A255067 -id:A255067 - cp's OEIS frontend

A255057 The trunk of number-of-runs beanstalk, halved: a(n) = A255056(n)/2.

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

Offset: 0

Antti Karttunen, Feb 14 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..16142

First differences: A255337.
Characteristic function: A255339.
Cf. A255056, A255058, A255067, A255122, A255125, A255126, A255328, A255329, A255330, A255338.

a(n) = A255056(n)/2.

a(n) = A255067(A255122(n)).

A255125 Number of times a multiple of four 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, 1, 1, 3, 6, 13, 26, 47, 81, 140, 253, 482, 949, 1875, 3666, 7088, 13614, 26100, 50082, 96246, 185131, 356123, 684758, 1316197, 2530257, 4868019, 9378335, 18096921, 34974646, 67669905, 130998912, 253565649, 490501587, 947992195, 1830664188, 3533571444

Offset: 0

Antti Karttunen, Feb 18 2015

nonn

Also the number of even numbers in range [A255062(n) .. A255061(n+1)] of A255057 (equally, in A255067). See the sum-formulas.

			For n=5 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, m(36) = 32 and finally m(32) = 30, which is (2^5)-2. Of the nine numbers encountered, only 60, 36 and 32 are multiples of four, thus a(5) = 3.

A. Karttunen, Ratio a(n)/A255126(n) plotted, with the help of OEIS-server

Cf. A005811, A059841, A133872, A236840, A255056, A255057, A255061, A255062, A255067, A255071, A255126, A255332.

Similar sequences: A218542, A255063.

A005811(n) = hammingweight(bitxor(n, n\2));
write_A255125_and_A255126_and_A255071(n) = { my(k, i, s25, s26); k = (2^(n+1))-2; i = 1; s25 = 0; s26 = 0; while(1, if((k%4),s26++,s25++); k = k - A005811(k); if(!bitand(k+1, k+2), break, i++)); write("b255125.txt", n, " ", s25); write("b255126.txt", n, " ", s26); write("b255071.txt", n, " ", i); };
for(n=1,42,write_A255125_and_A255126_and_A255071(n));

(define (A255125 n) (if (zero? n) 1 (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s 0)) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A133872 i))))))))
;; Alternatively:
(define (A255125 n) (add (COMPOSE A059841 A255057) (A255062 n) (A255061 (+ 1 n))))
(define (A255125 n) (add (COMPOSE A059841 A255067) (A255062 n) (A255061 (+ 1 n))))
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A059841(A255057(k)).
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A059841(A255067(k)).
a(n) = A255071(n) - A255126(n).

A255066 The trunk of number-of-runs beanstalk (A255056) with reversed subsections.

Original entry on oeis.org

0, 2, 6, 4, 14, 12, 10, 30, 28, 26, 22, 18, 62, 60, 58, 54, 50, 46, 42, 36, 32, 126, 124, 122, 118, 114, 110, 106, 100, 96, 94, 90, 84, 78, 74, 68, 64, 254, 252, 250, 246, 242, 238, 234, 228, 224, 222, 218, 212, 206, 202, 196, 192, 190, 186, 180, 174, 168, 162, 156, 152, 148, 142, 138, 132, 128, 510

Offset: 0

Antti Karttunen, Feb 14 2015

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with term x = (2^(n+1))-2 and subtract repeatedly the number of runs in binary representation of x to get successive x's, until the number that has already been listed (which is always (2^n)-2) is encountered, which is not listed second time, but instead, the current row is finished [and thus containing only terms of equal binary length, A000523(n) on row n]. The next row then starts with (2^(n+2))-2, with the same process repeated.

			Rows 0 - 5 of the array:
0;
2;
6, 4;
14, 12, 10;
30, 28, 26, 22, 18;
62, 60, 58, 54, 50, 46, 42, 36, 32;
After row 0, the length of row n is given by A255071(n).

Antti Karttunen, Table of n, a(n) for n = 0..16142; Rows 0..16, flattened

Cf. A255067 (same seq, terms divided by 2).
Cf. A255071 (gives row lengths).
Cf. A000523, A004755, A005811, A236840, A255056, A255122.
Analogous sequences: A218616, A230416.

a(0) = 0, a(1) = 2, a(2) = 6; and for n > 2, a(n) = A004755(A004755(A236840(a(n-1)))) if A236840(a(n-1))+2 is power of 2, otherwise just A236840(a(n-1)) [where A004755(x) adds one 1-bit to the left of the most significant bit of x].
In other words, for n > 2, let k = A236840(a(n-1)). Then, if k+2 is not a power of 2, a(n) = k, otherwise a(n) = k + (6 * (2^A000523(k))).
Other identities. For all n >= 0:
a(n) = A255056(A255122(n)).

A255126 Number of times a number of the form 4n+2 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, 2, 4, 6, 10, 16, 27, 50, 97, 188, 355, 652, 1177, 2126, 3886, 7204, 13501, 25465, 48192, 91411, 173851, 331821, 636035, 1224505, 2366662, 4588124, 8913418, 17338878, 33756650, 65766474, 128239805, 250346859, 489422205, 958304970, 1879145187, 3689012737

Offset: 0

Antti Karttunen, Feb 18 2015

nonn

Also the number of odd numbers in range [A255062(n) .. A255061(n+1)] of A255057 (equally, in A255067). See the sum-formulas.

			For n=5 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, m(36) = 32 and finally m(32) = 30, which is (2^5)-2. Of the nine numbers encountered, only 58, 54, 50, 46, 42 and 30 are of the form 4n+2, thus a(5) = 6. Note that the initial value 2^(n+1)-2 is not included in the cases, but the final (2^n) - 2 is.

Cf. A000035, A005811, A021913, A236840, A255057, A255061, A255062, A255067, A255071, A255125.

Similar sequences: A218543, A255064.

PARI
```
\\ Use the PARI-code given in A255125.
```

(define (A255126 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s 1)) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A021913 i))))))))
;; Alternatively:
(define (A255126 n) (add (COMPOSE A000035 A255057) (A255062 n) (A255061 (+ 1 n))))
(define (A255126 n) (add (COMPOSE A000035 A255067) (A255062 n) (A255061 (+ 1 n))))
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A000035(A255057(k)).
a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A000035(A255067(k)).
a(n) = A255071(n) - A255125(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

Antti Karttunen, Feb 14 2015

nonn

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

Antti Karttunen, Table of n, a(n) for n = 0..721
Index entries for sequences that are permutations of the natural numbers

Cf. A255061, A255062, A255071, A255120, A255121.
Cf. A255056, A255057, A255066, A255067.
Similar permutations: A218602, A230432.

(define (A255122 n) (- (A255061 (+ 1 (A255121 n))) (A255120 n)))

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

cp's OEIS Frontend

A255057 The trunk of number-of-runs beanstalk, halved: a(n) = A255056(n)/2.

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A255125 Number of times a multiple of four 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).

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A255066 The trunk of number-of-runs beanstalk (A255056) with reversed subsections.

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A255126 Number of times a number of the form 4n+2 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).

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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).

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