A255071 -id:A255071 - cp's OEIS frontend

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).

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).

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

Antti Karttunen, Feb 14 2015

nonn

			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.

Cf. A001969, A005811, A010059, A236840, A254113, A255056, A255064, A255066, A255071.

Similar sequences: A255125, A218542.

\\ 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));

(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)))))

(define (A255063 n) (add A254113 (A255062 n) (A255061 (+ 1 n))))

(define (A255063 n) (add (COMPOSE A010059 A255066) (A255062 n) (A255061 (+ 1 n))))

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

Antti Karttunen, Feb 14 2015

nonn

			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.

Cf. A000069, A005811, A010060, A236840, A254114, A255056, A255063, A255066, A255071.

Similar sequences: A255126, A218543.

PARI
```
\\ Use the Pari-code given in A255063.
```

(define (A255064 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s (modulo n 2))) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A010060 i))))))))

(define (A255064 n) (add A254114 (A255062 n) (A255061 (+ 1 n))))

(define (A255064 n) (add (COMPOSE A010060 A255066) (A255062 n) (A255061 (+ 1 n))))

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).

cp's OEIS Frontend

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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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).

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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).

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