cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A217112 Greatest number (in decimal representation) with n nonprime substrings in binary representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 3, 7, 6, 15, 14, 31, 29, 30, 63, 61, 62, 127, 54, 125, 126, 255, 117, 251, 254, 189, 511, 479, 509, 510, 379, 502, 1023, 1021, 1007, 1022, 958, 1018, 1014, 2047, 2045, 1791, 2046, 2042, 2027, 2037, 4091, 4095, 4063, 3069, 4094, 4090, 4085, 8159, 8187, 8191, 8189, 8127
Offset: 1

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Author

Hieronymus Fischer, Dec 20 2012

Keywords

Comments

There are no numbers with zero nonprime substrings in binary representation. For all bases > 2 there is always a number (=2) with zero nonprime substrings.
The set of numbers with n nonprime substrings is finite. Proof: Evidently, each 1-digit binary number represents 1 nonprime substring. Hence, each (n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 2^n, such that all numbers > b have more than n nonprime substrings.

Examples

			(1) = 1, since 1 = 1_2 (binary) is the greatest number with 1 nonprime substring.
a(2) = 3 = 11_2 has 3 substrings in binary representation (1, 1 and 11), two of them are nonprime substrings (1 and 1), and 11_2 = 3 is the only prime substrings. 3 is the greatest number with 2 nonprime substrings.
a(8) = 29 = 11101_2 has 15 substrings in binary representation (0, 1, 1, 1, 1, 11, 11, 10, 01, 111, 110, 101, 1110, 1101, 11101), exactly 8 of them are nonprime substrings (0, 1, 1, 1, 1, 01, 110, 1110). There is no greater number with 8 nonprime substrings in binary representation.
a(14) = 54 = 110110_2 has 21 substrings in binary representation, only 7 of them are prime substrings (10, 10, 11, 11, 101, 1011, 1101), which implies that exactly 14 substrings must be nonprime. There is no greater number with 14 nonprime substrings in binary representation.
		

Crossrefs

Formula

a(n) >= A217102(n).
a(n) >= A217302(A000217(A070939(a(n)))-n).
Example: a(9)=30=11110_2, A000217(A070939(31))=15, hence, a(9)>=A217302(15-9)=27.
a(n) <= 2^n.
a(n) <= 2^min(6 + n/6, 20*floor((n+125)/126)).
a(n) <= 64*2^(n/6).
With m := floor(log_2(a(n))) + 1:
a(n+m+1) >= 2*a(n), if a(n) is even.
a(n+m) >= 2*a(n), if a(n) is odd.

A217119 Greatest number (in decimal representation) with n nonprime substrings in base-9 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

47, 428, 1721, 6473, 14033, 35201, 58961, 58967, 465743, 530701, 530710, 1733741, 4250788, 4723108, 4776398, 25051529, 37327196, 42450640, 42986860, 42987589, 42996409, 225463817, 382055767, 382571822, 386888308, 386888419, 387356789
Offset: 0

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Author

Hieronymus Fischer, Dec 20 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n nonprime substrings is not empty and finite. Proof of existence: Define m(n):=2*sum_{j=i..k} 9^j, where k:=floor((sqrt(8n+1)-1)/2), i:= n-(k(k+1)/2). For n=0,1,2,3,... the m(n) in base-9 representation are 2, 22, 20, 222, 220, 200, 2222, 2220, 2200, 2000, 22222, 22220, .... m(n) has k+1 digits and (k-i+1) 2’s. Thus, the number of nonprime substrings of m(n) is ((k+1)(k+2)/2)-k-1+i=(k(k+1)/2)+i=n. This proves the statement of existence. Proof of finiteness: Each 3-digit base-9 number has at least 1 nonprime substring. Hence, each 3(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 9^(3n+2) such that all numbers > b have more than n nonprime substrings. It follows, that the set of numbers with n nonprime substrings is finite.

Examples

			a(0) = 47, since 47 = 52_9 (base-9) is the greatest number with zero nonprime substrings in base-9 representation.
a(1) = 428 = 525_9 has 1 nonprime substring in base-9 representation (= 525_9). All the other base-9 substrings (2, 5, 5, 25, 52) are prime substrings. 525_9 is the greatest number with 1 nonprime substring.
a(2) = 1721 = 2322_9 has 10 substrings in base-9 representation, exactly 2 of them are nonprime substrings (22_9 and 23_3=8), and there is no greater number with 2 nonprime substrings in base-9 representation.
a(7) = 58967= 88788_9 has 15 substrings in base-9 representation, exactly 7 of them are nonprime substrings (4-times 8, 2-times 88, and 8788), and there is no greater number with 7 nonprime substrings in base-9 representation.
		

