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.

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A219651 a(n) = n minus (sum of digits in factorial base expansion of n).

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 5, 5, 6, 6, 7, 7, 10, 10, 11, 11, 12, 12, 15, 15, 16, 16, 17, 17, 23, 23, 24, 24, 25, 25, 28, 28, 29, 29, 30, 30, 33, 33, 34, 34, 35, 35, 38, 38, 39, 39, 40, 40, 46, 46, 47, 47, 48, 48, 51, 51, 52, 52, 53, 53, 56, 56, 57, 57, 58, 58, 61, 61
Offset: 0

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

See A007623 for the factorial base number system representation.

Crossrefs

Bisection: A219650. Analogous sequence for binary system: A011371, for Zeckendorf expansion: A219641.

Programs

  • Mathematica
    (* First run program for A007623 to define factBaseIntDs *) Table[n - Plus@@factBaseIntDs[n], {n, 0, 99}] (* Alonso del Arte, Nov 25 2012 *)
  • Python
    from itertools import count
    def A219651(n):
        c, f = 0, 1
        for i in count(2):
            f *= i
            if f>n:
                break
            c += (i-1)*(n//f)
        return c # Chai Wah Wu, Oct 11 2024
  • Scheme
    (define (A219651 n) (- n (A034968 n)))
    

Formula

a(n) = n - A034968(n).

A328316 Iterates of A276086 starting from 0.

Original entry on oeis.org

0, 1, 2, 3, 6, 5, 18, 125, 43218, 258413198822535882125
Offset: 0

Views

Author

Antti Karttunen, Oct 14 2019

Keywords

Comments

The unique infinite sequence such that a(0) = 0, a(n) = A276085(a(n+1)) for n >= 0, and A129251(a(n)) = 0 for n >= 1, i.e., all nonzero terms must be in A048103.
a(10) is 240 decimal digits long (can be found in b-file), and a(11) is too big to fit even into a b-file as it is 32700 decimal digits long, but it can be found in the given a-file.

Crossrefs

Cf. A002110, A048103, A129251, A276085, A276086, A328317 (the smallest prime not dividing a(n)), A328318, A328319 (digit sum in primorial base), A328322 (max. digit), A328323.
Cf. A153013, and also A109162, A179016, A219666, A259934 for more or less analogous sequences.
Cf. also A328313.

Programs

Formula

a(0) = 0; and for n > 0, a(n) = A276086(a(n-1)).

A219652 Number of steps to reach 0 starting with n and using the iterated process: x -> x - (sum of digits in factorial expansion of x).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19
Offset: 0

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

See A007623 for the factorial number system representation.

Crossrefs

Analogous sequence for binary system: A071542, for Zeckendorf expansion: A219642. Cf. A007623, A034968, A219650, A219651, A219653-A219655, A219659, A219661, A219666.

Programs

  • Mathematica
    nn = 72; m = 1; While[Factorial@ m < nn, m++]; m; Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, n, # > 0 &] - 1, {n, 0, nn}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

Formula

a(0)=0; for n>0, a(n) = 1 + a(A219651(n)).

Extensions

Erroneous description corrected by Antti Karttunen, Dec 03 2012

A219648 The infinite trunk of Zeckendorf beanstalk. The only infinite sequence such that a(n-1) = a(n) - number of 1's in Zeckendorf representation of a(n).

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 9, 12, 14, 17, 20, 22, 24, 27, 29, 33, 35, 37, 40, 42, 45, 47, 50, 54, 56, 58, 61, 63, 67, 70, 74, 76, 79, 83, 88, 90, 92, 95, 97, 101, 104, 108, 110, 113, 117, 121, 123, 126, 130, 134, 138, 143, 145, 147, 150, 152, 156, 159, 163, 165, 168
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Comments

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "Zeckendorf beanstalk" from its root (zero).
There are many finite sequences such as 0,1,2; 0,1,2,4,5; etc. (see A219649) and as the length increases, so (necessarily) does the similarity to this infinite sequence.
There can be only one infinite trunk in "Zeckendorf beanstalk" as all paths downwards from numbers >= A000045(i) must pass through A000045(i)-1 (i.e. A000071(i)). This provides also a well-defined method to compute the sequence, for example, via a partially reversed version A261076.
See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Crossrefs

Cf. A000045, A000071, A007895, A014417, A219641, A219649, A261076, A261102. For all n, A219642(a(n)) = n and A219643(n) <= a(n) <= A219645(n). Cf. also A261083 & A261084.
Other similarly constructed sequences: A179016, A219666, A255056.

Programs

Formula

a(n) = A261076(A261102(n)).

A226061 Partial sums of A219661.

Original entry on oeis.org

0, 1, 3, 8, 27, 110, 538, 3149, 21622, 172348, 1549896, 15401144, 168011252, 2003304293, 25928878272, 361788001015, 5411160126367, 86353882249911, 1464841397585335, 26323224850512719, 499551889319197565
Offset: 1

Views

Author

Antti Karttunen, May 28 2013

Keywords

Comments

a(n) tells the position of (n!)-1 in A219666.

Crossrefs

One less than A219665.
Analogous sequence for binary system: A218600.
Cf. also A230410, A231719.

Programs

  • Mathematica
    Accumulate@ Table[Length@ NestWhileList[# - Total@ IntegerDigits[#,
    MixedRadix[Reverse@ Range[2, 120]]] &, (n + 1)! - 1, # > n! - 1 &] - 1, {n, 0, 8}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

Formula

a(n) = a(n-1) + A219661(n-1) with a(1) = 0.
a(n) = A219652(n!-1).
a(n) = A219665(n) - 1.

