A247648 -id:A247648 - cp's OEIS frontend

A265705 Triangle read by rows: T(n,k) = k IMPL n, 0 <= k <= n, bitwise logical IMPL.

0, 1, 1, 3, 2, 3, 3, 3, 3, 3, 7, 6, 5, 4, 7, 7, 7, 5, 5, 7, 7, 7, 6, 7, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 15, 14, 13, 12, 11, 10, 9, 8, 15, 15, 15, 13, 13, 11, 11, 9, 9, 15, 15, 15, 14, 15, 14, 11, 10, 11, 10, 15, 14, 15, 15, 15, 15, 15, 11, 11, 11, 11, 15

Offset: 0

Reinhard Zumkeller, Dec 15 2015

			.          10 | 1010                            12 | 1100
.           4 |  100                             6 |  110
.   ----------+-----                     ----------+-----
.   4 IMPL 10 | 1011 -> T(10,4)=11       6 IMPL 12 | 1101 -> T(12,6)=13
.
First 16 rows of the triangle, where non-symmetrical rows are marked, see comment concerning A158582 and A089633:
.   0:                                 0
.   1:                               1   1
.   2:                             3   2   3
.   3:                           3   3   3   3
.   4:                         7   6   5   4   7    X
.   5:                       7   7   5   5   7   7
.   6:                     7   6   7   6   7   6   7
.   7:                   7   7   7   7   7   7   7   7
.   8:                15  14  13  12  11  10   9   8  15    X
.   9:              15  15  13  13  11  11   9   9  15  15    X
.  10:            15  14  15  14  11  10  11  10  15  14  15    X
.  11:          15  15  15  15  11  11  11  11  15  15  15  15
.  12:        15  14  13  12  15  14  13  12  15  14  13  12  15    X
.  13:      15  15  13  13  15  15  13  13  15  15  13  13  15  15
.  14:    15  14  15  14  15  14  15  14  15  14  15  14  15  14  15
.  15:  15  15  15  15  15  15  15  15  15  15  15  15  15  15  15  15 .

Reinhard Zumkeller, Rows n = 0..255 of triangle, flattened
Eric Weisstein's World of Mathematics, Implies

Cf. A003817, A007088, A029578, A089633, A158582, A247648, A265716 (central terms), A265736 (row sums).
Cf. A053644, A265885, A327490.
Other triangles: A080099 (AND), A080098 (OR), A051933 (XOR), A102037 (CNIMPL).

a265705_tabl = map a265705_row [0..]
a265705_row n = map (a265705 n) [0..n]
a265705 n k = k `bimpl` n where
   bimpl 0 0 = 0
   bimpl p q = 2 * bimpl p' q' + if u <= v then 1 else 0
               where (p', u) = divMod p 2; (q', v) = divMod q 2

using IntegerSequences
for n in 0:15 println(n == 0 ? [0] : [Bits("IMP", k, n) for k in 0:n]) end  # Peter Luschny, Sep 25 2021

A265705 := (n, k) -> Bits:-Implies(k, n):
seq(seq(A265705(n, k), k=0..n), n=0..11); # Peter Luschny, Sep 23 2019

T[n_, k_] := If[n == 0, 0, BitOr[2^Length[IntegerDigits[n, 2]]-1-k, n]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 25 2021, after David A. Corneth's PARI code *)

T(n, k) = if(n==0,return(0)); bitor((2<David A. Corneth, Sep 24 2021

T(n,0) = T(n,n) = A003817(n).
T(2*n,n) = A265716(n).
Let m = A089633(n): T(m,k) = T(m,m-k), k = 0..m.
Let m = A158582(n): T(m,k) != T(m,m-k) for at least one k <= n.
Let m = A247648(n): T(2*m,m) = 2*m.
For n > 0: A029578(n+2) = number of odd terms in row n; no even terms in odd-indexed rows.
A265885(n) = T(prime(n),n).
A053644(n) = smallest k such that row k contains n.

A247649 Number of terms in expansion of f^n mod 2, where f = 1/x^2 + 1/x + 1 + x + x^2 mod 2.