Crossrefs

Formula

a(n) >= A217109(n).
a(n) >= A217309(A000217(num_digits_9(a(n)))-n), where num_digits_9(x)=floor(log_9(x))+1 is the number of digits of the base-9 representation of x.
a(n) <= 9^(n+2).
a(n) <= 9^min(n+2, 6*floor((n+7)/8)).
a(n) <= 9^((3/4)*(n + 3)).
a(n+m+1) >= 9*a(n), where m := floor(log_9(a(n))) + 1.

A213304 Smallest number with n nonprime substrings (Version 3: substrings with leading zeros are counted as nonprime if the corresponding number is not a prime).

Original entry on oeis.org

2, 1, 10, 14, 101, 104, 144, 1001, 1014, 1044, 1444, 10010, 10014, 10144, 10444, 14444, 100120, 100104, 100144, 101444, 104444, 144444, 1000144, 1001040, 1001044, 1001444, 1014444, 1044444, 1444444, 10001044, 10001444, 10010404, 10010444, 10014444, 10144444, 10444444, 14444444, 100010404, 100010444, 100014444, 100104044, 100104444, 100144444, 101444444, 104444444, 144444444
Offset: 0

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is well defined since for each n >= 0 there is a number with n nonprime substrings.
Different from A213303, first difference is at a(16).

Examples

			a(0)=2, since 2 is the least number with zero nonprime substrings.
a(1)=1, since 1 has 1 nonprime substrings.
a(2)=10, since 10 is the least number with 2 nonprime substrings, these are 1 and 10 ('0' will not be counted).
a(3)=14, since 14 is the least number with 3 nonprime substrings, these are 1 and 4 and 14. 10, 11 and 12 only have 2 such substrings.
		

Crossrefs

Formula

a(m(m+1)/2) = (13*10^(m-1)-4)/9, m>0.
With b(n):=floor((sqrt(8*n-7)-1)/2):
a(n) > 10^b(n), for n>2, a(n) = 10^b(n) for n=1,2.
a(n) >= 10^b(n)+4*10^(n-1-b(n)(b(n)+1)/2)-1)/9, equality holds if n or n+1 is a triangular number > 0 (cf. A000217).
a(n) >= A213303(n).
a(n) <= A213307(n).

A213303 Smallest number with n nonprime substrings (Version 2: substrings with leading zeros are counted as nonprime if the corresponding number is > 0).

Original entry on oeis.org

2, 1, 10, 14, 101, 104, 144, 1001, 1014, 1044, 1444, 10010, 10014, 10144, 10444, 14444, 100101, 100104, 100144, 101444, 104444, 144444, 1000144, 1001014, 1001044, 1001444, 1014444, 1044444, 1444444, 10001044, 10001444, 10010144, 10010444, 10014444, 10144444, 10444444, 14444444, 100010144
Offset: 0

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is well defined since for each n >= 0 there is a number with n nonprime substrings.
Different from A213304, first different term is a(16).

Examples

			a(0)=2, since 2 is the least number with zero nonprime substrings.
a(1)=1, since 1 has 1 nonprime substrings.
a(2)=10, since 10 is the least number with 2 nonprime substrings, these are 1 and 10 ('0' will not be counted).
a(3)=14, since 14 is the least number with 3 nonprime substrings, these are 1 and 4 and 14. 10, 11 and 12 only have 2 such substrings.
		

Crossrefs

Formula

a(m(m+1)/2) = (13*10^(m-1)-4)/9, m>0.
With b(n):=floor((sqrt(8*n-7)-1)/2):
a(n) > 10^b(n), for n>2, a(n) = 10^b(n) for n=1,2.
a(n) >= 10^b(n)+4*10^(n-1-b(n)(b(n)+1)/2)-1)/9, equality holds if n or n+1 is a triangular number > 0 (cf. A000217).
a(n) <= A213304(n).
a(n) <= A213306(n).

A213306 Minimal prime with n nonprime substrings (Version 2: substrings with leading zeros are counted as nonprime if the corresponding number is > 0).

Original entry on oeis.org

2, 13, 11, 103, 101, 149, 1009, 1021, 1049, 1481, 10039, 10069, 10169, 11681, 14669, 100109, 100189, 100169, 101681, 104681, 146669, 1000669, 1001041, 1001081, 1004669, 1014469, 1046849, 1468469, 10001081, 10004669, 10010851
Offset: 0

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Examples

			a(0) = 2, since 2 is the least number with zero nonprime substrings.
a(1) = 13, since 13 has 1 nonprime substring (=’1’).
a(2) = 11, since 11 is the least number with 2 nonprime substrings (= 2 times ‘1’).
a(3) = 103, since 103 is the least number with 3 nonprime substrings, these are ‘1’ and ‘10’ and ‘03’ (‘0’ is not a valid substring in version 2).
		

