 16, 25, 27, 32, 36, 49} but the 8 following integers can.
a(3) = 87 as 87 cannot be obtained by any combination of {9, 16, 25, 27, 32, 36, 49, 64, 81} but the 9 following integers can.

Martin Y. Champel, Table of n, a(n) for n = 1..281

Cf. A001597.

# Python version 2.7
from copy import *
from math import *
sol ={}
def a(n):
    global sol
    if n in sol:  return sol[n]
    k = n**2 + 100
    yt = sorted(list(set([b**a for a in range(2, 1+int(log(k)/log(2))) for b in range(1, 1+int(k**(1./a)))])))[n:]
    p0 = yt[0]
    if n-1 in sol and n > 28:  p1 = sol[n-1] + 2 * p0
    else: p1 = 7 * p0 + 400
    yt = sorted(list(set([b**a for a in range(2, 1+int(log(p1)/log(2))) for b in range(1, 1+int(p1**(1./a)))])))[n:]
    st = []
    while st != yt:
        st = deepcopy(yt)
        yt = sorted(list(set(yt + [i+j for i in yt for j in yt if i>=j if i+j < p1])))
    d = 0
    f = yt[0] + 1
    t = f
    for i in range(1,len(yt)):
        if yt[i] == f:
            d += 1
            f += 1
            if d == yt[0] + 1:
                yt = yt[:yt.index(t+1)]
                sol[n] = yt.pop() + 1
                return sol[n]
        else:
            t = f
            f = yt[i]+1
            d = 0

A263717 Number of partitions of n into perfect odd powers (1 being excluded).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 1, 1, 2, 2, 0, 3, 0, 5, 0, 1, 1, 2, 2, 0, 3, 0, 5, 0, 1, 1, 2, 2, 0, 3, 1, 6

Offset: 1

Martin Y. Champel, Oct 24 2015

nonn

			a(97) = #{8*9+25, 5*9+25+27, 2*9+25+2*27} = 3.

Martin Y. Champel, Table of n, a(n) for n = 1...999

Cf. A075109, A112344.

Needs["Combinatorica`"]; Length@ Select[Combinatorica`Partitions@ #, AllTrue[#, And[PrimePowerQ@ #, ! PrimeQ@ #, OddQ@ #] &, 1] &] & /@ Range[2, 52] (* Michael De Vlieger, Nov 05 2015, Version 10 *)

# Python version 2.7
def a(n):
    base = sorted(list(set([a**b for b in range(2,int(log(n)/log(2))) for a in range(3,1+int(n**(1./b)),2)])))
    lb = len(base)
    if lb == 0:
        return 0
    sol = 0
    s = [n // base[0]]
    if lb == 1:
        if n % base[0]  == 0: return 1
        return 0
    while True:
        k = s.pop()
        while k < 0:
            if s ==(lb-1)*[0]:
                return sol
            k = s.pop() - 1
        s.append(k)
        x = n - sum([s[i]*base[i] for i in range(len(s))])
        ls = len(s)
        if ls == lb:
            continue
        a = x // base[ls]
        b = x % base[ls]
        if b == 0:
            s.append(a)
            sol +=1
            if len(s) == lb:
                s.pop()
                s.append(-1)
            r = s.pop() - 2
            s.append(r)
        else:
            s.append(a-1)
            if a!=0:
                if len(s) == lb: s[lb-1]=-1

A256749 a(n) is the time expressed in seconds starting at 0:00:00 at which the second-hand of a classical three hands clock is passing over any of the two other hands considering the minute-hand turns of 6 degrees in one minute and the hour-hand of 6 degrees in 12 minutes.

Original entry on oeis.org

0, 60, 61, 120, 122, 180, 183, 240, 244, 300, 305, 360, 366, 420, 427, 480, 488, 540, 549, 600, 610, 660, 671, 721, 732, 781

Offset: 0

Martin Y. Champel, Apr 09 2015

The sequence is given for its first 1249 terms which corresponds to 12 hours. At 12:00:00 the clock is back to its initial state but as the time wheel keeps on turning a(1248) = 43200 + a(0) , a(1249) = 43200 + a(1), a(n + p*1248) = a(n) + p*43200 for any nonnegative integer p.