Original entry on oeis.org

1, 5, 5, 7, 5, 17, 7, 19, 5, 25, 17, 19, 7, 31, 19, 25, 5, 25, 25, 35, 17, 61, 19, 71, 7, 35, 31, 41, 19, 71, 25, 77, 5, 25, 25, 35, 25, 85, 35, 95, 17, 85, 61, 71, 19, 91, 71, 77, 7, 35, 35, 49, 31, 107, 41, 121, 19, 95, 71, 85, 25, 113, 77, 103

Offset: 0

N. J. A. Sloane, Sep 25 2014

nonn

This is the number of cells that are ON after n generations in a one-dimensional cellular automaton defined by the odd-neighbor rule where the neighborhood consists of 5 contiguous cells.

a(n) is also the number of odd entries in row n of A035343. - Leon Rische, Feb 02 2023

			The first few generations are:
..........X..........
........XXXXX........
......X.X.X.X.X......
....XX..X.X.X..XX.... (f^3)
..X...X...X...X...X..
XXXX.XXX.XXX.XXX.XXXX
...
f^3 mod 2 = x^6 + x^5 + x^2 + 1/x^2 + 1/x^5 + 1/x^6 + 1 has 7 terms, so a(3) = 7.
From _Omar E. Pol_, Mar 02 2015: (Start)
Also, written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
  1;
  5;
  5, 7;
  5,17, 7,19;
  5,25,17,19, 7,31,19,25;
  5,25,25,35,17,61,19,71, 7,35,31,41,19,71,25,77;
  5,25,25,35,25,85,35,95,17,85,61,71,19,91,71,77,7,35,35,49,31,107,41,121,19, ...
(End)
It follows from the Generalized Run Length Transform result mentioned in the comments that in each row the first quarter of the terms (and no more) are equal to 5 times the beginning of the sequence itself. It cannot be said that the rows converge (in any meaningful sense) to five times the sequence. - _N. J. A. Sloane_, Mar 03 2015

Chai Wah Wu, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for sequences related to cellular automata

Cf. A071053, A247647, A247648, A253085, A255490.

Partial sums are in A255654.

import sympy
from functools import reduce
from operator import mul
x = sympy.symbols('x')
f = 1/x**2+1/x+1+x+x**2
A247649_list, g = [1], 1
for n in range(1,1001):
    s = [int(d,2) for d in bin(n)[2:].split('00') if d != '']
    g = (g*f).expand(modulus=2)
    if len(s) == 1:
        A247649_list.append(g.subs(x,1))
    else:
        A247649_list.append(reduce(mul,(A247649_list[d] for d in s)))
# Chai Wah Wu, Sep 25 2014

The values of a(n) for n in A247647 (or A247648) determine all the values, as follows. Parse the binary expansion of n into terms from A247647 separated by at least two zeros: m_1 0...0 m_2 0...0 m_3 ... m_r 0...0. Ignore any number (one or more) of trailing zeros. Then a(n) = a(m_1)*a(m_2)*...*a(m_r). For example, n = 37_10 = 100101_2 is parsed into 1.00.101, and so a(37) = a(1)*a(5) = 5*17 = 85. This is a generalization of the Run Length Transform.

A277020 Unary-binary representation of Stern polynomials: a(n) = A156552(A260443(n)).

Original entry on oeis.org

0, 1, 2, 5, 4, 13, 10, 21, 8, 45, 26, 93, 20, 109, 42, 85, 16, 173, 90, 477, 52, 957, 186, 733, 40, 749, 218, 1501, 84, 877, 170, 341, 32, 685, 346, 3549, 180, 12221, 954, 7133, 104, 14269, 1914, 49021, 372, 28605, 1466, 5853, 80, 5869, 1498, 30685, 436, 61373, 3002, 23517, 168, 12013, 1754, 24029, 340, 7021, 682, 1365

Offset: 0

Antti Karttunen, Oct 07 2016

Sequence encodes Stern polynomials (see A125184, A260443) with "unary-binary method" where any nonzero coefficient c > 0 is encoded as a run of c 1-bits, separated from neighboring 1-runs by exactly one zero (this follows because A260442 is a subsequence of A073491). See the examples.

Terms which are not multiples of 4 form a subset of A003754, or in other words, each term is 2^k * {a term from a certain subsequence of A247648}, which subsequence remains to be determined.

			n    Stern polynomial       Encoded as              a(n)
                            "unary-binary" number   (-> decimal)
----------------------------------------------------------------
0    B_0(x) = 0                     "0"               0
1    B_1(x) = 1                     "1"               1
2    B_2(x) = x                    "10"               2
3    B_3(x) = x + 1               "101"               5
4    B_4(x) = x^2                 "100"               4
5    B_5(x) = 2x + 1             "1101"              13
6    B_6(x) = x^2 + x            "1010"              10
7    B_7(x) = x^2 + x + 1       "10101"              21
8    B_8(x) = x^3                "1000"               8
9    B_9(x) = x^2 + 2x + 1     "101101"              45

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

Cf. A087808 (a left inverse), A156552, A260443, A277189 (odd bisection).

