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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A079066 "Memory" of prime(n): the number of distinct (previous) primes contained as substrings in prime(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 1, 2, 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 1, 0, 1, 0, 1, 1, 0, 3, 2, 3, 4, 2, 0, 1, 2, 1, 2, 4, 3, 0, 1, 2, 3, 1, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 1, 3, 2, 1, 3, 3, 2, 3, 3, 4, 2, 3, 3, 1, 3, 3, 3, 4, 4, 2, 2, 3, 0, 0, 2, 1, 3, 2, 2, 2, 0, 2, 1, 1, 2, 3, 1, 0, 0, 2, 1, 2, 4, 2, 3, 2, 2, 1, 3
Offset: 1

Views

Author

Joseph L. Pe, Feb 02 2003

Keywords

Examples

			The primes contained as substrings in prime(3) = 113 are 3, 11, 13. Hence a(30) = 3. 113 is the smallest prime with memory = 3.
		

Crossrefs

Programs

  • Haskell
    import Data.List (isInfixOf)
    a079066 n =
       length $ filter (`isInfixOf` (primesDec !! n)) $ take n primesDec
    primesDec = "_" : map show a000040_list
    -- Reinhard Zumkeller, Jul 19 2011
  • Maple
    a:= n-> (s-> -1+nops(select(t -> t[1]<>"0" and isprime(parse(t)),
            {seq(seq(s[i..j], i=1..j), j=1..length(s))})))(""||(ithprime(n))):
    seq(a(n), n=1..105);  # Alois P. Heinz, Jul 29 2025
  • Mathematica
    ub = 105; tprime = Table[ToString[Prime[i]], {i, 1, ub}]; a = {}; For[i = 1, i <= ub, i++, m = 0; For[j = 1, j < i, j++, If[Length[StringPosition[tprime[[i]], tprime[[j]]]] > 0, m = m + 1]]; a = Append[a, m]]; a

Formula

a(n) = A039997(prime(n)) - 1.
a(n) = A039996(n) - 1. - Alois P. Heinz, Jul 29 2025

Extensions

Edited by Robert G. Wilson v, Feb 25 2003
Name clarified by Sean A. Irvine, Jul 29 2025

A213302 Smallest number with n nonprime substrings (Version 1: substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

2, 1, 11, 10, 103, 101, 100, 1017, 1011, 1002, 1000, 10037, 10023, 10007, 10002, 10000, 100137, 100073, 100023, 100003, 100002, 100000, 1000313, 1000037, 1000033, 1000023, 1000003, 1000002, 1000000, 10000337, 10000223, 10000137, 10000037, 10000023, 10000013, 10000002, 10000000, 100001733
Offset: 0

Views

Author

Hieronymus Fischer, Aug 26 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} 10^j, where k=floor((sqrt(8*n+1)-1)/2), i:= n-A000217(k). For n=0,1,2,3,… the m(n) 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.
The 3 versions according to A213302 - A213304 are quite different. Example: 1002 has 9 nonprime substrings in version 1 (0, 0, 00, 02, 002, 1, 10 100, 1002), in version 2 there are 6 nonprime substrings (02, 002, 1, 10, 100, 1002) and there are 4 nonprime substrings in version 3 (1, 10, 100, 1002).

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)=11, since 11 is the least number with 2 nonprime substrings.
a(3)=10, since 10 is the least number with 3 nonprime substrings, these are 1, 0 and 10 (‘0’ will be counted).
		

Crossrefs

Formula

a(n) >= 10^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n is a triangular number > 0 (cf. A000217).
a(A000217(n)) = 10^(n-1), n>0.
a(A000217(n)-k) >= 10^(n-1)+k, n>0, 0<=k
a(A000217(n)-1) = 10^(n-1)+2, n>3, provided 10^(n-1)+1 is not a prime (which is proved to be true for all n-1 <= 50000 (cf. A185121) except n-1=16384 and is generally true for n-1 unequal to a power of 2).
a(A000217(n)-k) = 10^(n-1)+p, where p is the minimal number such that 10^(n-1) + p, has k prime substrings, n>0, 0<=k
Min(a(A000217(n)-k-i), 0<=i<=m) <= 10^(n-1)+p, where p is the minimal number with k prime substrings and m is the number of digits of p, and k+m
Min(a(A000217(n)-k-i), 0<=i<=A055642(A035244(k)) <= 10^(n-1)+A035244(k).
a(A000217(n)-k) <= 10^(n-1)+max(p(i), k<=i<=k+m), where p(i) is the minimal number with i prime substrings and m is the number of digits of p(i), and k+m
a(A000217(n)-k) <= 10^(n-1)+max(A035244(i), k<=i<=k+ A055642(i).
a(n) <= A213305(n).