			a(0) = 0 as the second-hand is over the minute-hand and the hour-hand at 0:00:00,
a(1) = 60 as the second-hand is over the hour-hand at 0:01:00,
a(2) = 61 as the second-hand is over the minute-hand at 0:01:01,...
a(23) = 721 as the second-hand is over the hour-hand at 0:12:01.

m0 = 0
s = [0]
for i in range(1, 60*12+1):
    m0 += 1
    m = m0 % 60
    h = (m0 // 12) % 60
    a, b = i * 60 + h, i * 60 + m
    s.append(a)
    s.append(b)
s = list(sorted(set(s)))
for i in range(len(s)):
    print ('A256749('+str(i)+') = '+str(s[i]))

A256589 Triangle read by rows: T(n,k) divided by (n-k+1)! is the expected value of number of possible subsets in a partition of a set of n elements with no subsets of cardinality smaller than k.

Original entry on oeis.org

1, 3, 1, 11, 2, 1, 50, 8, 2, 1, 274, 36, 6, 2, 1, 1764, 200, 30, 6, 2, 1, 13068, 1300, 168, 24, 6, 2, 1, 109584, 9720, 1080, 144, 24, 6, 2, 1, 1026576, 82180, 8100, 960, 120, 24, 6, 2, 1

Offset: 1

Martin Y. Champel, Apr 03 2015

This sequence can be seen as a generalization of A233744 which is the particular case where minimum subset cardinality is 2 (k=2).
T(n,k) / A004736(n,k)! = 1 + 1 / ( A002024(n,k) - A002260(n,k) + 1) * (Sum of (T(n,p) / A004736(n,p)!) for p starting at A002260(n,k) up to A002024(n) - A002260(n) if 2 * A002260(n) <= A002024(n)).
The triangle below but including the diagonal is A166350 because there is only one possible partition of subsets of cardinality >= k in any set whose cardinality is between k and 2*k-1.

			The triangle T(n, k) starts:
n\k  1   2   3  4  5  6  ...
1:   1
2:   3   1
3:  11   2   1
4:  50   8   2  1
5: 274  36   6  2  1
6:1764 200  36  6  2  1

Martin Y. Champel, Table of n, a(n) for n = 1..10000

Column 2 is A233744.

Cf. A002024, A002260, A004736, A026791, A166350, A135010.

A255400 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's.

Original entry on oeis.org

0, 5, 10, 15, 20, 264, 25, 30, 35, 40, 45, 101805, 50, 55, 60, 65, 70

Offset: 0

Martin Y. Champel, Feb 22 2015

Most multiples of 5 belong to the sequence (if not all).
All terms whose indices are included in A000966 are far bigger than their neighboring terms whose indices are multiples of 5.
a(11) is a multiple of 5, we can verify a(11) = a(25448).

			a(0) = 0 as 0! = 1 does not contain '0'.
a(1) = 5 as 5! = 120 contains '0'.
a(2) = 10 as 10! = 3628800 contains '00' and 10 is the smallest integer for which the condition is met.

Cf. A027868, A000966.
Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254716, A254717.
Cf. A252652.

\\ uses is() from A000966
f(k, special, sz, sz1) = my(f=k!); if (special, s=Str(f/10^valuation(f, 10)), s=Str(k!)); #strsplit(s, sz) - #strsplit(s, sz1);
a(n) = if (n==0, return(0)); my(sz= concat(vector(n, k, "0")), sz1=concat(sz, "0"), k=1,special=is(n)); while (f(k, special, sz, sz1) != 1, k++); k; \\ Michel Marcus, Oct 25 2023

# Python version 2.7
from math import factorial as fct
def trailing_zero(n):
    k=0
    while n!=0:
        n/=5
        k+=n
    return k
def A255400():
    index = 1
    f = 1
    while True:
        if trailing_zero(f) == index:
            print("A255400("+str(index)+") = " +str(f))
            index += 1
        elif trailing_zero(f) > index:
            while True:
                clnzer = str(fct(f))[:-trailing_zero(f)]
                if index*'0' in clnzer and (index+1)*'0' not in clnzer:
                    print("A255400("+str(index)+") = " +str(f))
                    index += 1
                    f = 0
                    break
                f +=1
        f +=1
    return

import re
def A255400(n):
    f, i, s = 1, 0, re.compile('[0-9]*[1-9]0{'+str(n)+'}[1-9][0-9]*')
    while s.match(str(f)+'1') is None:
        i += 1
        f *= i
    return i # Chai Wah Wu, Apr 02 2015

A252652 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's, not including trailing 0's.

