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.

Showing 1-8 of 8 results.

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.

A217105 Minimal number (in decimal representation) with n nonprime substrings in base-5 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

2, 1, 5, 6, 27, 25, 34, 127, 128, 125, 170, 636, 632, 627, 625, 850, 3162, 3137, 3132, 3127, 3125, 4250, 15686, 15661, 15638, 15632, 15627, 15625, 21250, 78192, 78163, 78162, 78137, 78132, 78127, 78125, 106250, 390818, 390692, 390686, 390662, 390638, 390632
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} 5^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-5 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-5 representation), p != 1 (mod 5), m>=d, than b := p*5^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_5 is the least number with zero nonprime substrings in base-4 representation.
a(1) = 1, since 1 = 1_5 is the least number with 1 nonprime substring in base-5 representation.
a(2) = 5, since 5 = 10_5 is the least number with 2 nonprime substrings in base-5 representation (0 and 1).
a(3) = 6, since 6 = 11_5 is the least number with 3 nonprime substrings in base-5 representation (2-times 1 and 11).
a(4) = 27, since 27 = 102_5 is the least number with 4 nonprime substrings in base-5 representation, these are 0, 1, 02, and 102 (remember, that substrings with leading zeros are considered to be nonprime).
a(6) = 34, since 34 = 114_5 is the least number with 6 nonprime substrings in base-5 representation, these are 1, 1, 4, 11, 14, and 114.
		

Crossrefs

Formula

a(n) >= 5^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(A000217(n)-1) = 5^(n-1), n>1.
a(A000217(n)) = floor(34 * 5^(n-3)), n>0.
a(A000217(n)) = 114000...000_5 (with n digits), n>0.
a(A000217(n)-k) >= 5^(n-1) + k-1, 1<=k<=n, n>1.
a(A000217(n)-k) = 5^(n-1) + p, where p is the minimal number >= 0 such that 5^(n-1) + p, has k prime substrings in base-5 representation, 1<=k<=n, n>1.

A217113 Greatest 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, 23, 71, 26, 77, 233, 239, 719, 701, 647, 725, 2159, 2177, 2158, 2157, 5822, 5741, 6551, 6476, 6532, 6531, 18944, 19436, 19655, 19601, 19673, 19653, 58310, 58309, 58316, 58967, 59021, 58964, 157211, 157217, 174950, 176408, 176407, 176903, 177065, 177064, 471653, 511511
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} 3^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-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. This proves the statement of existence. Proof of finiteness: Each 2-digit base-3 number has at least 1 nonprime substring. Hence, each 2(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 3^(2n+1) 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) = 2, since 2 = 2_3 (base-3) is the greatest number with zero nonprime substrings in base-3 representation.
a(1) = 23 = 212_3 has 1 substring in base-3 representation (= 1). All the other base-3 substrings (2, 2, 21, 12, 212) are prime substrings. 23 is the greatest number with 1 nonprime substring.
a(2) = 71 = 2122_3 has 10 substrings in base-3 representation (1, 2, 2, 2, 12, 21, 22, 122, 212, 2122), exactly 2 of them are nonprime substrings (1 and 22_3=8), and there is no greater number with 2 nonprime substrings in base-3 representation.
a(3) = 26 = 222_3 has 6 substrings in base-3 representation, only 3 of them are prime substrings (2, 2, 2) which implies that exactly 3 substrings must be nonprime, and there is no greater number with 3 nonprime substrings in base-3 representation.
		

Crossrefs

Formula

a(n) >= A217103(n).
a(n) >= A217303(A000217(A081604(a(n)))-n).
Example: a(12)=2177=2222122_3, A000217(A081604(2177))=28, hence a(12)>=A217303(28-12)=1934.
a(n) <= 3^min(n + 2, 5*floor((n+4)/5)).
a(n) <= 3^(n + 2).
a(n) <= 3^min((n + 11)/3, 11*floor((n+32)/33)).
a(n) <= 3^((1/3)*(n + 11)).
With m := floor(log_3(a(n))) + 1:
a(n+m+1) >= 3*a(n), if a(n)!=1 (mod 3).
a(n+m) >= 3*a(n), if a(n)=1 (mod 3).

