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.
Since 9 is 1001 in binary, the 9th row is 1,2,1. Since 11 is 1011 in binary, the 11th row is 1,1,2. Triangle begins: 1; 1,1; 2; 1,2; 1,1,1; 2,1; 3; 1,3;
import Data.List (group) a101211 n k = a101211_tabf !! (n-1) !! (k-1) a101211_row n = a101211_tabf !! (n-1) a101211_tabf = map (reverse . map length . group) $ tail a030308_tabf -- Reinhard Zumkeller, Dec 16 2013
# Maple program due to W. Edwin Clark: Runs := proc (L) local j, r, i, k; j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: # Row n is obtained with the command c(n). - Emeric Deutsch, Jul 03 2017 # Maple program due to W. Edwin Clark, yielding the integer ind corresponding to a given composition (the index of the composition): ind := proc (x) local X, j, i: X := NULL: for j to nops(x) do if type(j, odd) then X := X, seq(1, i = 1 .. x[j]) end if: if type(j, even) then X := X, seq(0, i = 1 .. x[j]) end if end do: X := [X]: add(X[i]*2^(nops(X)-i), i = 1 .. nops(X)) end proc; # Clearly, ind(c(n))= n. - Emeric Deutsch, Jan 23 2018
Table[Length /@ Split@ IntegerDigits[n, 2], {n, 38}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)
apply( {A101211_row(n)=Vecrev((n=vecextract([-1..exponent(n)], bitxor(2*n, bitor(n,1))))[^1]-n[^-1])}, [1..19]) \\ replacing older code by M. F. Hasler, Mar 24 2025
from itertools import groupby
def arow(n): return [len(list(g)) for k, g in groupby(bin(n)[2:])]
def auptorow(rows):
alst = []
for i in range(1, rows+1): alst.extend(arow(i))
return alst
print(auptorow(38)) # Michael S. Branicky, Oct 02 2021
a(0) = 1, as zero has no runs of 1's, and an empty product is 1. a(1) = 1, as 1 is "1" in binary, and the length of that only 1-run is 1. a(2) = 1, as 2 is "10" in binary, and again there is only one run of 1-bits, of length 1. a(3) = 2, as 3 is "11" in binary, and there is one run of two 1-bits. a(55) = 6, as 55 is "110111" in binary, and 2 * 3 = 6. a(119) = 9, as 119 is "1110111" in binary, and 3 * 3 = 9. From _Omar E. Pol_, Feb 10 2015: (Start) Written as an irregular triangle in which row lengths is A011782: 1; 1; 1,2; 1,1,2,3; 1,1,1,2,2,2,3,4; 1,1,1,2,1,1,2,3,2,2,2,4,3,3,4,5; 1,1,1,2,1,1,2,3,1,1,1,2,2,2,3,4,2,2,2,4,2,2,4,6,3,3,3,6,4,4,5,6; ... Right border gives A028310: 1 together with the positive integers. (End) From _Omar E. Pol_, Mar 19 2015: (Start) Also, the sequence can be written as an irregular tetrahedron T(s, r, k) as shown below: 1; .. 1; .. 1; 2; .... 1,1; 2; 3; ........ 1,1,1,2; 2,2; 3; 4; ................ 1,1,1,2,1,1,2,3; 2,2,2,4; 3,3; 4; 5; ................................ 1,1,1,2,1,1,2,3,1,1,1,2,2,2,3,4; 2,2,2,4,2,2,4,6; 3,3,3,6; 4,4; 5; 6; ... Apart from the initial 1, we have that T(s, r, k) = T(s+1, r, k). (End)
a:= proc(n) local i, m, r; m, r:= n, 1;
while m>0 do
while irem(m, 2, 'h')=0 do m:=h od;
for i from 0 while irem(m, 2, 'h')=1 do m:=h od;
r:= r*i
od; r
end:
seq(a(n), n=0..100); # Alois P. Heinz, Jul 11 2013
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(i, i in lis);
ans:=[op(ans), a];
od:
ans; # N. J. A. Sloane, Sep 05 2014
onBitRunLenProd[n_] := Times @@ Length /@ Select[Split @ IntegerDigits[n, 2], #[[1]] == 1 & ]; Array[onBitRunLenProd, 100, 0] (* Jean-François Alcover, Mar 02 2016 *)
from operator import mul from functools import reduce from re import split def A227349(n): return reduce(mul, (len(d) for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014
# uses[RLT from A246660] A227349_list = lambda len: RLT(lambda n: n, len) A227349_list(88) # Peter Luschny, Sep 07 2014
(define (A227349 n) (apply * (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2))))) (define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z))))) (define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