Crossrefs

Formula

a(n) > 10^floor((sqrt(8*n+1)-1)/2), for n>2.
a(n) >= A213303(n).
a(n) <= A213307(n).

A213307 Minimal prime with n nonprime substrings (Version 3: substrings with leading zeros are counted as nonprime if the corresponding number is not a prime).

Original entry on oeis.org

2, 13, 11, 127, 101, 149, 1009, 1063, 1049, 1481, 10091, 10069, 10169, 11681, 14669, 100129, 100189, 100169, 101681, 104681, 146669, 1000669, 1001219, 1001081, 1004669, 1014469, 1046849, 1468469, 10001081, 10004669, 10010851
Offset: 0

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Examples

			a(0) = 2, since 2 is the least number with zero nonprime substrings.
a(1) = 13, since 13 there is one nonprime substring (=1).
a(2) = 11, since 11 is the least number with 2 nonprime substrings (2 times ‘1’).
a(3) = 127, since 127 is the least number with 3 nonprime substrings, these are 1 and 12 and 27 (according to version 3).
		

Crossrefs

Formula

a(n) > 10^floor((sqrt(8*n+1)-1)/2), for n>2.
a(n) >= A213304(n).
a(n) >= A213306(n).

A217103 Minimal number (in decimal representation) with n nonprime substrings in base-3 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

2, 1, 3, 4, 14, 9, 34, 29, 30, 27, 89, 88, 83, 84, 81, 268, 251, 250, 248, 245, 243, 752, 754, 746, 740, 734, 731, 729, 2237, 2239, 2210, 2203, 2198, 2192, 2189, 2187, 6632, 6611, 6614, 6584, 6577, 6569, 6563, 6564, 6561, 19814, 19754, 19733, 19736, 19706
Offset: 0

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Author

Hieronymus Fischer, Dec 12 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n nonprime substrings is not empty. Proof: Define m(n):=2*sum_{j=i..k} 3^j, where k:=floor((sqrt(8*n+1)-1)/2), i:= n-A000217(k). For n=0,1,2,3,… the m(n) in base-3 representation are 2, 22, 20, 222, 220, 200, 2222, 2220, 2200, 2000, 22222, 22220, .... m(n) has k+1 digits and (k-i+1) 2’s, thus, the number of nonprime substrings of m(n) is ((k+1)*(k+2)/2)-k-1+i = (k*(k+1)/2)+i = n, which proves the statement.
If p is a number with k prime substrings and d digits (in base-3 representation), p != 1 (mod 3), m>=d, than b := p*3^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_3 is the least number with zero nonprime substrings in base-3 representation.
a(1) = 1, since 1 = 1_3 is the least number with 1 nonprime substring in base-3 representation.
a(2) = 3, since 3 = 10_3 is the least number with 2 nonprime substrings in base-3 representation (0 and 1).
a(3) = 4, since 4 = 11_3 is the least number with 3 nonprime substrings in base-3 representation (1, 1 and 11).
a(4) = 14, since 14 = 112_3 is the least number with 4 nonprime substrings in base-3 representation, these are 1, 1, 11 and 112 (remember, that substrings with leading zeros are considered to be nonprime).
a(7) = 29, since 29 = 1002_3 is the least number with 7 nonprime substrings in base-3 representation, these are 0, 0, 1, 00, 02, 002 and 100 (remember, that substrings with leading zeros are considered to be nonprime, 2_3 = 2, 10_3 = 3 and 1002_3 = 29 are base-3 prime substrings).
		

Crossrefs

Formula

a(n) >= 3^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n=1 or n+1 is a triangular number (cf. A000217).
a(n) >= 3^floor((sqrt(8*n+1)-1)/2) for n>3, equality holds if n+1 is a triangular number.
a(A000217(n)-1) = 3^(n-1), n>1.
a(A000217(n)-k) >= 3^(n-1) + k-1, 1<=k<=n, n>1.
a(A000217(n)-k) = 3^(n-1) + p, where p is the minimal number >= 0 such that 3^(n-1) + p, has k prime substrings in base-3 representation, 1<=k<=n, n>1.

A217303 Minimal natural number (in decimal representation) with n prime substrings in base-3 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 2, 5, 11, 17, 23, 50, 104, 71, 152, 215, 395, 476, 701, 719, 1367, 1934, 1448, 4127, 4121, 4346, 5822, 12302, 12383, 17468, 25505, 32066, 39113, 51749, 91040, 111509, 110798, 117359, 157211, 332396, 334358, 465092, 333791, 819386, 865232, 1001375, 1396673
Offset: 0

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Author

Hieronymus Fischer, Nov 22 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n prime substrings is not empty. Proof: Define m(0):=1, m(1):=2 and m(n+1):=3*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 3^j = 3^n - 1 or m(n)=1, 2, 22, 222, 2222, 22222, …,for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base-3 representation. This is why every substring of m(n) with more than one digit is a product of two integers > 1 (by definition) and can therefore not be prime number.
No term is divisible by 3.