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

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).
		

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

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).
		

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.

A078826 Number of distinct primes contained as binary substrings in binary representation of n.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 2, 1, 1, 2, 4, 2, 4, 3, 2, 1, 2, 1, 3, 2, 2, 4, 6, 2, 2, 4, 5, 3, 6, 3, 3, 1, 1, 2, 3, 1, 3, 3, 4, 2, 3, 2, 5, 4, 5, 6, 7, 2, 3, 2, 3, 4, 5, 5, 7, 3, 3, 6, 8, 3, 7, 4, 3, 1, 1, 1, 3, 2, 3, 3, 5, 1, 2, 3, 5, 3, 5, 4, 5, 2, 3, 3, 6, 2, 2, 5, 7, 4, 5, 5, 5, 6, 8, 7, 8, 2, 3, 3, 3, 2, 5, 3, 5, 4
Offset: 0

Author

Reinhard Zumkeller, Dec 08 2002

Keywords

Comments

A143792(n) <= a(n) for n > 0. - Reinhard Zumkeller, Sep 08 2008
For n > 1: number of primes in n-th row of A165416, lengths in n-th row of A225243. - Reinhard Zumkeller, Jul 17 2015, Aug 14 2013

Examples

			n=7 -> '111' contains 2 different binary substrings which are primes: '11' (11b or b11) and '111' itself, therefore a(7)=2.
		

Programs

  • Haskell
    a078826 n | n <= 1 = 0
              | otherwise = length $ a225243_row n
    -- Reinhard Zumkeller, Aug 14 2013
  • Mathematica
    a[n_] := (bits = IntegerDigits[n, 2]; lg = Length[bits]; Reap[Do[If[PrimeQ[p = FromDigits[bits[[i ;; j]], 2]], Sow[p]], {i, 1, lg-1}, {j, i+1, lg}]][[2, 1]] // Union // Length); a[0] = a[1] = 0; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, May 23 2013 *)

A039997 Number of distinct primes which occur as substrings of the digits of n.

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 2, 0, 1, 0, 2, 0, 1, 1, 1, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 3, 1, 1, 0, 1, 1, 2, 0, 1, 0, 2, 0, 0, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 0, 1, 1, 1, 0, 1, 0, 2, 0, 0, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 0, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 2, 0, 0, 0, 1, 1, 2, 0, 1
Offset: 1

Keywords

Examples

			a(22) = 1 because 22 has two substrings which are prime but they are identical. a(103) = 2, since the primes 3 and 103 occur as substrings.
		

Crossrefs

Different from A039995 after the 100th term. Cf. A035232.

Programs

  • Haskell
    import Data.List (isInfixOf)
    a039997 n = length [p | p <- takeWhile (<= n) a000040_list,
                            show p `isInfixOf` show n]
    a039997_list = map a039997 [1..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    a:= n-> (s-> nops(select(t -> t[1]<>"0" and isprime(parse(t)),
            {seq(seq(s[i..j], i=1..j), j=1..length(s))})))(""||n):
    seq(a(n), n=1..100);  # Alois P. Heinz, Aug 09 2022
  • Mathematica
    a[n_] := Block[{s = IntegerDigits[n], c = 0, d = {}}, l = Length[s]; t = Flatten[ Table[ Take[s, {i, j}], {i, 1, l}, {j, i, l}], 1]; k = l(l + 1)/2; While[k > 0, If[ t[[k]][[1]] != 0, d = Append[d, FromDigits[ t[[k]] ]]]; k-- ]; Count[ PrimeQ[ Union[d]], True]]; Table[ a[n], {n, 1, 105}]
  • PARI
    dp(n)=if(n<12,return(if(isprime(n),[n],[])));my(v=vecsort(select(isprime, eval(Vec(Str(n)))),,8),t);while(n>9,if(gcd(n%10,10)>1,n\=10;next);t=10; while((t*=10)Charles R Greathouse IV, Jul 10 2012
    