Cf. also A000079, A000120, A000225, A002450, A002487, A003754, A073491, A247648, A260442, A277010, A277012, A276081.

(define (A277020 n) (A156552 (A260443 n)))
;; Another implementation, more practical to run:
(define (A277020 n) (list_of_numbers_to_unary_binary_representation (A260443as_index_lists n)))
(definec (A260443as_index_lists n) (cond ((zero? n) (list)) ((= 1 n) (list 1)) ((even? n) (cons 0 (A260443as_index_lists (/ n 2)))) (else (add_two_lists (A260443as_index_lists (/ (- n 1) 2)) (A260443as_index_lists (/ (+ n 1) 2))))))
(define (add_two_lists nums1 nums2) (let ((len1 (length nums1)) (len2 (length nums2))) (cond ((< len1 len2) (add_two_lists nums2 nums1)) (else (map + nums1 (append nums2 (make-list (- len1 len2) 0)))))))
(define (list_of_numbers_to_unary_binary_representation nums) (let loop ((s 0) (nums nums) (b 1)) (cond ((null? nums) s) (else (loop (+ s (* (A000225 (car nums)) b)) (cdr nums) (* (A000079 (+ 1 (car nums))) b))))))

a(n) = A156552(A260443(n)).
Other identities. For all n >= 0:
A087808(a(n)) = n.
A000120(a(n)) = A002487(n).
a(2n) = 2*a(n).
a(2^n) = 2^n.
a(A000225(n)) = A002450(n).

A385816 The number k such that the k-th composition in standard order lists the maximal anti-run lengths of the binary indices of n. Standard composition number of row n of A384877.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 6, 3, 5, 7, 15, 1, 2, 2, 6, 2, 4, 6, 14, 3, 5, 5, 13, 7, 11, 15, 31, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14, 30, 3, 5, 5, 13, 5, 9, 13, 29, 7, 11, 11, 27, 15, 23, 31, 63, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14

Offset: 0

Gus Wiseman, Jul 15 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
If the k-th composition in standard order is y, then the standard composition number of y is defined to be k.

			The binary indices of 181 are {1,3,5,6,8}, with maximal anti-runs ((1,3,5),(6,8)), with lengths (3,2), which is the 18th composition in standard order, so a(181) = 18.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The reverse version is A209859.
Sorted positions of first appearances are A247648.
These are standard composition numbers of rows of A384877 (duplicates removed A385886).
For runs instead of anti-runs the reverse is A385887 (duplicates removed A232559).
For runs instead of anti-runs we have A385889 (duplicates removed A385818).
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
Cf. A000120, A029931, A044813, A048793, A052499, A069010, A348366, A384175, A384878, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 5, 3, 6, 7, 15, 1, 2, 2, 5, 2, 4, 5, 11, 3, 6, 6, 13, 7, 14, 15, 31, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 3, 6, 6, 13, 6, 12, 13, 27, 7, 14, 14, 29, 15, 30, 31, 63, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 2, 4, 4, 9, 4, 8, 9, 19, 5, 10, 10, 21, 11, 22, 23, 47, 3, 6, 6, 13, 6, 12, 13, 27, 6, 12, 12, 25, 13

Offset: 0

Antti Karttunen, Mar 24 2012

This is the number k such that the k-th composition in standard order is the reversed sequence of lengths of the maximal anti-runs of the binary indices of n. Here, the binary indices of n are row n of A048793, and the k-th composition in standard order is row k of A066099. For example, the binary indices of 100 are {3,6,7}, with maximal anti-runs ((3,6),(7)), with reversed lengths (1,2), which is the 6th composition in standard order, so a(100) = 6. - Gus Wiseman, Jul 27 2025

			102 in binary is 1100110, we rewrite it from the left so that first two 1's stay same ("11"), then "001" is rewritten to "0", the last 1 to "1", and we ignore the last 0, thus getting 1101, which is binary expansion of 13, thus a(102) = 13.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