Examples

			a(1) = 2 = 2_3, since 2 is the least number with 1 prime substring in base-3 representation.
a(2) = 5 = 12_3, since 5 is the least number with 2 prime substrings in base-3 representation (2_3 and 12_3).
a(3) = 11 = 102_3, since 11 is the least number with 3 prime substrings in base-3 representation (2_3, 10_3, and 102_3).
a(5) = 23 = 212_3, since 23 is the least number with 5 prime substrings in base-3 representation (2 times 2_3, 12_3=5, 21_3=19, and 212_3=23).
a(7) = 104 = 10212_3, since 104 is the least number with 7 prime substrings in base-3 representation (2 times 2_3, 10_3=3, 12_3=5, 21_3=19, 102_3=11, and 212_3=23).
		

Crossrefs

Formula

a(n) > 3^floor(sqrt(8*n+1)-1)/2), for n>1.
a(n) <= 3^n - 1.
a(n+1) <= 3a(n)+2.

A217308 Minimal natural number (in decimal representation) with n prime substrings in base-8 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 2, 11, 19, 83, 107, 157, 669, 751, 1259, 4957, 6879, 6011, 14303, 47071, 48093, 65371, 188143, 327515, 440287, 384751, 1029883, 2604783, 2948955, 3602299, 6946651, 20304733, 23846747, 23937003, 23723867, 57278299, 167689071, 175479547, 191496027, 233824091
Offset: 0

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Author

Hieronymus Fischer, Nov 22 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n prime substrings is not empty. Proof: Define m(0):=1, m(1):=2 and m(n+1):=8*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 8^j = 2*(8^n - 1)/7 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base-8) for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base-8 representation. This is why every substring of m(n) with more than one digit is a product of two integers > 1 (by definition) and can therefore not be prime number.
No term is divisible by 8.

Examples

			a(1) = 2 = 2_8, since 2 is the least number with 1 prime substring in base-8 representation.
a(2) = 11 = 13_8, since 11 is the least number with 2 prime substrings in base-8 representation (3_8 and 13_8).
a(3) = 19 = 23_8, since 19 is the least number with 3 prime substrings in base-8 representation (2_8, 3_8, and 23_8).
a(4) = 83 = 123_8, since 83 is the least number with 4 prime substrings in base-8 representation (2_8, 3_8, 23_8=19, and 123_8=83).
a(8) = 751 = 1357_8, since 751 is the least number with 8 prime substrings in base-8 representation (3_8, 5_8, 7_8, 13_8=11, 35_8=29, 57_8=47, 357_8=239, and 1357_8=751).
		

Crossrefs

Formula

a(n) > 8^floor(sqrt(8*n-7)-1)/2), for n>0.
a(n) <= 2*(8^n - 1)/7, n>0.
a(n+1) <= 8*a(n)+2.

A213299 Partial sums of A211681.

Original entry on oeis.org

2, 5, 10, 17, 40, 77, 130, 203, 440, 813, 1350, 2087, 4460, 8197, 13570, 20943, 44680, 82053, 135790, 209527, 446900, 820637, 1358010, 2095383, 4469120, 8206493, 13580230, 20953967, 44691340
Offset: 1

Views

Author

Hieronymus Fischer, Jun 08 2012

Keywords

Comments

The terms are primes for n = 1, 2, 4, 12, 22, 32 and possibly further n’s (Question).

Crossrefs

Formula

a(n) = ((3982 + 2709*k + 567*k^2 + 54*k^3)*10^m - 1980*m - 2200 - 495*k + 162*((n+1) mod 2) * (-1)^m * (-1)^floor(n/2))/891, where m=floor((n-1)/4), k=(n-1) mod 4.
G.f.: (2*x*(1+x^10) + 3*x^2*(1 + x^3 + x^5 + x^6) + 5*x^3*(1+x^6) + 7*x^4*(1+x^2))/((1-x)*(1-10*x^4)*(1-x^8)).
From Chai Wah Wu, Feb 08 2023: (Start)
a(n) = 2*a(n-1) - a(n-2) + 9*a(n-4) - 18*a(n-5) + 9*a(n-6) + 10*a(n-8) - 20*a(n-9) + 10*a(n-10) for n > 10.
G.f.: x*(-2*x^7 + 2*x^6 - 5*x^5 + 2*x^4 - 2*x^3 - 2*x^2 - x - 2)/((x - 1)^2*(x^4 + 1)*(10*x^4 - 1)). (End)

Extensions

Typo in g.f. corrected by Hieronymus Fischer, Sep 03 2012
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