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.
A215244(k) = 4 for k = 7,8,12,14,18,22, so a(4) = 22.
The values of A215244(k) for k=8 through 15 are (4, 3, 3, 3, 4, 3, 4, 8), with minimal value a(3) = 3.
A215245 := proc(n) local a,k ; a := A215244(2^n) ; for k from 2^n+1 to 2^(n+1)-1 do a := min(a,A215244(k)) ; end do: a ; end proc: # R. J. Mathar, Aug 07 2012
palQ[L_] := SameQ[L, Reverse[L]];
b[L_] := b[L] = Module[{a = palQ[L] // Boole, c}, For[c = 1, c < Length[L], c++, If[palQ[L[[;; c]]], a = a + b[L[[c+1 ;;]]]]]; a];
a215244[n_] := If[n == 1, 1, b[IntegerDigits[n, 2]]];
a215245[n_] := Module[{a, k}, a = a215244[2^n]; For[k = 2^n+1, k <= 2^(n+1) - 1, k++, a = Min[a, a215244[k]]]; a];
a215245 /@ Range[0, 20] (* Jean-François Alcover, Oct 28 2019 *)
If the numbers are written under each other, there is a suggestion of a pattern (see A215255 for the most obvious pattern). It would be interesting to have more terms to see if the pattern continues. 0 1 1 1 10 10 2 100 100 3 1001 1001 4 10010 10010 5 100101 a 6 1001101 b1 7 10010110 a10 8 101001101 10b1 9 1001011001 a1001 10 10010110010 a10010 11 100101100101 aa 12 1001101001101 bb1 13 10010110010110 aa10 14 101001101001101 10bb1 15 1001011001011001 aa1001 16 10010110010110010 aa10010 17 100101100101100101 aaa 18 1001101001101001101 bbb1 19 10010110010110010110 aaa10 20 101001101001101001101 10bbb1 21 1001011001011001011001 aaa1001 22 10010110010110010110010 aaa10010 23 100101100101100101100101 aaaa 24 1001101001101001101001101 bbbb1 25 10010110010110010110010110 aaaa10 26 101001101001101001101001101 10bbbb1 The rightmost column is obtained by substituting a=100101 and b=100110. A period of 6 is apparent. - _Lars Blomberg_, May 18 2019
A215244 takes the value 4 exactly 6 times, so a(4) = 6.
(11 base 2) = 1011, containing the palindrome 101, therefore a(11) = 3.
import Data.Char (intToDigit, digitToInt)
import Numeric (showIntAtBase)
a050430 n = a050430_list !! (n-1)
a050430_list = f 1 where
f n = g (showIntAtBase 2 intToDigit n "") : f (n+1)
g zs | zs == reverse zs = length zs
| otherwise = max (h $ init zs) (h $ tail zs)
h zs@('0':_) = g zs
h zs@('1':_) = a050430 $ foldl (\v d -> digitToInt d + 2*v) 0 zs
-- Reinhard Zumkeller, Jul 16 2011
# A050430 Length of longest palindromic factor of n for n in [M1..M2] - from N. J. A. Sloane, Aug 07 2012, revised Aug 11 2012 isPal := proc(L) local d ; for d from 1 to nops(L)/2 do if op(d, L) <> op(-d, L) then return false; end if; end do: return true; end proc: # start of main program ans:=[]; M1:=0; M2:=64; for n from M1 to M2 do t1:=convert(n,base,2); rec:=0: l1:=nops(t1); for j1 from 0 to l1-1 do for j2 from j1+1 to l1 do F1 := [op(j1+1..j2,t1)]; if (isPal(F1) and j2-j1>rec) then rec:=j2-j1; fi; od: od: ans:=[op(ans),rec]: od: ans;
f[n_] := Block[{id = IntegerDigits[n, 2]}, k = Length@ id; While[ Union[# == Reverse@# & /@ Partition[id, k, 1]][[-1]] != True, k--]; k]; Array[f, 105] (* Robert G. Wilson v, Jul 16 2011 *)
As the INVERT transform of A006138, (1, 2, 5, 11, 26, 59, ...); a(4) = 104 = (26, 11, 5, 2, 1) dot (1, 1, 3, 10, 32) = (26 + 11 + 15 + 20 + 32).
a := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(2) fi: 2*a(n-1) + 4*a(n-2); end: for n from 1 to 50 do printf(`%d,`,a(n)+a(n-1)) od: f:=n-> simplify(expand((1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1 -sqrt(5))^n )); # N. J. A. Sloane, Aug 10 2012
a[n_]:=(MatrixPower[{{1,5},{1,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2, 4}, {1, 3}, 100] (* G. C. Greubel, Feb 18 2017 *)
{ for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); if (n, write("b063782.txt", n, " ", a + a2)) ) } \\ Harry J. Smith, Aug 31 2009
a(1) = 1, since 1 = 1_2 is the only palindromic bit pattern; a(4) = 4, since 4 = 100_2 and there are the following palindromic bit patterns: 1, 0, 0, 00; a(5) = 4, since 5 = 101_2 and there are the following palindromic bit patterns: 1, 0, 1, 101; a(8) = 7, since 8 = 1000_2 and there are the following palindromic bit patterns: 1, 0, 0, 0, 00, 00, 000.
import Data.Map (fromList, (!), insert)
import Data.List (inits, tails)
a206925 n = a206925_list !! (n-1)
a206925_list = 1 : f [0, 1] (fromList [(Bin [0], 1), (Bin [1], 1)]) where
f bs'@(b:bs) m = y : f (succ bs') (insert (Bin bs') y m) where
y = m ! (Bin bs) +
length (filter (\ds -> ds == reverse ds) $ tail $ inits bs')
succ [] = [1]; succ (0:ds) = 1 : ds; succ (1:ds) = 0 : succ ds
-- Reinhard Zumkeller, Dec 17 2012
a(n)=n=binary(n);sum(k=0,#n-1,sum(i=1,#n-k,prod(j=0, k\2,n[i+j]==n[i+k-j]))) \\ Charles R Greathouse IV, Mar 21 2012
def A206925(n): s = bin(n)[2:] k = len(s) return sum(1 for i in range(k) for j in range(i+1,k+1) if s[i:j] == s[j-1:i-1-k:-1]) # Chai Wah Wu, Jan 31 2023
A206925 "Answers the number of symmetric bit patterns of n as a binary." | m p q n numSym | n := self. n < 2 ifTrue: [^1]. m := n integerFloorLog: 2. p := n printStringRadix: 2. numSym := 0. 1 to: m + 1 do: [:k | 1 to: k do: [:j | q := p copyFrom: j to: k. q = q reverse ifTrue: [numSym := numSym + 1]]]. ^numSym // Hieronymus Fischer, Feb 16 2013
For n = 19: the binary expansion of 19, "10011", can be split in 11 ways into nonempty substrings with no two consecutive equal substrings: - (10011), - (1001)(1), - (100)(11), - (10)(011), - (10)(01)(1), - (10)(0)(11), - (1)(0011), - (1)(001)(1), - (1)(00)(11), - (1)(0)(011), - (1)(0)(01)(1). Hence a(19) = 11.
a(n{, pp=0}) = if (n==0, return (1), my (v=0, p=1); while (n, p=(p*2) + (n%2); n\=2; if (p!=pp, v+=a(n, p))); return (v))
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