This is an "inverse" of A071162, i.e. a(A071162(n)) = n for all n. Bisection: A209639. Used to construct permutation A209862.
Removing duplicates appears to give A358654.
Sorted positions of firsts appearances appear to be A247648+1.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A023758, A029931, A044813, A048793, A052499, A069010, A348366, A384878, A384879, A384893.
Cf. A384877, A385816, A385887, A385889.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Reverse[Length/@Split[bpe[n],#2!=#1+1&]]],{n,0,100}] (* Gus Wiseman, Jul 25 2025 *)

import re
def a(n): return int(re.sub("0+1", "0", bin(n)[2:].rstrip("0")), 2) if n else 0
print([a(n) for n in range(109)])  # Michael S. Branicky, Jul 25 2025

(define (A209859 n) (let loop ((n n) (s 0) (i (A053644 n))) (cond ((zero? n) s) ((> i n) (if (> (/ i 2) n) (loop n s (/ i 2)) (loop (- n (/ i 2)) (* 2 s) (/ i 4)))) (else (loop (- n i) (+ (* 2 s) 1) (/ i 2))))))

a(n) = a(A000265(n)).

A385889 The number k such that the k-th composition in standard order is the sequence of lengths of maximal runs of binary indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 4, 1, 3, 3, 5, 2, 6, 4, 8, 1, 3, 3, 5, 3, 7, 5, 9, 2, 6, 6, 10, 4, 12, 8, 16, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 2, 6, 6, 10, 6, 14, 10, 18, 4, 12, 12, 20, 8, 24, 16, 32, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 3

Offset: 0

Gus Wiseman, Jul 16 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 27 are {1,2,4,5}, with maximal runs ((1,2),(4,5)), with lengths (2,2), which is the 10th composition in standard order, so a(27) = 10.
The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with lengths (1,2), which is the 6th composition in standard order, so a(100) = 6.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

Sorted positions of firsts appearances appear to be A247648+1.
After removing duplicates we get A385818.
The reverse version is A385887.
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
A384877 lists anti-run lengths of binary indices (ranks A385816), reverse A209859.
Cf. A000120, A023758, A029931, A044813, A048793, A069010, A348366, A384175, A384878, A385817.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Length/@Split[bpe[n],#2==#1+1&]],{n,0,100}]

A247647 Binary numbers that begin and end with 1 and do not contain two adjacent zeros.

Original entry on oeis.org

1, 11, 101, 111, 1011, 1101, 1111, 10101, 10111, 11011, 11101, 11111, 101011, 101101, 101111, 110101, 110111, 111011, 111101, 111111, 1010101, 1010111, 1011011, 1011101, 1011111, 1101011, 1101101, 1101111, 1110101, 1110111, 1111011, 1111101, 1111111, 10101011, 10101101, 10101111, 10110101, 10110111, 10111011, 10111101

Offset: 1

N. J. A. Sloane, Sep 25 2014

Chai Wah Wu, Table of n, a(n) for n = 1..1000

See A247648 for the decimal equivalents.

Cf. A007088, A247649, A253085.

a247647 = a007088 . a247648  -- Reinhard Zumkeller, Sep 25 2014

With[{upto=500},Map[FromDigits,Select[IntegerString[Range[1,upto,2],2],StringFreeQ[#,"00"]&]]] (* Paolo Xausa, Dec 06 2023 *)

A247647_list = [int(bin(n)[2:]) for n in range(1,10**5) if n % 2 and not '00' in bin(n)]
# Chai Wah Wu, Sep 25 2014

a(n) = A007088(A247648(n)).

A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

Original entry on oeis.org

1, 3, 7, 10, 15, 22, 26, 31, 36, 46, 54, 58, 63, 76, 84, 94, 100, 110, 118, 122, 127, 136, 156, 172, 180, 190, 204, 212, 222, 228, 238, 246, 250, 255, 280, 296, 316, 328, 348, 364, 372, 382, 392, 412, 428, 436, 446, 460, 468, 478, 484, 494, 502, 506, 511, 528, 568

Offset: 1

Mikhail Kurkov, Jul 15 2022

Numbers k such that A007814(k) = A086784(k).
To reproduce the sequence through itself, use the following rule: if binary 1xyz is a term then so are 11xyz and 10xyz0 (except for 1 alone where 100 is not a term).
The number of terms with bit length k is equal to Fibonacci(k-1) for k > 1.
Conjecture: 2*A247648(n-1) + 1 with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n) without trailing 0 bits in binary.
Odd terms are positive Mersenne numbers (A000225), because there is no 0 in their binary expansion. - Bernard Schott, Oct 12 2022

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A000120, A000523, A007814, A010056, A025480, A030109, A054429, A060142, A072649, A084471, A086784, A091892, A132665, A133512, A200650, A213911, A232559, A280514, A343152, A348366.
Cf. A356385 (first differences).
Subsequences with same number k of trailing 0 bits and other 0 bits: A000225 (k=0), 2*A190620 (k=1), 4*A357773 (k=2), 8*A360573 (k=3).

