A219654 -id:A219654 - cp's OEIS frontend

A219655 Greatest inverse of A219652; a(n) = maximal i such that A219652(i) = n.

0, 1, 3, 5, 7, 11, 15, 19, 23, 25, 29, 33, 37, 41, 47, 51, 55, 59, 65, 71, 77, 83, 89, 95, 101, 107, 115, 119, 121, 125, 129, 133, 137, 143, 147, 151, 155, 161, 167, 173, 179, 185, 191, 197, 203, 211, 217, 225, 233, 239, 243, 247, 251, 257, 263, 269, 275, 281

Offset: 0

Antti Karttunen, Nov 25 2012

nonn

A. Karttunen, Table of n, a(n) for n = 0..10080

Cf. A219653 for the least inverse. A219654 gives the first differences.

This sequence is based on factorial number system: A007623. Analogous sequence for binary system: A173601, for Zeckendorf expansion: A219645.

a(n) = A219653(n) + A219654(n) - 1.

A219653 Least inverse of A219652; a(n) = minimal i such that A219652(i) = n.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 12, 16, 20, 24, 26, 30, 34, 38, 42, 48, 52, 56, 60, 66, 72, 78, 84, 90, 96, 102, 108, 116, 120, 122, 126, 130, 134, 138, 144, 148, 152, 156, 162, 168, 174, 180, 186, 192, 198, 204, 212, 218, 226, 234, 240, 244, 248, 252, 258, 264, 270, 276

Offset: 0

Antti Karttunen, Nov 25 2012

nonn

A. Karttunen, Table of n, a(n) for n = 0..10080

Cf. A219655 for the greatest inverse. A219654 gives the first differences.

This sequence is based on Factorial number system: A007623. Analogous sequence for binary system: A213708 and for Zeckendorf expansion: A219643. Cf. A219652, A219659, A219666.

A255054 Run lengths in A255072.

Original entry on oeis.org

1, 2, 3, 1, 4, 3, 1, 4, 4, 4, 3, 1, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 7, 6, 7, 8, 5, 6, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4

Offset: 0

Antti Karttunen, Feb 14 2015

nonn

Number of integers k which require exactly n steps to reach 0, when starting from k and iterating the map: x -> x - (number of runs in binary representation of x).

			0 is the only number reached from 0 in zero steps, thus a(0) = 1.
Both 1 and 2, in binary '1' and '10', when the number of runs (A005811) is subtracted from them, result zero: 1-1 = 2-2 = 0, and these are only such numbers where the zero is reached with one step, thus a(1) = 2.
For 3, 4 and 5, in binary '11', '100' and '101', subtracting the number of runs results 2 in all cases, thus two steps are requires to reach zero, and as there are no other such cases, a(2) = 3.
For 6, in binary '110', subtracting A005811 repeatedly gives -> 6-2 = 4, 4-2 = 2, 2-2 = 0, three steps in total, and as 6 is the only such number requiring three steps, a(3) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..16143

Cf. A005811, A236840, A255053, A255055, A255072, A255123, A255124, A255056.
Cf. A255059 (positions of odd terms), A255060 (positions of even terms), A255061 (apart from its second term 1, gives positions of ones here).
Analogous sequences: A086876, A219644, A219654.

a(n) = A255053(n+1) - A255053(n).
a(n) = 1 + A255055(n) - A255053(n).
Other identities. For all n >= 0:
a(n) = 1 + A255123(n) + A255124(n).

A219644 Run lengths in A219642.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 3, 2, 3, 3, 2, 3, 3, 3, 2, 2, 3, 3, 3, 2, 3, 4, 1, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 3, 4, 4, 4, 5, 1, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 3, 4, 4, 4, 5, 3, 3, 3, 5, 5, 3, 5, 5, 3, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3

Offset: 0

Antti Karttunen, Nov 24 2012

nonn

a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219642 (if going around in 0->0 loop is disallowed).

A. Karttunen, Table of n, a(n) for n = 0..10000

a(n) = 1+(A219645(n)-A219643(n)).

