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.

Showing 1-7 of 7 results.

A255067 Terms of A255066 halved.

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 5, 15, 14, 13, 11, 9, 31, 30, 29, 27, 25, 23, 21, 18, 16, 63, 62, 61, 59, 57, 55, 53, 50, 48, 47, 45, 42, 39, 37, 34, 32, 127, 126, 125, 123, 121, 119, 117, 114, 112, 111, 109, 106, 103, 101, 98, 96, 95, 93, 90, 87, 84, 81, 78, 76, 74, 71, 69, 66, 64, 255
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Crossrefs

Programs

Formula

a(n) = A255066(n)/2.
Other identities. For all n >= 0:
a(n) = A255057(A255122(n)).

A255056 Trunk of number-of-runs beanstalk: The unique infinite sequence such that a(n-1) = a(n) - number of runs in binary representation of a(n).

Original entry on oeis.org

0, 2, 4, 6, 10, 12, 14, 18, 22, 26, 28, 30, 32, 36, 42, 46, 50, 54, 58, 60, 62, 64, 68, 74, 78, 84, 90, 94, 96, 100, 106, 110, 114, 118, 122, 124, 126, 128, 132, 138, 142, 148, 152, 156, 162, 168, 174, 180, 186, 190, 192, 196, 202, 206, 212, 218, 222, 224, 228, 234, 238, 242, 246, 250, 252, 254
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Comments

All numbers of the form (2^n)-2 are present, which guarantees the uniqueness and also provides a well-defined method to compute the sequence, for example, via a partially reversed version A255066.
The sequence was inspired by a similar "binary weight beanstalk", A179016, sharing some general properties with it (like its partly self-copying behavior, see A255071), but also differing in some aspects. For example, here the branching degree is not the constant 2, but can vary from 1 to 4. (Cf. A255058.)

Crossrefs

First differences: A255336.
Terms halved: A255057.
Cf. A255053 & A255055 (the lower & upper bound for a(n)) and also A255123, A255124 (distances to those limits).
Cf. A255327, A255058 (branching degree for node n), A255330 (number of nodes in the finite subtrees branching from the node n), A255331, A255332
Subsequence: A000918 (except for -1).
Similar "beanstalk's trunk" sequences using some other subtracting map than A236840: A179016, A219648, A219666.

Programs

Formula

a(n) = A255066(A255122(n)).
Other identities and observations. For all n >= 0:
a(n) = 2*A255057(n).
A255072(a(n)) = n.
A255053(n) <= a(n) <= A255055(n).

A236840 n minus number of runs in the binary expansion of n: a(n) = n - A005811(n).

Original entry on oeis.org

0, 0, 0, 2, 2, 2, 4, 6, 6, 6, 6, 8, 10, 10, 12, 14, 14, 14, 14, 16, 16, 16, 18, 20, 22, 22, 22, 24, 26, 26, 28, 30, 30, 30, 30, 32, 32, 32, 34, 36, 36, 36, 36, 38, 40, 40, 42, 44, 46, 46, 46, 48, 48, 48, 50, 52, 54, 54, 54, 56, 58, 58, 60, 62, 62, 62, 62, 64, 64, 64
Offset: 0

Views

Author

Antti Karttunen, Apr 18 2014

Keywords

Comments

All terms are even. Used by the "number-of-runs beanstalk" sequence A255056 and many of its associated sequences.

Crossrefs

Cf. A091067 (the positions of records), A106836 (run lengths).
Cf. A255070 (terms divided by 2).

Programs

  • Maple
    A236840 := proc(n) local i, b; if n=0 then 0 else b := convert(n, base, 2); select(i -> (b[i-1]<>b[i]), [$2..nops(b)]); n-1-nops(%) fi end: seq(A236840(i), i=0..69); # Peter Luschny, Apr 19 2014
  • Mathematica
    a[n_] := n - Length@ Split[IntegerDigits[n, 2]]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jul 16 2023 *)
  • Scheme
    (define (A236840 n)  (- n (A005811 n)))

Formula

a(n) = n - A005811(n) = n - A000120(A003188(n)).
a(n) = 2*A255070(n).

A255063 Number of times an evil number is encountered when iterating from 2^(n+1)-2 to (2^n)-2 with the map x -> x - (number of runs in binary representation of x).

Original entry on oeis.org

1, 0, 1, 2, 2, 5, 7, 14, 24, 52, 84, 173, 290, 586, 1038, 2025, 3740, 7177, 13498, 25832, 49027, 93918, 179291, 344128, 660058, 1270590, 2447944, 4728357, 9145214, 17718039, 34365068, 66717630, 129619518, 251953756, 489964171, 953141850, 1854911347
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Examples

			For n=0 we count the evil numbers (A001969) found in range A255056(0..0), and A255056(0) = 0 is an evil number, thus a(0) = 1.
For n=1 we count the evil numbers in range A255056(1..1), and A255056(1) = 2 is not an evil number, thus a(1) = 0.
For n=2 we look at the numbers in range A255056(2..3), i.e. 4 and 6 and while 4 is not an evil number, 6 is, thus a(2) = 1.
For n=5 we look at the numbers in range A255056(12..20) which are (32, 36, 42, 46, 50, 54, 58, 60, 62). Or we take them in the order they come when iterating A236840 (as in A255066(12..20): 62, 60, 58, 54, 50, 46, 42, 36, 32), that is, we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32 which is (2^5). Of the nine numbers encountered, only 60, 58, 54, 46 and 36 are evil numbers, thus a(5) = 5.
		