N:= 10: # for terms <= 2^N
S:= {1};
for d from 1 to N do
  for k from 0 to d/2-1 do
    B:= combinat:-choose([$k+1..d-2],k);
    S:= S union convert(map(proc(t) local s; 2^d - 2^k - add(2^(s),s=t) end proc,B),set);
od od:
sort(convert(S,list)); # Robert Israel, Sep 21 2023

Join[{1}, Select[Range[2, 600], IntegerExponent[#, 2] == Floor[Log2[# - 1]] - DigitCount[# - 1, 2, 1] &]] (* Amiram Eldar, Jul 16 2022 *)

isok(k) = if (k==1, 1, (logint(k-1, 2)-hammingweight(k-1) == valuation(k, 2))); \\ Michel Marcus, Jul 16 2022

from itertools import islice, count
def A353654_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:(m:=(~n & n-1).bit_length()) == bin(n>>m)[2:].count('0'),count(max(startvalue,1)))
A353654_list = list(islice(A353654_gen(),30)) # Chai Wah Wu, Oct 14 2022

a(n) = a(n-1) + A356385(n-1) for n > 1 with a(1) = 1.
Conjectured formulas: (Start)
a(n) = 2^g(n-1)*(h(n-1) + 2^A000523(h(n-1))*(2 - g(n-1))) for n > 2 with a(1) = 1, a(2) = 3 where f(n) = n - A130312(n), g(n) = [n > 2*f(n)] and where h(n) = a(f(n) + 1).
a(n) = 1 + 2^r(n-1) + Sum_{k=1..r(n-1)} (1 - g(s(n-1, k)))*2^(r(n-1) - k) for n > 1 with a(1) = 1 where r(n) = A072649(n) and where s(n, k) = f(s(n, k-1)) for n > 0, k > 1 with s(n, 1) = n.
a(n) = 2*(2 + Sum_{k=1..n-2} 2^(A213911(A280514(k)-1) + 1)) - 2^A200650(n) for n > 1 with a(1) = 1.
A025480(a(n)-1) = A348366(A343152(n-1)) for n > 1.
a(A000045(n)) = 2^(n-1) - 1 for n > 1. (End)

A358654 a(n) = A025480(A353654(n+1) - 1).

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 9, 10, 23, 12, 27, 29, 30, 63, 8, 19, 21, 22, 47, 25, 26, 55, 28, 59, 61, 62, 127, 17, 18, 39, 20, 43, 45, 46, 95, 24, 51, 53, 54, 111, 57, 58, 119, 60, 123, 125, 126, 255, 16, 35, 37, 38, 79, 41, 42, 87, 44, 91, 93

Offset: 0

Mikhail Kurkov, Nov 25 2022

Permutation of the nonnegative integers.

Conjecture: A247648(n) with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n).

Cf. A025480, A048679, A247648, A343152, A348366, A353654, A355489.

Conjecture: a(n) = A348366(A343152(n)) for n > 0 with a(0) = 1.

A277189 Odd bisection of A277020: a(n) = A277020(2n+1).

Original entry on oeis.org

1, 5, 13, 21, 45, 93, 109, 85, 173, 477, 957, 733, 749, 1501, 877, 341, 685, 3549, 12221, 7133, 14269, 49021, 28605, 5853, 5869, 30685, 61373, 23517, 12013, 24029, 7021, 1365, 2733, 28125, 192445, 97245, 384957, 2031485, 1032125, 113629, 227261, 4128637, 16252669, 3112829, 1564605, 6225789, 913341, 46813, 46829, 915421

Offset: 0

Antti Karttunen, Oct 08 2016

Antti Karttunen, Table of n, a(n) for n = 0..8191

Cf. A156552, A277020, A277324.

Terms form a proper subset of A247648.

(define (A277189 n) (A277020 (+ n n 1)))

a(n) = A277020(2*n + 1).
a(n) = A156552(A277324(n)).
Other identities. For all n >= 0:
a(n) = 1 mod 4.

Offset changed from 1 to 0 by Antti Karttunen, Oct 11 2016

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