Table begins as: Row n in Terms on n binary that row 1 1 1; 2 10 1,1; 3 11 2; 4 100 2,1; 5 101 1,1,1; 6 110 1,2; 7 111 3; 8 1000 3,1; 9 1001 1,2,1; 10 1010 1,1,1,1; 11 1011 2,1,1; 12 1100 2,2; 13 1101 1,1,2; 14 1110 1,3; 15 1111 4; 16 10000 4,1; etc. with the terms of row n appearing in reverse order compared how the runs of the same length appear in the binary expansion of n (Cf. A101211). From _Omar E. Pol_, Sep 08 2013: (Start) Illustration of initial terms: --------------------------------------- k m Diagram Composition --------------------------------------- . _ 1 1 |_|_ 1; 2 1 |_| | 1, 1, 2 2 |_ _|_ 2; 3 1 |_ | | 2, 1, 3 2 |_|_| | 1, 1, 1, 3 3 |_| | 1, 2, 3 4 |_ _ _|_ 3; 4 1 |_ | | 3, 1, 4 2 |_|_ | | 1, 2, 1, 4 3 |_| | | | 1, 1, 1, 1, 4 4 |_ _|_| | 2, 1, 1, 4 5 |_ | | 2, 2, 4 6 |_|_| | 1, 1, 2, 4 7 |_| | 1, 3, 4 8 |_ _ _ _|_ 4; 5 1 |_ | | 4, 1, 5 2 |_|_ | | 1, 3, 1, 5 3 |_| | | | 1, 1, 2, 1, 5 4 |_ _|_ | | 2, 2, 1, 5 5 |_ | | | | 2, 1, 1, 1, 5 6 |_|_| | | | 1, 1, 1, 1, 1, 5 7 |_| | | | 1, 2, 1, 1, 5 8 |_ _ _|_| | 3, 1, 1, 5 9 |_ | | 3, 2, 5 10 |_|_ | | 1, 2, 2, 5 11 |_| | | | 1, 1, 1, 2, 5 12 |_ _|_| | 2, 1, 2, 5 13 |_ | | 2, 3, 5 14 |_|_| | 1, 1, 3, 5 15 |_| | 1, 4, 5 16 |_ _ _ _ _| 5; . Also irregular triangle read by rows in which row k lists the compositions of k, k >= 1. Triangle begins: [1]; [1,1], [2]; [2,1], [1,1,1], [1,2],[3]; [3,1], [1,2,1], [1,1,1,1], [2,1,1], [2,2], [1,1,2], [1,3], [4]; [4,1], [1,3,1], [1,1,2,1], [2,2,1], [2,1,1,1], [1,1,1,1,1], [1,2,1,1], [3,1,1], [3,2], [1,2,2], [1,1,1,2], [2,1,2], [2,3], [1,1,3], [1,4], [5]; Row k has length A001792(k-1). Row sums give A001787(k), k >= 1. (End)
import Data.List (group) a227736 n k = a227736_tabf !! (n-1) !! (k-1) a227736_row n = a227736_tabf !! (n-1) a227736_tabf = map (map length . group) $ tail a030308_tabf -- Reinhard Zumkeller, Aug 11 2014
Array[Length /@ Reverse@ Split@ IntegerDigits[#, 2] &, 34] // Flatten (* Michael De Vlieger, Dec 11 2020 *)
apply( {A227736_row(n, r=[1], b=n%2)=while(n\=2, n%2==b && r[#r]++ || [b=1-b, r=concat(r,1)]); r}, [1..22]) \\ M. F. Hasler, Mar 11 2025
def A227736_row(n): return[len(list(g))for _,g in groupby(bin(n)[:1:-1])] from itertools import groupby # M. F. Hasler, Mar 11 2025
(define (A227736 n) (A227186bi (A227737 n) (A227740 n))) ;; The Scheme-function for A227186bi has been given in A227186.
8 has binary expansion "1000", whose runs have lengths [3,1] when arranged from the least significant to the most significant end. Taking partial sums of 3 and 0, we get 3 and 3, whose product is 9, thus a(8) = 9. For 44, in binary "101100", the run lengths are [2,2,1,1] (from the least significant end), and subtracting one from all terms except the first one, we get [2,1,0,0], whose partial sums are [2,3,3,3], and 2*3*3*3 = 54, thus a(44)=54.
Table[Function[b, Times @@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 0, 75}] // Flatten (* Michael De Vlieger, May 09 2017 *)
def A227184(n): '''Product of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n.''' p = 1 b = n%2 i = 1 while (n != 0): n >>= 1 if ((n%2) == b): i += 1 else: b = n%2 p *= i return(p)
(define (A227184 n) (if (zero? n) 1 (apply * (binexp_to_ascpart n)))) (define (binexp_to_ascpart n) (let ((runlist (reverse! (binexp->runcount1list n)))) (PARTSUMS (cons (car runlist) (map -1+ (cdr runlist)))))) (define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2))))))) (define (PARTSUMS a) (cdr (reverse! (fold-left (lambda (psums n) (cons (+ n (car psums)) psums)) (list 0) a))))
import Data.List (group)
a246588 = product . map (a000120 . length) .
