A277012 -id:A277012 - cp's OEIS frontend

A156552 Unary-encoded compressed factorization of natural numbers.

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144

Offset: 1

Leonid Broukhis, Feb 09 2009

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1024 terms from Antti Karttunen)
Hans Havermann, Factorization of the first 10000 terms, in format [[primes], [exponents]]
A puzzle by Sergey Orlov (in Russian)
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences computed from indices in prime factorization

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. A000079, A000120, A001222, A052126, A054429, A061395, A064216, A064989, A003188, A243071, A243065-A243066, A244153, A243354, A112798, A125106, A056239, A161511.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.
Other related permutations: A253551, A253792, A253564, A253791, A277195, A297163, A297164, A297165, A297166, A302023, A305418, A322863, A322864.

Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)

a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019

# Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
# However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
# Note that the correct answer for n=137 is A156552(137) = 4294967296.
$max = $ARGV[0];
$pow = 0;
foreach $i (2..$max) {
@a = split(/ /, `factor $i`);
shift @a;
$shift = 0;
$cur = 0;
while ($n = int shift @a) {
$prime{$n} = 1 << $pow++ if !defined($prime{$n});
$cur |= $prime{$n} << $shift++;
}
print "$cur, ";
}
print "\n";
(Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
(definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
(definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
;; Antti Karttunen, Jun 26 2014

from sympy import primepi, factorint
def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

More terms from Antti Karttunen, Jun 28 2014

A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 1, 2, 2, 3, 2, 3, 2, 3, 3

Offset: 0

Antti Karttunen, Mar 02 2001

nonn

			19 = 3*(3!) + 0*(2!) + 1*(1!), thus it is written as "301" in factorial base (A007623). The count of nonzero digits in that representation is 2, so a(19) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A007623, A034968, A055091, A060117, A060118, A060128, A060129, A060131, A060502, A257687, A275734, A275735, A276076.
Cf. A227130 (positions of even terms), A227132 (of odd terms).
Cf. also A225901, A232094, A257694, A257695.
The topmost row and the leftmost column in array A230415, the left edge of triangle A230417.
Differs from similar A267263 for the first time at n=30.

A060130(n) = count_nonfixed(convert(PermUnrank3R(n), 'disjcyc'))-nops(convert(PermUnrank3R(n), 'disjcyc')) or nops(fac_base(n))-nops(positions(0, fac_base(n)))
fac_base := n -> fac_base_aux(n, 2); fac_base_aux := proc(n, i) if(0 = n) then RETURN([]); else RETURN([op(fac_base_aux(floor(n/i), i+1)), (n mod i)]); fi; end;
count_nonfixed := l -> convert(map(nops, l), `+`);
positions := proc(e, ll) local a, k, l, m; l := ll; m := 1; a := []; while(member(e, l[m..nops(l)], 'k')) do a := [op(a), (k+m-1)]; m := k+m; od; RETURN(a); end;
# For procedure PermUnrank3R see A060117

Block[{nn = 105, r}, r = MixedRadix[Reverse@ Range[2, -1 + SelectFirst[Range@ 12, #! > nn &]]]; Array[Count[IntegerDigits[#, r], k_ /; k > 0] &, nn, 0]] (* Michael De Vlieger, Dec 30 2017 *)

(define (A060130 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (quotient n i) (+ 1 i) (+ s (if (zero? (remainder n i)) 0 1)))))))
;; Two other implementations, that use memoization-macro definec:
(definec (A060130 n) (if (zero? n) n (+ 1 (A060130 (A257687 n)))))
(definec (A060130 n) (if (zero? n) n (+ (A257511 n) (A060130 (A257684 n)))))
;; Antti Karttunen, Dec 30 2017

a(0) = 0; for n > 0, a(n) = 1 + a(A257687(n)).
a(0) = 0; for n > 0, a(n) = A257511(n) + a(A257684(n)).
a(n) = A060129(n) - A060128(n).
a(n) = A084558(n) - A257510(n).
a(n) = A275946(n) + A275962(n).
a(n) = A275948(n) + A275964(n).
a(n) = A055091(A060119(n)).
a(n) = A069010(A277012(n)) = A000120(A275727(n)).
a(n) = A001221(A275733(n)) = A001222(A275733(n)).
a(n) = A001222(A275734(n)) = A001222(A275735(n)) = A001221(A276076(n)).
a(n) = A046660(A275725(n)).
a(A225901(n)) = a(n).
A257511(n) <= a(n) <= A034968(n).
A275806(n) <= a(n).
a(A275804(n)) = A060502(A275804(n)). [A275804 gives all the positions where this coincides with A060502.]
a(A276091(n)) = A260736(A276091(n)). [A276091 gives all the positions where this coincides with A260736.]