Crossrefs

Programs

  • PARI
    \\ Compute sequences A255063, A255064 and A255071 at the same time, starting from n=1:
    A005811(n) = hammingweight(bitxor(n, n\2));
    write_A255063_and_A255064_and_A255071(n) = { my(k, i, s63, s64); k = (2^(n+1))-2; i = 1; s63 = 0; s64 = 0; while(1, if((hammingweight(k)%2),s64++,s63++); k = k - A005811(k); if(!bitand(k+1, k+2), break, i++)); write("b255063.txt", n, " ", s63); write("b255064.txt", n, " ", s64); write("b255071.txt", n, " ", i); };
    for(n=1,36,write_A255063_and_A255064_and_A255071(n));
    
  • Scheme
    (define (A255063 n) (if (zero? n) 1 (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s (modulo (+ 1 n) 2))) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A010059 i))))))))
    (define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1)))))
    
  • Scheme
    (define (A255063 n) (add A254113 (A255062 n) (A255061 (+ 1 n))))
    
  • Scheme
    (define (A255063 n) (add (COMPOSE A010059 A255066) (A255062 n) (A255061 (+ 1 n))))

Formula

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A254113(k).
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A010059(A255066(k)).
Other identities. For all n >= 1:
a(n) = A255071(n) - A255064(n).

A255064 Number of times an odious number is encountered when iterating from 2^(n+1)-2 to (2^n)-2 with the map x -> x - (number of runs in binary representation of x).

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 9, 15, 29, 45, 94, 155, 318, 548, 1088, 1976, 3812, 7115, 13617, 25733, 49247, 93739, 179691, 343816, 660735, 1270112, 2448975, 4727786, 9146539, 17717760, 34366228, 66718749, 129619199, 251958752, 489959621, 953155315, 1854898028
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Examples

			For n=0 we count the odious numbers (A000069) found in range A255056(0..0), and A255056(0) = 0 is not an odious number, thus a(0) = 0.
For n=1 we count the odious numbers in range A255056(1..1), and A255056(1) = 2 is an odious number, thus a(1) = 1.
For n=2 we look at the numbers in range A255056(2..3), i.e. 4 and 6 and while 4 is an odious number, 6 is not, thus a(2) = 1.
For n=5 we look at the numbers in range A255056(12..20) which are (32, 36, 42, 46, 50, 54, 58, 60, 62), or if we take them in the order the come when iterating A236840 (as in A255066(12..20): 62, 60, 58, 54, 50, 46, 42, 36, 32), that is, we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32 which is (2^5). Of the nine numbers encountered, only 62, 50, 42 and 32 are odious numbers, thus a(5) = 4.
		

Crossrefs

Programs

Formula

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A254114(k).
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A010060(A255066(k)).
Other identities. For all n >= 1:
a(n) = A255071(n) - A255063(n).

A255122 Simple self-inverse permutation of natural numbers: after zero, list each block of A255071(n) numbers in reverse order, from A255061(n+1) to A255062(n).

Original entry on oeis.org

0, 1, 3, 2, 6, 5, 4, 11, 10, 9, 8, 7, 20, 19, 18, 17, 16, 15, 14, 13, 12, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 118
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Comments

Maps between A255056 and A255066. (Equally, between A255057 and A255067.)

Crossrefs

Programs

Formula

a(n) = A255061(1+A255121(n)) - A255120(n).

A255069 First differences of A255071.

Original entry on oeis.org

1, 1, 2, 4, 7, 13, 24, 44, 81, 150, 280, 526, 992, 1875, 3551, 6740, 12823, 24450, 46709, 89383, 171325, 328962, 632849, 1219909, 2356217, 4559224, 8835610, 17144046, 33295497, 64705083, 125802338, 244673791, 476011284, 926373373, 1803512210, 3512774806
Offset: 1

Views

Author

Antti Karttunen, Feb 21 2015

Keywords

Comments

Also, a(n) = the number of times a number whose binary expansion begins with 10... (cf. A004754) is encountered when iterating from 2^(n+2)-2 to (2^(n+1))-2 with the map x -> x - (number of runs in binary representation of x), i.e., with m(n) = A236840(n). For example, when starting from the initial value (2^(4+2))-2 = 62, we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32, which is (2^(4+1)). Of the nine numbers encountered, only 46, 42, 36 and 32 (in binary: 101110, 101010, 100100 and 100000) are in A004754, thus a(4) = 5.

Crossrefs

First differences of A255071.
Analogous sequence: A226060.

Programs

Formula

a(n) = A255071(n+1) - A255071(n).
For n > 1, a(n-1) = Sum_{k = A255062(n) .. A255061(n+1)}(1-secondmsb(A255056(k))).
Here secondmsb is implemented by the starting offset 2 version of A079944, and effectively gives the second most significant bit in the binary expansion of n. The formula follows from the semi-regular nature of number-of-runs beanstalk, see comments above and at A255071.
Showing 1-7 of 7 results.