Extensions

Terms a(16) - a(21) computed from the new terms of A219661 by Antti Karttunen, Jun 27 2016

A219658 Complement of A219650. Natural numbers that do not occur in A219651.

Original entry on oeis.org

3, 4, 8, 9, 13, 14, 18, 19, 20, 21, 22, 26, 27, 31, 32, 36, 37, 41, 42, 43, 44, 45, 49, 50, 54, 55, 59, 60, 64, 65, 66, 67, 68, 72, 73, 77, 78, 82, 83, 87, 88, 89, 90, 91, 95, 96, 100, 101, 105, 106, 110, 111, 112, 113, 114, 115, 116, 117, 118, 122, 123, 127
Offset: 1

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

These are positive integers i for which there does not exist any k such that A034968(i+k) = k.

Crossrefs

A219650, A219651, A219666. Analogous sequence for binary system: A055938, for Zeckendorf expansion: A219638.

A276623 The infinite trunk of ternary beanstalk: The only infinite sequence such that a(n-1) = a(n) - A053735(a(n)), where A053735(n) = base-3 digit sum of n.

Original entry on oeis.org

0, 2, 4, 8, 10, 12, 16, 20, 26, 28, 30, 34, 38, 42, 46, 52, 56, 62, 68, 72, 80, 82, 84, 88, 92, 96, 100, 106, 110, 116, 122, 126, 134, 140, 144, 152, 160, 164, 170, 176, 180, 188, 194, 198, 204, 212, 216, 224, 232, 242, 244, 246, 250, 254, 258, 262, 268, 272, 278, 284, 288, 296, 302, 306, 314, 322, 326, 332, 338, 342, 350, 356, 360
Offset: 0

Views

Author

Antti Karttunen, Sep 11 2016

Keywords

Crossrefs

Cf. A004128, A024023, A053735, A054861, A261231 (left inverse), A261233, A276622, A276624, A276603 (terms divided by 2), A276604 (first differences).
Cf. A179016, A219648, A219666, A255056, A259934, A276573, A276583, A276613 for similar constructions.
Cf. also A263273.

Programs

Formula

a(n) = A276624(A276622(n)).
Other identities. For all n >= 0:
A261231(a(n)) = n.
a(A261233(n)) = A024023(n) = 3^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

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Crossrefs

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.

A219659 Irregular table where row n (n >= 0) starts with n, the next term is A219651(n), and the successive terms are obtained by repeatedly subtracting the sum of digits in the previous term's factorial expansion, until zero is reached, after which the next row starts with one larger n.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 1, 0, 4, 2, 1, 0, 5, 2, 1, 0, 6, 5, 2, 1, 0, 7, 5, 2, 1, 0, 8, 6, 5, 2, 1, 0, 9, 6, 5, 2, 1, 0, 10, 7, 5, 2, 1, 0, 11, 7, 5, 2, 1, 0, 12, 10, 7, 5, 2, 1, 0, 13, 10, 7, 5, 2, 1, 0, 14, 11, 7, 5, 2, 1, 0, 15, 11, 7, 5, 2, 1, 0, 16, 12, 10, 7, 5, 2, 1, 0
Offset: 0

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

Rows converge towards A219666 (reversed).
See A007623 for the Factorial number system representation.

Crossrefs

Cf. A007623, A034968, A219651, A219657. Analogous sequence for binary system: A218254, for Zeckendorf expansion: A219649.

A219662 Number of times an even number is encountered, when going from (n+1)!-1 to n!-1 using the iterative process described in A219652.

Original entry on oeis.org

1, 1, 2, 10, 49, 268, 1505, 9667, 81891, 779193, 7726623, 80770479, 921442854, 11621384700, 159894957124
Offset: 1

Views

Author

Antti Karttunen, Dec 03 2012

Keywords

Comments

At least for n=7, 8, 9 and 10, a(n) is equal to a(n+1) when taken modulo n.

Examples

			(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1. Zero is an even number, so a(1)=1.
(2!)-1 (1) is reached from (3!)-1 (5) with two steps by first subtracting A034968(5) from 5 -> 2, and then subtracting A034968(2) from 2 -> 1. Two is an even number, but one is not, so a(2)=1.
(3!)-1 (5) is reached from (4!)-1 (23) with five steps by repeatedly subtracting the sum of digits in factorial expansion as: 23 - 6 = 17, 17 - 5 = 12, 12 - 2 = 10, 10 - 3 = 7, 7 - 2 = 5. Of these only 12 and 10 are even numbers, so a(3)=2.
		

Crossrefs

Programs

  • Scheme
    (definec (A219662 n) (if (< n 2) n (let loop ((i (- (A000142 (1+ n)) (A000217 n) 1)) (s 0)) (cond ((isA000142? (1+ i)) (+ s (- 1 (modulo i 2)))) (else (loop (A219651 i) (+ s (- 1 (modulo i 2)))))))))
    (define (isA000142? n) (and (> n 0) (let loop ((n n) (i 2)) (cond ((= 1 n) #t) ((not (zero? (modulo n i))) #f) (else (loop (/ n i) (1+ i)))))))

Formula

a(n) = A219661(n) - A219663(n).
Previous Showing 21-30 of 39 results. Next