Example-section added, name edited, the old Maple-code moved away from the formula-section, and replaced with all the new formulas by Antti Karttunen, Dec 30 2017

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).

A277022 Primorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

Original entry on oeis.org

0, 1, 2, 5, 6, 13, 4, 9, 10, 21, 22, 45, 12, 25, 26, 53, 54, 109, 28, 57, 58, 117, 118, 237, 60, 121, 122, 245, 246, 493, 8, 17, 18, 37, 38, 77, 20, 41, 42, 85, 86, 173, 44, 89, 90, 181, 182, 365, 92, 185, 186, 373, 374, 749, 188, 377, 378, 757, 758, 1517, 24, 49, 50, 101, 102, 205, 52, 105, 106, 213, 214, 429, 108, 217, 218, 437, 438, 877, 220

Offset: 0

Antti Karttunen, Sep 26 2016

			9 = "111" in primorial base (A002110(0) + A002110(1) + A002110(2) = 9) is converted to three 1-bits, with separating zeros, in binary as "10101" = A007088(21), thus a(9) = 21.
91 = "3001" in primorial base (91 = 3*A002110(3) + A002110(0)) is converted to binary number "1110001" = A007088(113), thus a(91) = 113. Note how two of the zeros come from the primorial base representation and the third zero is an extra separating zero inserted after each run of 1-bits apart from the rightmost 1-run.
120 = "4000" in primorial base (120 = 4*A002110(3)) is converted to the binary number "1111000" = A007088(120), thus a(120) = 120.

Cf. A000035, A000079, A000120, A000225, A005940, A007088, A049345, A156552, A267263, A276084, A276086, A276088, A276093, A276150.
Cf. A277018 (terms sorted into ascending order).
Cf. A277021 (a left inverse).
Differs from analogous A277012 for the first time at n=24, where a(24) = 60, while A277012(24) = 8.

a(0) = 0; for n >= 1, a(n) = A000225(A276088(n))*A000079(A276084(n)) + A000079(A276088(n))*a(A276093(n)).
a(n) = A156552(A276086(n)).
Other identities. For all n >= 0:
A277021(a(n)) = n.
A005940(1+a(n)) = A276086(n).
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A000120(a(n)) = A276150(n).
A069010(a(n)) = A267263(n).

A277007 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run > 1 + the total number of zeros anywhere right of that run.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0

Offset: 0

Antti Karttunen, Sep 25 2016

			For n=3, "11" in binary, the only maximal run of 1-bits is of length 2, and 2 > 0+1 (where 0 is the total number of zeros to the right of it), thus a(3) = 1.
For n=59, "111011" in binary, both the length of run "11" at the least significant end exceeds the limit (see case n=3 above), and also the length of run "111" exceeds 1 + the total number of 0's to the right of it, thus a(59) = 1+1 = 2.
For n=60, "111100" in binary, the length of only run of 1's is 4, and 4 > 2+1, thus a(60) = 1.
For n=118, "1110110" in binary, the length of rightmost run of 1-bits is 2, but that is not > 1+1 (one more than the number of 0-bits right to it). Also, the length of the leftmost run of 1-bits is 3, but that is not > 1+2, thus a(118) = 0.
For n=246, "11110110" in binary, the rightmost run of 1-bits does not contribute, but the leftmost run of 1-bits has now length 4, which is more than 1+2 (where 2 is the total number of 0-bits right of it), thus a(246) = 1.

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

Cf. A277008 (positions of zeros), A277009 (of nonzeros).
Cf. A005940, A277011, A277012, A276077.
Differs from the similar entry A277017 for the first time at n=60, where a(60)=1, while A277017(60)=0.

;; Standalone iterative implementation:
(define (A277007 n) (let loop ((e 0) (n n) (z 0) (r 0)) (cond ((zero? n) (+ e (if (> r (+ 1 z)) 1 0))) ((even? n) (loop (+ e (if (> r (+ 1 z)) 1 0)) (/ n 2) (+ 1 z) 0)) (else (loop e (/ (- n 1) 2) z (+ 1 r))))))

a(n) = A276077(A005940(1+n)).

cp's OEIS Frontend

A277009 Numbers not in range of A277012: numbers such that at least one run of 1-bits in their binary expansion is longer than 1 + the total number of 0-bits anywhere right of that run.

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A277011 Left inverse of A277012.

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A156552 Unary-encoded compressed factorization of natural numbers.

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Mathematica

PARI

PARI

Perl

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A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

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Maple

Mathematica

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

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A277022 Primorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

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A277007 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run > 1 + the total number of zeros anywhere right of that run.

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A277008 Numbers k such that in the binary expansion of k no run of 1-bits is longer than 1 + the total number of 0-bits anywhere to the right of that run.

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