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 1-4 of 4 results.

A219666 The infinite trunk of factorial expansion beanstalk. The only infinite sequence such that a(n-1) = a(n) - sum of digits in factorial expansion of a(n).

Original entry on oeis.org

0, 1, 2, 5, 7, 10, 12, 17, 23, 25, 28, 30, 35, 40, 46, 48, 52, 57, 63, 70, 74, 79, 85, 92, 97, 102, 109, 119, 121, 124, 126, 131, 136, 142, 144, 148, 153, 159, 166, 170, 175, 181, 188, 193, 198, 204, 213, 221, 228, 238, 240, 244, 249, 255, 262, 266, 271, 277
Offset: 0

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Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "factorial beanstalk" from its root (zero).
There are many finite sequences such as 0,1,2,4; 0,1,2,5,6; etc. obeying the same condition (see A219659) and as the length increases, so (necessarily) does the similarity to this infinite sequence.
See A007623 for the factorial number system representation.

Crossrefs

Cf. A007623, A034968, A219651, A230411, A226061. For all n, A219652(a(n)) = n and A219653(n) <= a(n) <= A219655(n).
Characteristic function: Χ_A219666(n) = A230418(n+1)-A230418(n).
The first differences: A230406.
Subsets: A230428 & A230429.
Analogous sequence for binary system: A179016, for Fibonacci number system: A219648.

Programs

  • Mathematica
    nn = 10^3; m = 1; While[m! < Floor[6 nn/5], m++]; m; t = TakeWhile[Reverse@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, Floor[6 nn/5], # > 0 &], # <= nn &] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)
  • Scheme
    ;; Memoizing definec-macro from Antti Karttunen's IntSeq-library
    (definec (A219666 n) (cond ((<= n 2) n) ((= (A226061 (A230411 n)) n) (- (A000142 (A230411 n)) 1)) (else (- (A219666 (+ n 1)) (A034968 (A219666 (+ n 1)))))))
    ;; Another variant, utilizing A230416 (which gives a more convenient way to compute large number of terms of this sequence):
    (define (A219666 n) (A230416 (A230432 n)))
    ;; This function is for checking whether n belongs to this sequence:
    (define (inA219666? n) (or (zero? n) (= 1 (- (A230418 (+ 1 n)) (A230418 n)))))

Formula

a(0) = 0, a(1) = 1, and for n>1, if A226061(A230411(n)) = n then a(n) = A230411(n)!-1, otherwise a(n) = a(n+1) - A034968(a(n+1)).
a(n) = A230416(A230432(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

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Comments

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.

Examples

			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).
		

Crossrefs

Cf. A255067 (same seq, terms divided by 2).
Cf. A255071 (gives row lengths).
Analogous sequences: A218616, A230416.

Formula

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)).

A230432 Simple self-inverse permutation of natural numbers: after zero, list each block of A219661(n) numbers in reverse order, from A226061(n+1) to A219665(n).

Original entry on oeis.org

0, 1, 3, 2, 8, 7, 6, 5, 4, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 110, 109, 108, 107, 106, 105, 104, 103, 102, 101, 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73
Offset: 0

Views

Author

Antti Karttunen, Oct 22 2013

Keywords

Comments

This permutation can be used to map between the sequences A219666 and A230416. E.g. A219666(n) = A230416(a(n)) and vice versa: A230416(n) = A219666(a(n)).

Crossrefs

Analogous sequence for binary system: A218602.

Programs

Formula

a(n) = A219665(A230411(n+1)) - A230431(n) - 1.

A230411 a(n) = minimal k for which A219665(k) >= n; a(n) = one more than the factorial base width (A084558) of the (n-1)th term in the infinite trunk of factorial beanstalk (A219666).

Original entry on oeis.org

1, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
Offset: 1

Views

Author

Antti Karttunen, Oct 22 2013

Keywords

Comments

a(1)=1, after which each term n occurs A219661(n-1) times.
Auxiliary function for computing A219666, A230431 and A230432.

Crossrefs

Analogous sequence for binary system: A213711.

Formula

a(n) = 1 + A084558(A219666(n-1)) = 1 + A084558(A230416(n-1)). [Each a(n) is one more than the number of digits needed in factorial base to write the (n-1)-th term in the infinite trunk of factorial beanstalk]
Showing 1-4 of 4 results.