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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A213316 Numbers with exactly 9 nonprime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1002, 1003, 1005, 1007, 1009, 1010, 1014, 1016, 1018, 1020, 1024, 1026, 1028, 1041, 1042, 1045, 1049, 1050, 1054, 1056, 1058, 1062, 1065, 1069, 1082, 1085, 1089, 1090, 1094, 1096, 1098, 1099, 1100, 1104, 1106, 1108, 1140, 1144, 1146, 1148
Offset: 1

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is finite. Proof: Each 8-digit number has at least 10 nonprime substrings. Thus each number with more than 8 digits has >= 10 nonprime substrings, too. Consequently there is a boundary b<10^7 such that all numbers > b have more than 9 nonprime substrings.
The first term is a(1) = 1002 = A213302(9). The last term is a(12411) = 9973331 = A213300(9).

Examples

			a(1) = 1002 is in the sequence, since 1002 has 9 nonprime substrings (0,  0, 1, 00, 02, 10, 002, 100, 1002).
a(12411) = 9973331 is in the sequence since there are 9 nonprime substrings (1, 9, 9, 33, 33, 99, 333, 973, 97333).
		

Crossrefs

A213317 Numbers with exactly 10 nonprime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1000, 1001, 1004, 1006, 1008, 1040, 1044, 1046, 1048, 1060, 1064, 1066, 1068, 1080, 1081, 1084, 1086, 1088, 1400, 1404, 1406, 1408, 1440, 1444, 1446, 1448, 1460, 1464, 1466, 1468, 1469, 1480, 1484, 1486, 1488, 1600
Offset: 1

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is finite. Proof: Each 9-digit number has at least 15 nonprime substrings. Thus, each number with more than 9 digits has >= 15 nonprime substrings, too. Consequently, there is a boundary b<10^9, such that all numbers > b have more than 10 nonprime substrings.
The first term is a(1)=1000=A213302(10). The last term is a(20230)=37337397=A213300(10).

Examples

			a(1)=1000, since 1000 has 10 nonprime substrings (0, 0, 0, 1, 00, 00, 10, 000, 100, 1000).
a(20230)= 37337397, since there are 10 nonprime substrings (9, 33, 39, 7337, 7397, 73373, 373373, 733739, 7337397, 37337397).
		

Crossrefs

A213318 Numbers with exactly 11 nonprime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

10037, 10103, 10111, 10117, 10123, 10127, 10130, 10134, 10136, 10138, 10151, 10153, 10157, 10159, 10163, 10167, 10171, 10172, 10175, 10191, 10192, 10195, 10199, 10213, 10217, 10227, 10229, 10231, 10232, 10235, 10239, 10243
Offset: 1

Views

Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is finite. Proof: Each 9-digit number has at least 15 nonprime substrings. Thus, each number with more than 9 digits has >= 15 nonprime substrings, too. Consequently, there is a boundary b<10^9, such that all numbers > b have more than 11 nonprime substrings.
The first term is a(1)=10037=A213302(11). The last term is a(32869)=82337397=A213300(11).

Examples

			a(1)= 10037, since 10037 has 11 nonprime substrings (0, 0, 1, 00, 03, 10, 003, 037, 100, 0037, 1003).
a(32869)= 82337397, since there are 11 nonprime substrings (8, 9, 33, 39, 82, 2337, 7397, 23373, 82337, 233739, 82337397).
		

Crossrefs

A213319 Numbers with exactly 12 nonprime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

10023, 10053, 10067, 10073, 10079, 10093, 10097, 10107, 10112, 10115, 10119, 10122, 10125, 10129, 10141, 10143, 10147, 10152, 10155, 10170, 10174, 10176, 10178, 10181, 10183, 10190, 10194, 10196, 10198, 10212, 10215, 10219
Offset: 1

Views

Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is finite. Proof: Each 9-digit number has at least 15 nonprime substrings. Thus, each number with more than 9 digits has >= 15 nonprime substrings, too. Consequently, there is a boundary b<10^9, such that all numbers > b have more than 12 nonprime substrings.
The first term is a(1)=10023=A213302(12). The last term is a(51477)=99733313=A213300(12).

Examples

			a(1)=10023, since 10023 has 12 nonprime substrings (0, 0, 1, 00, 02, 10, 002, 023, 100, 0023, 1002, 10023).
a(51477)=99733313, since there are 11 nonprime substrings (1, 9, 9, 33, 33, 99, 333, 973, 33313, 97333, 733313, 99733313).
		

Crossrefs

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

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} 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).
Previous Showing 41-48 of 48 results.