  • Python
    from sympy import isprime
    def a(n):
        s = str(n)
        ss = (int(s[i:j]) for i in range(len(s)) for j in range(i+1, len(s)+1))
        return len(set(k for k in ss if isprime(k)))
    print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 07 2022

Formula

a(A062115(n)) = 0; a(A093301(n)) = n and a(m) <> n for m < A093301(n). - Reinhard Zumkeller, Jul 16 2007
a(A163753(n)) > 0; a(A205667(n)) = 1. [Reinhard Zumkeller, Jan 31 2012]

Extensions

Edited by Robert G. Wilson v, Feb 24 2003

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

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.
		

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

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.
		

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.

A039995 Number of distinct primes which occur as subsequences of the sequence of digits of n.

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 2, 0, 1, 0, 2, 0, 1, 1, 1, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 3, 1, 1, 0, 1, 1, 2, 0, 1, 0, 2, 0, 0, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 0, 1, 1, 1, 0, 1, 0, 2, 0, 0, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 0, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 2, 0, 0, 0, 2, 1, 3, 0, 1
Offset: 1

Keywords

Comments

a(n) counts subsequences of digits of n which denote primes.

Examples

			a(103) = 3; the 3 primes are 3, 13 and 103.
		

Crossrefs

A039997 counts only the primes which occur as substrings, i.e. contiguous subsequences. Cf. A035232.
Cf. A010051.

Programs

  • Haskell
    import Data.List (subsequences, nub)
    a039995 n = sum $
       map a010051 $ nub $ map read (tail $ subsequences $ show n)
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Mathematica
    cnt[n_] := Module[{d = IntegerDigits[n]}, Length[Union[Select[FromDigits /@ Subsets[d], PrimeQ]]]]; Table[cnt[n], {n, 105}] (* T. D. Noe, Jan 31 2012 *)
  • Python
    from sympy import isprime
    from itertools import chain, combinations as combs
    def powerset(s): # nonempty subsets of s
        return chain.from_iterable(combs(s, r) for r in range(1, len(s)+1))
    def a(n):
        ss = set(int("".join(s)) for s in powerset(str(n)))
        return sum(1 for k in ss if isprime(k))
    print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 07 2022

Formula

a(A094535(n)) = n and a(m) < n for m < A094535(n); A039995(39467139) = 100, cf. A205956. - Reinhard Zumkeller, Feb 01 2012

A085557 Numbers that have more prime digits than nonprime digits.

Original entry on oeis.org

2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77, 122, 123, 125, 127, 132, 133, 135, 137, 152, 153, 155, 157, 172, 173, 175, 177, 202, 203, 205, 207, 212, 213, 215, 217, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232
Offset: 1

Author

Jason Earls, Jul 04 2003

Keywords

Comments

Begins to differ from A046034 at the 21st term (which is the first 3-digit term).

Examples

			133 is in the sequence as the prime digits are 3 and 3 (those are two digits; counted with multiplicity) and one nonprime digit 1 and so there are more prime digits than nonprime digits. - _David A. Corneth_, Sep 06 2020
		

Programs

  • PARI
    is(n) = my(d = digits(n), c = 0); for(i = 1, #d, if(isprime(d[i]), c++)); c<<1 > #d \\ David A. Corneth, Sep 06 2020
    
  • Python
    from itertools import count, islice
    def A085557_gen(startvalue=1): # generator of terms
        return filter(lambda n:len(s:=str(n))<(sum(1 for d in s if d in {'2','3','5','7'})<<1),count(max(startvalue,1)))
    A085557_list = list(islice(A085557_gen(),20)) # Chai Wah Wu, Feb 08 2023
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