A217118 Greatest number (in decimal representation) with n nonprime substrings in base-8 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

491, 3933, 24303, 32603, 188143, 253789, 261117, 1555423, 2030319, 2088797, 2088943, 16185163, 16710383, 16710381, 16768991, 99606365, 129884143, 133683069, 134150015, 134209503, 770611067, 1039073149, 1069408239, 1073209071, 1073209083, 1073676029, 5065578363
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} 8^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-8 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 4-digit base-8 number has at least 1 nonprime substring. Hence, each 4(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 8^(4n+3) 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) = 491, since 491 = 753_8 (base-8) is the greatest number with zero nonprime substrings in base-8 representation.
a(1) = 3933 = 7535_8 has 1 nonprime substring in base-8 representation (=7535_8). All the other base-8 substrings are prime substrings. 3933 is the greatest such number with 1 nonprime substring.
a(2) = 24303 = 57357_8 has 15 substrings in base-8 representation, exactly 2 of them are nonprime substrings (57357_8 and 735_8), and there is no greater number with 2 nonprime substrings in base-3 representation.
a(3) = 32603 = 77533_8 has 15 substrings in base-8 representation, only 3 of them are nonprime substrings (33_8, 77_8, and 7753_8), and there is no greater number with 3 nonprime substrings in base-8 representation.
		

Crossrefs

Formula

a(n) >= A217108(n).
a(n) >= A217308(A000217(num_digits_8(a(n)))-n), where num_digits_8(x) is the number of digits of the base-8 representation of x.
a(n) <= 8^min(n+3, 7*floor((n+6)/7)).
a(n) <= 512*8^n.
a(n+m+1) >= 8*a(n), where m := floor(log_8(a(n))) + 1.

A217114 Greatest number (in decimal representation) with n nonprime substrings in base-4 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

11, 59, 239, 251, 751, 1007, 1019, 3823, 4079, 4055, 16111, 16087, 16319, 16367, 48991, 64351, 65263, 65269, 65471, 253919, 260959, 261079, 261847, 261871, 916319, 1043839, 1047391, 1044463, 1047511, 3665279, 3140991, 4189567, 4118519, 4177759, 4189565, 4193239, 14661117
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} 4^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-4 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-4 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 < 4^(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) = 11, since 11 = 23_4 (base-4) is the greatest number with zero nonprime substrings in base-4 representation.
a(1) = 59 = 323_4 has 6 substrings in base-4 representation (2, 3, 3, 23, 32 and 323), only 32_4=14 is a nonprime substring. 59 is the greatest such number with 1 nonprime substring.
a(2) = 239 = 3233_4 has 10 substrings in base-4 representation (2, 3, 3, 23, 32, 323, 233 and 3233), exactly 2 of them are nonprime substrings (32_4=14 and 33_4=15), and there is no greater number with 2 nonprime substrings in base-4 representation.
a(11) = 16087 = 3323113_4 has 28 substrings in base-4 representation. The base-4 nonprime substrings are 1, 1, 32, 33, 231, 332, 3113, 3231, 32311, 33321 and 323113. There is no greater number with 11 nonprime substrings in base-4 representation.
		

Crossrefs

Formula

a(n) >= A217104(n).
a(n) >= A217304(A000217(A110591(a(n)))-n).
a(n) <= 4^(n+2).
a(n) <= 4^min((n + 6)/2, 9*floor((n+18)/19)).
a(n) <= 64*4^(n/2).
a(n+m+1) >= 4*a(n), where m := floor(log_4(a(n))) + 1.

A217115 Greatest number (in decimal representation) with n nonprime substrings in base-5 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

67, 88, 442, 567, 2213, 2837, 3067, 11068, 14713, 15337, 15338, 57943, 73568, 77213, 76697, 289717, 280338, 370443, 386068, 386587, 389713, 1852217, 1524067, 1898442, 1930342, 1932943, 1948568, 7242943, 9261088, 9664717, 9586567, 9654712, 9710942, 9742849, 46305443
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} 5^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-5 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 4-digit base-5 number has at least 2 nonprime substrings. Hence, each m := 4*floor((n+2)/2)-digit number has at least 2*(m/4) = 2*floor((n+2)/2) >= n+1 nonprime substrings. Consequently, there is a boundary b < 5^(m-1) 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) = 67, since 67 = 232_5 (base-5) is the greatest number with zero nonprime substrings in base-5 representation.
a(1) = 88 = 323_5 has 6 substrings in base-5 representation (2, 2, 3, 23, 32, 323), the only nonprime substring is 323_5. 88 is the greatest number with 1 nonprime substring.
a(2) = 442 = 3232_5 has 10 substrings in base-5 representation (2, 2, 3, 3, 23, 32, 32, 232, 323 and 3232), exactly 2 of them are nonprime substrings (323_5=88 and 3232_5=442), and there is no greater number with 2 nonprime substrings in base-5 representation.
a(5) = 2837 = 42322_5 has 5 nonprime substrings in base-5 representation, these are 4, 22, 42, 322 and 4232, all the other substrings are prime. There is no greater number with 5 nonprime substrings in base-5 representation.
		