filter ((== 1) . head) . group . a030308_row
-- Reinhard Zumkeller, Feb 13 2015, Sep 05 2014
A000120 := proc(n) local w, m, i; w := 0; m :=n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120; ans:=[]; for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0; for i from 1 to L1 do if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1; elif out1 = 0 and t1[i] = 1 then c:=c+1; elif out1 = 1 and t1[i] = 0 then c:=c; elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0; fi; if i = L1 and c>0 then lis:=[c, op(lis)]; fi; od: a:=mul(wt(i), i in lis); ans:=[op(ans), a]; od: ans;
f[n_] := DigitCount[n, 2, 1]; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)
from operator import mul from functools import reduce from re import split def A246588(n): return reduce(mul,(bin(len(d)).count('1') for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014
# uses[RLT from A246660] A246588_list = lambda len: RLT(lambda n: sum(Integer(n).digits(2)), len) A246588_list(88) # Peter Luschny, Sep 07 2014
For 8, "1000" in binary, the run lengths are 1 and 3, 1*3=3, and 8-3 = 5, thus a(8)=5. For 24, "11000" in binary, the run lengths are 2 and 3, 2*3=6, and 24-6 = 18, thus a(24)=18.
a227190 n = n - a167489 n -- Reinhard Zumkeller, Jul 05 2013
Table[n-Times@@(Length/@Split[IntegerDigits[n,2]]),{n,70}] (* Harvey P. Dale, Aug 02 2013 *)
(define (A227190 n) (- n (A167489 n))) ;; The Scheme-program for A167489 is found under that entry.
The top-left corner of the array: 0, 0, 0, 0, 0, ... (0, in binary 0, has no runs (by convention), thus at first we have all-0 sequence) 1, 0, 0, 0, 0, ... (1, in binary 1, has one run of length 1) 1, 1, 0, 0, 0, ... (2, in binary 10, has two runs of length 1 both) 2, 0, 0, 0, 0, ... (3, in binary 11, has one run of length 2) 2, 1, 0, 0, 0, ... (4, in binary 100, the rightmost run of length 2 given first, then the second run of length 1) 1, 1, 1, 0, 0, ... (5, in binary 101, has three runs of one bit each) 1, 2, 0, 0, 0, ... 3, 0, 0, 0, 0, ... 3, 1, 0, 0, 0, ... 1, 2, 1, 0, 0, ... 1, 1, 1, 1, 0, ... 2, 1, 1, 0, 0, ... 2, 2, 0, 0, 0, ... 1, 1, 2, 0, 0, ... 1, 3, 0, 0, 0, ... 4, 0, 0, 0, 0, ...
A227186 := proc(n,k) local bdgs,ru,i,b,a; bdgs := convert(n,base,2) ; if nops(bdgs) = 0 then return 0 ; end if; ru := 0 ; i := 1 ; b := op(i,bdgs) ; a := 1 ; for i from 2 to nops(bdgs) do if op(i,bdgs) <> op(i-1,bdgs) then if ru = k then return a; end if; a := 1 ; ru := ru+1 ; else a := a+1 ; end if; end do: if ru =k then a ; else 0 ; end if; end proc: # R. J. Mathar, Jul 23 2013
A227186(n,k)=while(k>=0,for(c=1,n,bittest(n,0)==bittest(n\=2,0)&next;k&break;return(c));n||return;k--) \\ To let A(0,0)=1 add "!n||!" in front of while(...). TO DO: add default value k=-1 and implement "flattened" sequence, such that A227186(n) yields a(n). M. Hasler, Jul 21 2013
(define (A227186 n) (A227186bi (A002262 n) (A025581 n))) (define (A227186bi n k) (cond ((< (A005811 n) (+ 1 k)) 0) ((zero? k) (A136480 n)) (else (A227186bi (A163575 n) (- k 1)))))
a(0) = 1, regardless whether one considers the binary representation of zero to have one zero or no digits at all, because also the empty product is 1. Similarly for any numbers of form (2^k)-1, where no nonleading 0-bits occur, the result of an empty product is always 1. a(8) = 3, as 8 is "1000" in binary, and there is one run of three 0-bits present. a(68) = 6, 68 is "1000100" in binary, there are one run of three 0-bits and one run of two 0-bits, thus 2*3 = 6.
a:= proc(n) local i, m, r; m, r:= n, 1;
while m>0 do
for i from 0 while irem(m, 2, 'h')=0 do m:=h od;
while irem(m, 2, 'h')=1 do m:=h od;
r:= r*max(i, 1)
od; r
end:
seq(a(n), n=0..100); # Alois P. Heinz, Jul 11 2013
a[n_] := Times @@ Length /@ Select[Split @ IntegerDigits[n, 2], #[[1]] == 0 &]; Array[a, 100, 0] (* Jean-François Alcover, Mar 03 2016 *)
(define (A227350 n) (apply * (bisect (reverse (binexp->runcount1list n)) (modulo n 2)))) (define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z))))) (define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
For n=12, A007088(12) = "1100" in binary, the run lengths are [2,2], thus a(12) = lcm(2,2) = 2.
from math import lcm from itertools import groupby def a(n): return lcm(*(len(list(g)) for k, g in groupby(bin(n)[2:]))) print([a(n) for n in range(87)]) # Michael S. Branicky, Oct 15 2022
(define (A284559 n) (apply lcm (binexp->runcount1list n))) ;; Or: (define (A284559 n) (reduce lcm 1 (binexp->runcount1list n))) ;; For binexp->runcount1list, see the Program section of A227349.
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