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 1-10 of 23 results. Next

A217302 Minimal natural number (in decimal representation) with n prime substrings in binary representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 2, 5, 7, 11, 15, 27, 23, 31, 55, 47, 63, 111, 95, 187, 127, 223, 191, 381, 255, 447, 503, 383, 511, 1015, 895, 767, 1023, 1533, 1791, 1535, 1919, 3039, 3069, 3067, 3839, 3967, 6079, 6139, 6135, 7679, 8063, 8159, 12159, 12271, 15359, 16127
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 in binary representation is not empty. Proof: A000975(n+1) has exactly n prime substrings in binary representation (see A000975).
All terms with n > 1 are odd.

Examples

			a(1) = 2 = 10_2, since 2 is the least number with 1 prime substring (=10_2) in binary representation.
a(2) = 5 = 101_2, since 5 is the least number with 2 prime substrings in binary representation (10_2 and 101_2).
a(4) = 11 = 1011_2, since 11 is the least number with 4 prime substrings in binary representation (10_2, 11_2, 101_2 and 1011_2).
a(8) = 31 = 11111_2, since 31 is the least number with 8 prime substrings in binary representation (4 times 11_2, 3 times 111_2, and 11111_2).
a(9) = 47 = 101111_2, since 47 is the least number with 9 prime substrings in binary representation (10_2, 3 times 11_2, 101_2, 2 times 111_2, 1011_2, and 10111_2).
		

Crossrefs

Formula

a(n) >= 2^ceiling(sqrt(8*n+1)-1)/2).
a(n) <= A000975(n+1).
a(n+1) <= 2*a(n)+1.

A217102 Minimal 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, 2, 7, 5, 4, 11, 10, 12, 8, 22, 21, 19, 17, 16, 60, 39, 37, 34, 36, 32, 83, 71, 74, 69, 67, 66, 64, 143, 139, 141, 135, 134, 131, 130, 128, 283, 271, 269, 263, 267, 262, 261, 257, 256, 541, 539, 527, 526, 523, 533, 519, 514, 516, 512, 1055, 1053, 1047, 1067
Offset: 1

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Author

Hieronymus Fischer, Dec 12 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 (Cf. A217103-A217109, A213302).
If p is a number with k prime substrings and d digits (in binary representation), p even, m>=d, than b := p*2^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(1) = 1, since 1 = 1_2 is the least number with 1 nonprime substring in binary representation.
a(2) = 2, since 2 = 10_2 is the least number with 2 nonprime substrings in binary representation (0 and 1).
a(3) = 7, since 7 = 111_2 is the least number with 3 nonprime substrings in binary representation (3-times 1, the prime substrings are 2-times 11 and 111).
a(10) = 22, since 22 = 10110_2 is the least number with 10 nonprime substrings in binary representation, these are 0, 0, 1, 1, 1, 01, 011, 110, 0110 and 10110 (remember, that substrings with leading zeros are considered to be nonprime).
		

Crossrefs

Formula

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

A217109 Minimal 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

2, 1, 12, 9, 83, 84, 81, 748, 740, 731, 729, 6653, 6581, 6563, 6564, 6561, 59222, 59069, 59068, 59051, 59052, 59049, 531614, 531569, 531464, 531460, 531452, 531443, 531441, 4784122, 4783142, 4783147, 4783070, 4782989, 4782971, 4782972, 4782969, 43048283
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} 9^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-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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-9 representation), m>=d, than b := p*9^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_9 is the least number with zero nonprime substrings in base-9 representation.
a(1) = 1, since 1 = 1_9 is the least number with 1 nonprime substring in base-9 representation.
a(2) = 12, since 12 = 13_9 is the least number with 2 nonprime substrings in base-9 representation (1 and 13).
a(3) = 9, since 9 = 10_9 is the least number with 3 nonprime substrings in base-9 representation (0, 1 and 10).
a(4) = 83, since 83 = 102_9 is the least number with 4 nonprime substrings in base-9 representation, these are 0, 1, 10, and 02 (remember, that substrings with leading zeros are considered to be nonprime).
		

Crossrefs

Formula

a(n) >= 9^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n is a triangular number (cf. A000217).
a(A000217(n)) = 9^(n-1), n>0.
a(A000217(n)-k) >= 9^(n-1) + k, 0<=k0.
a(A000217(n)-k) = 9^(n-1) + p, where p is the minimal number >= 0 such that 9^(n-1) + p, has k prime substrings in base-9 representation, 0<=k0.

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.

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

Views

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.

A217104 Minimal 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

2, 1, 5, 4, 19, 17, 16, 75, 67, 66, 64, 269, 263, 266, 257, 256, 1053, 1031, 1035, 1029, 1026, 1024, 4125, 4119, 4123, 4107, 4099, 4098, 4096, 16479, 16427, 16431, 16407, 16395, 16391, 16386, 16384, 65709, 65629, 65579, 65581, 65559, 65543, 65539, 65537, 65536
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} 4^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-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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-4 representation), m>=d, than b := p*4^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_4 is the least number with zero nonprime substrings in base-4 representation.
a(1) = 1, since 1 = 1_4 is the least number with 1 nonprime substring in base-4 representation.
a(2) = 5, since 5 = 11_4 is the least number with 2 nonprime substrings in base-4 representation (these are 2-times 1).
a(3) = 4, since 4 = 10_4 is the least number with 3 nonprime substrings in base-4 representation (these are 0, 1 and 10).
a(4) = 19, since 19 = 103_4 is the least number with 4 nonprime substrings in base-4 representation, these are 0, 1, 10, and 03 (remember, that substrings with leading zeros are considered to be nonprime).
a(7) = 75, since 75 = 1023_4 is the least number with 7 nonprime substrings in base-4 representation, these are 0, 1, 10, 02, 023, 102 and 1023 (remember, that substrings with leading zeros are considered to be nonprime: 2_4 = 2, 3_4 = 3 and 23_4 = 11 are the only base-4 prime substrings of 75).
		

Crossrefs

Formula

a(n) >= 4^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n is a triangular number (cf. A000217).
a(A000217(n)) = 4^(n-1), n>0.
a(A000217(n)-k) >= 4^(n-1) + k, 0<=k0.
a(A000217(n)-k) = 4^(n-1) + p, where p is the minimal number >= 0 such that 4^(n-1) + p, has k prime substrings in base-4 representation, 0<=k0.

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

Views

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.
Showing 1-10 of 23 results. Next