Crossrefs

Formula

a(n) >= A217105(n).
a(n) >= A217305(A000217(A110592(a(n)))-n).
a(n) <= 5^(n+3).
a(n) <= 5^(4*floor(n/2)), n>1.
a(n) <= 5^min((n + 6)/2, 9*floor((n+20)/21)).
a(n) <= 125*5^(n/2).
With m := floor(log_5(a(n))) + 1:
a(n+m+1) >= 5*a(n), if a(n)!=1 (mod 5).
a(n+m) >= 5*a(n), if a(n)=1 (mod 5).

A217116 Greatest number (in decimal representation) with n nonprime substrings in base-6 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

23, 839, 5039, 6983, 7127, 42743, 45863, 46199, 275183, 279143, 277199, 1088531, 1674863, 1651103, 1674859, 6713711, 9906599, 10045559, 10072943, 39190247, 40278647, 60273359, 60295079, 60294239, 60437659, 241671887, 342609527, 359245007, 361640159, 362625959
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} 6^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-6 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-6 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 < 6^(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

			(0) = 23, since 23 = 35_6 (base-6) is the greatest number with zero nonprime substrings in base-6 representation.
a(1) = 839 = 3515_6 has 1 nonprime substring in base-6 representation (= 1). All the other base-6 substrings (3, 5, 15_6=11, 35_6=23, 51_6=31, 351_6=139, 515_6=191 and 3515_6=839) are prime substrings. 839 is the greatest number with 1 nonprime substring.
a(2) = 5039 = 35155_6 has 2 nonprime substrings in base-6 representation (1 and 55_6=35), and there is no greater number with 2 nonprime substrings in base-6 representation.
a(5) = 42743 = 525515_6 has 5 nonprime substrings in base-6 representation, these are 1, 52_6=32, 55_6=35, 5515_6=1271 and 52551_6=7123, and there is no greater number with 5 nonprime substrings in base-6 representation.
		

Crossrefs

Formula

a(n) >= A217106(n).
a(n) >= A217306(A000217(num_digits_6(a(n)))-n), where num_digits_6(x) is the number of digits of the base-6 representation of x.
a(n) <= 6^min(n+3, 7*floor((n+7)/8)).
a(n) <= 216*6^n.
a(n+m+1) >= 6*a(n), where m := floor(log_6(a(n))) + 1.

A217117 Greatest number (in decimal representation) with n nonprime substrings in base-7 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

37, 331, 317, 2322, 2389, 15259, 16260, 16728, 100291, 113825, 116101, 117109, 796777, 796781, 819719, 823003, 4753901, 5577444, 5738035, 5738039, 5761027, 31150219, 39041113, 39336580, 40166250, 40326841, 40336249, 218051538, 273271861
Offset: 0

Views

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} 7^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-7 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-7 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 < 7^(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) = 37, since 37 = 52_7 (base-7) is the greatest number with zero nonprime substrings in base-7 representation.
a(1) = 331 = 652_7 has 1 nonprime substring in base-7 representation (= 6). All the other base-7 substrings (2, 5, 52_7=37, 65_7=47 and 652_7=331) are prime substrings. 331 is the greatest number with 1 nonprime substring.
a(2) = 317 = 632_7 has 6 substrings in base-7 representation (2, 3, 6, 32, 63 and 632), exactly 2 of them are nonprime substrings (6 and 32_6=20), and there is no greater number with 2 nonprime substrings in base-7 representation.
a(8) = 100291 = 565252_3 has 8 nonprime substrings in base-7 representation, these are 6, 252_7, 525_7, 565_7, 5252_7, 5652_7, 6525_7 and 65252_7. There is no greater number with 8 nonprime substrings in base-7 representation.
		

Crossrefs

Formula

a(n) >= A217107(n).
a(n) >= A217307(A000217(num_digits_7(a(n)))-n), where num_digits_7(x) is the number of digits of the base-7 representation of x.
a(n) <= 7^min(n+2, 5*floor((n+4)/5)).
a(n) <= 7^(n+2).
a(n) <= 7^min(3 + n/2, 8*floor((n+15)/16)).
a(n) <= 343*7^(n/2).
With m := floor(log_7(a(n))) + 1:
a(n+m+1) >= 7*a(n), if a(n)!=1 (mod 7).
a(n+m) >= 7*a(n), if a(n)=1 (mod 7).
Showing 1-8 of 8 results.