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 15 results. Next

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.

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.

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

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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.

A217106 Minimal 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

2, 1, 7, 6, 41, 37, 36, 223, 224, 218, 216, 1319, 1307, 1301, 1297, 1296, 7829, 7793, 7787, 7783, 7778, 7776, 46703, 46709, 46679, 46673, 46663, 46658, 46656, 280205, 280075, 279983, 279979, 279949, 279941, 279938, 279936, 1679879, 1679807, 1679699, 1679669
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} 6^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-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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-6 representation), m>=d, than b := p*6^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_6 is the least number with zero nonprime substrings in base-6 representation.
a(1) = 1, since 1 = 1_6 is the least number with 1 nonprime substring in base-6 representation.
a(2) = 7, since 7 = 11_6 is the least number with 2 nonprime substrings in base-6 representation (1 and 1).
a(3) = 6, since 6 = 10_6 is the least number with 3 nonprime substrings in base-6 representation (0, 1 and 10).
a(4) = 41, since 41 = 105_6 is the least number with 4 nonprime substrings in base-6 representation, these are 0, 1, 10, and 05 (remember, that substrings with leading zeros are considered to be nonprime).
		

Crossrefs

Formula

a(n) >= 6^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n is a triangular number (cf. A000217).
a(A000217(n)) = 6^(n-1), n>0.
a(A000217(n)-k) >= 6^(n-1) + k, 0<=k0.
a(A000217(n)-k) = 6^(n-1) + p, where p is the minimal number >= 0 such that 6^(n-1) + p, has k prime substrings in base-6 representation, 0<=k0.

A217107 Minimal 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

2, 1, 7, 8, 51, 49, 57, 353, 345, 343, 400, 2417, 2411, 2403, 2401, 9604, 16880, 16823, 16829, 16809, 16807, 67228, 117763, 117721, 117666, 117659, 117651, 117649, 470596, 823709, 823664, 823615, 823560, 823553, 823545, 823543, 3294172, 5765310, 5765063
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} 7^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-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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-7 representation), p != 1 (mod 7), m>=d, than b := p*7^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_7 is the least number with zero nonprime substrings in base-7 representation.
a(1) = 1, since 1 = 1_7 is the least number with 1 nonprime substring in base-7 representation.
a(2) = 7, since 7 = 10_7 is the least number with 2 nonprime substrings in base-7 representation (these are 0 and 1).
a(3) = 8, since 8 = 11_7 is the least number with 3 nonprime substrings in base-7 representation (1, 1 and 11).
a(4) = 51, since 51 = 102_7 is the least number with 4 nonprime substrings in base-7 representation, these are 0, 1, 02, and 102 (remember, that substrings with leading zeros are considered to be nonprime).
		

Crossrefs

Formula

a(n) >= 7^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) = 7^(n-1), n>1.
a(A000217(n)) = floor(400 * 7^(n-4)), n>0.
a(A000217(n)) = 111…111_7 (with n digits), n>0.
a(A000217(n)-k) >= 7^(n-1) + k-1, 1<=k<=n, n>1.
a(A000217(n)-k) = 7^(n-1) + p, where p is the minimal number >= 0 such that 7^(n-1) + p, has k prime substrings in base-7 representation, 1<=k<=n, n>1.

A217108 Minimal 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

2, 1, 10, 8, 67, 66, 64, 523, 525, 514, 512, 4127, 4115, 4099, 4098, 4096, 32797, 32799, 32779, 32771, 32770, 32768, 262237, 262239, 262173, 262163, 262147, 262146, 262144, 2097391, 2097259, 2097211, 2097181, 2097169, 2097163, 2097154, 2097152, 16777695
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} 8^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-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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-8 representation), m>=d, than b := p*8^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

Examples

			a(0) = 2, since 2 = 2_8 is the least number with zero nonprime substrings in base-8 representation.
a(1) = 1, since 1 = 1_8 is the least number with 1 nonprime substring in base-8 representation.
a(2) = 10, since 10 = 12_8 is the least number with 2 nonprime substrings in base-8 representation (1 and 12).
a(3) = 8, since 8 = 10_8 is the least number with 3 nonprime substrings in base-8 representation (0, 1 and 10).
a(4) = 67, since 67 = 103_8 is the least number with 4 nonprime substrings in base-8 representation, these are 0, 1, 10, and 03 (remember, that substrings with leading zeros are considered to be nonprime).
		

Crossrefs

Formula

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

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).
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