Original entry on oeis.org

0, 7, 12, 22, 107, 264, 812, 4919, 12154, 24612, 75705, 101805, 236441, 1946174

Offset: 0

Jon E. Schoenfield, Mar 22 2015, at the suggestion of Martin Y. Champel

			a(0) = 0 since 0! = 1, which does not contain a 0.
a(1) = 7 since 7! = 5040, which contains a 0 other than the trailing 0, and no integer smaller than 7 satisfies this requirement. (a(1) is not 5; 5! = 120, which has no 0 digits other than the trailing 0.)
a(2) = 12 since 12! = 479001600; discarding the trailing 0's leaves 4790016, which contains a string of exactly two consecutive 0's, and no integer smaller than 12 satisfies this requirement.

Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254716, A254717.

A252652[n_] := Module[{m = 0, s, t},
   If[n == 0, While[MemberQ[IntegerDigits[m!], 0], m++]; m,
    t = Table[0, n];
    While[s = Split[IntegerDigits[m!]];
     If[MemberQ[Last[s], 0], s = Delete[s, -1]]; ! MemberQ[s, t],
     m++]; m]];
Table[A252652[n], {n, 0, 13}] (* Robert Price, Mar 21 2019 *)

f(k, sz, sz1) = my(f=k!, s=Str(f/10^valuation(f, 10))); #strsplit(s, sz) - #strsplit(s, sz1);
a(n) = if (n==0, return(0)); my(sz=concat(vector(n, k, "0")), sz1=concat(sz, "0"), k=1); while (f(k, sz, sz1) != 1, k++); k; \\ Michel Marcus, Oct 25 2023

import re
def A252652(n):
    if n == 0:
        return 0
    f, i, s = 1, 0, re.compile('[0-9]*[1-9]0{'+str(n)+'}[1-9][0-9]*')
    while s.match(str(f)) == None:
        i += 1
        f *= i
    return i # Chai Wah Wu, Dec 29 2015

a(13) from Lars Blomberg, Apr 05 2015

A254717 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 9's.

Original entry on oeis.org

0, 12, 11, 36, 99, 207, 629, 3982, 13216, 24090, 65698, 131076, 176801, 2074822, 5203944, 3716991

Offset: 0

Martin Y. Champel, Feb 06 2015

			a(1) = 12 since 12! = 479001600 contains '9' and 12 is the smallest integer for which the condition is met,
a(2) = 11 since 11! = 39916800 contains '99' and 11 is the smallest integer for which the condition is met.

Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254716.

A254717[n_] := Module[{m = 0},
   If[n == 0, While[MemberQ[IntegerDigits[m!], 9], m++]; m,
    t = Table[9, n];
    While[! MemberQ[Split[IntegerDigits[m!]], t], m++]; m]];
Table[A254717[n], {n, 0, 7}] (* Robert Price, Mar 21 2019 *)

a(n)=k=0;while(k<10^4,d=digits(2*10^(#(digits(k!))+1)+10*k!);for(j=1,#d-n+1,c=0;for(i=j,j+n-1,if(d[i]==9,c++);if(d[i]!=9,c=0;break));if(c==n&&d[j+n]!=9&&d[j-1]!=9,return(k)));if(c==n,return(k));if(c!=n,k++))
for(n=1,6,print1(a(n),", ")) \\ Derek Orr, Feb 06 2015

a(12) from Jon E. Schoenfield, Feb 21 2015

a(13)-a(15) from Bert Dobbelaere, Oct 29 2018

A254716 a(n) is the smallest nonnegative integer m such that m! contains a string of exactly n consecutive 8's, or -1 if no such m exists.

Original entry on oeis.org

0, 11, 9, 16, 27, 482, 532, 4731, 2061, 22402, 50381, 187611, 757618, 591042, 5157267, 9003765

Offset: 0

Martin Y. Champel, Feb 06 2015

			a(1) = 11 since 11! = 39916800 contains '8' and 11 is the smallest integer for which the condition is met. (In 9! the '8's occur in a substring of length 2.)
a(2) = 9 since 9! = 362880 contains '88' and 9 is the smallest integer for which this condition is met.

Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254717.

f[n_] := Block[{k = 0, str = ToString[ 8(10^n - 1)/9]}, While[ Length@ StringPosition[ ToString[ k!], str] != 1, k++]; k]; f[0] = 0; Array[f, 14, 0] (* Robert G. Wilson v, Mar 10 2015 *)

a(n)=k=0; while(k<10^4, d=digits(2*10^(#(digits(k!))+1)+10*k!); for(j=1, #d-n+1, c=0; for(i=j, j+n-1, if(d[i]==8, c++); if(d[i]!=8, c=0; break)); if(c==n&&d[j+n]!=8&&d[j-1]!=8, return(k))); if(c==n, return(k)); if(c!=n, k++))
for(n=1,6,print1(a(n),", ")) \\ Derek Orr, Feb 06 2015

a(0)=0 added by M. F. Hasler, Feb 10 2015
a(11) from Jon E. Schoenfield, Feb 21 2015
a(13) from Jon E. Schoenfield, Feb 28 2015
a(12) from Jon E. Schoenfield, Mar 09 2015
a(14)-a(15) from Bert Dobbelaere, Oct 29 2018

A254502 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 7's.

Original entry on oeis.org

0, 6, 31, 48, 22, 599, 1102, 5280, 4667, 1753, 48861, 150336, 223254, 644487, 7016773, 9588848

Offset: 0

Martin Y. Champel, Jan 31 2015

			a(1) = 6 since 6! equals 720, which contains '7'.

Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254716, A254717.

A254452[n_] := Module[{m = 0},
   If[n == 0, While[MemberQ[IntegerDigits[m!], 7], m++]; m,
    t = Table[7, n];
    While[! MemberQ[Split[IntegerDigits[m!]], t], m++]; m]];
Table[A254452[n], {n, 0, 14}] (* Robert Price, Mar 21 2019 *)

a(11), a(12) from Jon E. Schoenfield, Feb 21 2015
a(0) prepended by Jon E. Schoenfield, Mar 01 2015
a(13) from Jon E. Schoenfield, Mar 06 2015
a(14)-a(15) from Bert Dobbelaere, Oct 29 2018

A254501 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 6's.

Original entry on oeis.org

0, 3, 20, 38, 35, 213, 1122, 3415, 10214, 32430, 145197, 351679, 666779, 813843, 3765934

Offset: 0

Martin Y. Champel, Jan 31 2015

			a(1) = 3 since 3! equals 6 and contains '6'.
a(2) = 20 since 20! contains '66' and 20 is the smallest integer for which the condition is met.

Cf. A254042, A254447, A254448, A254449, A254500, A254502, A254716, A254717.

A254451[n_] := Module[{m = 0},
   If[n == 0, While[MemberQ[IntegerDigits[m!], 6], m++]; m,
    t = Table[6, n];
    While[! MemberQ[Split[IntegerDigits[m!]], t], m++]; m]];
Table[A254451[n], {n, 0, 7}] (* Robert Price, Mar 21 2019 *)

a(10), a(11) from Jon E. Schoenfield, Feb 21 2015, Feb 23 2015
a(0) prepended by Jon E. Schoenfield, Mar 01 2015
a(12), a(13) from Jon E. Schoenfield, Mar 07 2015, Mar 10 2015
a(14) from Bert Dobbelaere, Oct 29 2018

cp's OEIS Frontend

User: Martin Y. Champel

A264117 Largest integer which cannot be partitioned using only parts from the set {perfect powers excluding the n smallest}.

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A263717 Number of partitions of n into perfect odd powers (1 being excluded).

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A256749 a(n) is the time expressed in seconds starting at 0:00:00 at which the second-hand of a classical three hands clock is passing over any of the two other hands considering the minute-hand turns of 6 degrees in one minute and the hour-hand of 6 degrees in 12 minutes.

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A256589 Triangle read by rows: T(n,k) divided by (n-k+1)! is the expected value of number of possible subsets in a partition of a set of n elements with no subsets of cardinality smaller than k.

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A255400 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's.

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A252652 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's, not including trailing 0's.

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A254717 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 9's.

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A254716 a(n) is the smallest nonnegative integer m such that m! contains a string of exactly n consecutive 8's, or -1 if no such m exists.

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