This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. Analogous sequence for binary system: A086876, for factorial number system: A219654.

a(n) = A219643(n+1)-A219643(n). (The first differences of A219643).

A231723 a(n) = the difference between the n-th node of the infinite trunk of the factorial beanstalk (A219666(n)) and the smallest integer (A219653(n)) which is as many A219651-iteration steps distanced from the root (zero); a(n) = A219666(n) - A219653(n).

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 0, 1, 3, 1, 2, 0, 1, 2, 4, 0, 0, 1, 3, 4, 2, 1, 1, 2, 1, 0, 1, 3, 1, 2, 0, 1, 2, 4, 0, 0, 1, 3, 4, 2, 1, 1, 2, 1, 0, 0, 1, 3, 2, 4, 0, 0, 1, 3, 4, 2, 1, 1, 2, 1, 0, 0, 0, 0, 0, 0, 1, 3, 4, 2, 1, 1, 2, 1, 0, 0, 0, 0, 0, 0, 1, 3, 5, 7, 8, 1, 0

Offset: 0

Antti Karttunen, Nov 13 2013

nonn

For all n, the following holds: A219653(n) <= A219666(n) <= A219655(n). This sequence gives the distance of the node n in the infinite trunk of factorial beanstalk (A219666(n)) from the left (lesser) edge of the A219654(n) wide window which it at that point must pass through.

This sequence relates to the factorial base representation (A007623) in the same way as A218603 relates to the binary system and similar remarks apply here.

Antti Karttunen, Table of n, a(n) for n = 0..3149

Cf. A231724, A230409, A219662 & A219663.

(define (A231723 n) (- (A219666 n) (A219653 n)))

a(n) = A219666(n) - A219653(n).

A219654(n) = a(n) + A231724(n) + 1.

A231724 a(n) = the difference between the n-th node of the infinite trunk of the factorial beanstalk (A219666(n)) and the greatest integer (A219655(n)) which is as many A219651-iteration steps distanced from the root (zero); a(n) = A219655(n) - A219666(n).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 3, 2, 0, 0, 1, 3, 2, 1, 1, 3, 3, 2, 2, 1, 3, 4, 4, 3, 4, 5, 6, 0, 0, 1, 3, 2, 1, 1, 3, 3, 2, 2, 1, 3, 4, 4, 3, 4, 5, 7, 4, 4, 5, 1, 3, 3, 2, 2, 1, 3, 4, 4, 3, 4, 5, 7, 5, 7, 7, 5, 6, 6, 1, 3, 4, 4, 3, 4, 5, 7, 5, 7, 7, 5, 6, 6, 2, 2, 3, 4, 5

Offset: 0

Antti Karttunen, Nov 13 2013

nonn

For all n, the following holds: A219653(n) <= A219666(n) <= A219655(n). This sequence gives the distance of the node n in the infinite trunk of factorial beanstalk (A219666(n)) from the right (greater) edge of the A219654(n) wide window which it at that point must pass through.

This sequence relates to the factorial base representation (A007623) in the same way as A218604 relates to the binary system and similar remarks apply here.

Antti Karttunen, Table of n, a(n) for n = 0..3149

Cf. A231723, A230409, A219662 & A219663.

(define (A231724 n) (- (A219655 n) (A219666 n)))

a(n) = A219655(n) - A219666(n).

A219654(n) = a(n) + A231723(n) + 1.

cp's OEIS Frontend

A219655 Greatest inverse of A219652; a(n) = maximal i such that A219652(i) = n.

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A219653 Least inverse of A219652; a(n) = minimal i such that A219652(i) = n.

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A255054 Run lengths in A255072.

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A219644 Run lengths in A219642.

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A231723 a(n) = the difference between the n-th node of the infinite trunk of the factorial beanstalk (A219666(n)) and the smallest integer (A219653(n)) which is as many A219651-iteration steps distanced from the root (zero); a(n) = A219666(n) - A219653(n).

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A231724 a(n) = the difference between the n-th node of the infinite trunk of the factorial beanstalk (A219666(n)) and the greatest integer (A219655(n)) which is as many A219651-iteration steps distanced from the root (zero); a(n) = A219655(n) - A219666(n).

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