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.

Previous Showing 21-27 of 27 results.

A276802 Non-repdigit numbers n such that A045876(n) ends with n.

Original entry on oeis.org

554, 3328, 55553, 77764, 222221, 444442, 666663, 888865, 888884, 5555552, 6666595, 9999840, 33332680, 55555526, 66666557, 99998670, 333332176, 333333312, 555555551, 666665752, 666666624, 999997536, 999999936, 9999976480, 9999997844, 9999999668, 9999999923, 11111111110
Offset: 1

Views

Author

Altug Alkan, Sep 17 2016

Keywords

Examples

			554 is a term because 455+545+554 = 1554 that ends with 554.
2338 is the least term having its digits. For all permutations p of digits of n, in this case 2338, (without leading zeros if any), A045876(n) = A045876(p). A045876(2338) = 53328. It contains the digits of 2338 and ends with its digits permuted. 2338 has 4 digits, as has 53328 mod 10^4 so 53328 mod 10^4 == 3328 is a term. - _David A. Corneth_, Oct 04 2016
		

Crossrefs

Cf. A045876, A139819 (non-repdigits), A179239.

Programs

Extensions

More terms from David A. Corneth, Oct 06 2016

A309415 Number of different numbers that are formed by permuting digits of n!.

Original entry on oeis.org

1, 1, 1, 1, 2, 6, 6, 12, 60, 360, 1260, 10080, 15120, 25200, 1247400, 32432400, 12612600, 6810804000, 7264857600, 185253868800, 1005663859200, 1117404288000, 4839757322400, 93504111468768000, 37401644587507200, 160787493266400000, 13023786954578400000
Offset: 0

Views

Author

Hauke Löffler, Jul 30 2019

Keywords

Comments

a(0) = 1 because 0! = 1 has one permutation (1).
a(4) = 2 because 4! = 24 has two permutations (24, 42).
a(5) = 6 because 5! = 120 has 6 permutations (012, 021, 102, 120, 201, 210).

Crossrefs

Programs

  • Haskell
    import Data.List (permutations, nub)
    factorial n = product [1..n]
    a309415 n = length $ nub $ permutations $ show $ factorial n
    map a309415 [0..]
  • Maple
    a:= n-> (l-> combinat[multinomial](add(i, i=l), l[])
          )([coeffs(add(x^i, i=convert(n!, base, 10)))]):
    seq(a(n), n=0..30);  # Alois P. Heinz, Jul 30 2019
  • Mathematica
    a[n_] := Block[{w = IntegerDigits[n!]}, Length[w]! / Times @@ ((Last /@ Tally[w])!)]; Array[a, 26, 0] (* Giovanni Resta, Jul 30 2019 *)

Extensions

More terms from Giovanni Resta, Jul 30 2019

A276596 Least k such that A276502(k) = n.

Original entry on oeis.org

1, 10, 100, 1000, 10000, 101, 1000000, 1001
Offset: 1

Views

Author

Altug Alkan, Sep 11 2016

Keywords

Crossrefs

Cf. A276502.

Programs

A276739 Least k such that A045876(k) is divisible by 10^n.

Original entry on oeis.org

1, 19, 10699, 102589, 10000112389
Offset: 0

Views

Author

Altug Alkan, Sep 16 2016

Keywords

Comments

Corresponding values of A045876(k) are 1, 110, 3333300, 333333000, ...
Sequence is infinite.
a(5) > 10^18. - Giovanni Resta, Sep 27 2016
Subsequence of A179239. - David A. Corneth, Oct 01 2016

Examples

			a(1) = 19 because 19+91 = 110 is divisible by 10.
		

Crossrefs

Cf. A045876.

Programs

  • Mathematica
    Table[k = 1; While[! Divisible[Total[FromDigits /@ Permutations@ IntegerDigits@ k], 10^n], k++]; k, {n, 0, 3}] (* Michael De Vlieger, Sep 16 2016 *)
  • PARI
    A047726(n) = n=digits(n); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
    A055642(n) = #Str(n);
    A007953(n) = sumdigits(n);
    A045876(n) = ((10^A055642(n)-1)/9)*(A047726(n)*A007953(n)/A055642(n));
    a(n) = {my(k = 1); while (A045876(k) % (10^n), k++); k;}

Extensions

a(4) from Giovanni Resta, Sep 27 2016

A276758 Numbers n such that A045876(n) = A045876(n+1).

Original entry on oeis.org

10, 1010, 1100, 1119, 1339, 1519, 3139, 5119, 8899, 27799, 46699, 48499, 50559, 55059, 64699, 72799, 84499, 100110, 101010, 101100, 110010, 110100, 111000, 111229, 112129, 117799, 121129, 136699, 147499, 163699, 168199, 171799, 174499, 177199, 186199
Offset: 1

Views

Author

Altug Alkan, Sep 17 2016

Keywords

Comments

A138147 is a subsequence. Therefore, the sequence is infinite. - David A. Corneth, Sep 17 2016
Suppose a term is of the form SDN, where S is a sequence of digits without leading zeros, D is a digit less than 9 and N is a sequence of digits 9 (possibly 0 nines; terms from A002283) and SDN is a concatenation of S, D and N. Let S' be a permutation of digits of S without leading zeros. Then S'DN is also in the sequence. To search terms one may choose S from A179239. - David A. Corneth, Sep 18 2016
Since (n + 8*k) = (n - k + 1)*(n - k) has solutions that are n = k + 3*sqrt(k) and n = k - 3*sqrt(k), for square values of k there are infinitely many terms such that: 1119, 1111119999, 111111111999999999, ...

Examples

			1339 is a term because A045876(1339) = A045876(1340).
See 2nd comment. As 27799 is in the sequence, we can see S = 27, D = 7 and N = 99. Now all permutations S' (distinct) of S without leading zeros give terms. They are 72, giving term 72799. - _David A. Corneth_, Sep 18 2016
		

Crossrefs

Programs

A276810 Numbers n such that A045876(n) has distinct decimal digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 39, 48, 49, 57, 58, 59, 67, 68, 69, 75, 76, 78, 79, 84, 85, 86, 87, 89, 93, 94, 95, 96, 97, 98, 149, 158, 167, 176, 185, 194, 199, 239, 248, 257, 275, 284, 289, 293, 298, 329, 347, 356, 365, 374, 379, 388, 392, 397, 419, 428, 437, 469, 473, 478, 482
Offset: 1

Views

Author

Altug Alkan, Sep 18 2016

Keywords

Comments

This sequence contains 146 elements. The largest is 991. No more terms below 10^10. As A045876(n) >= n, for all n >= 10^10, A045876(n) will have at least one digit not distinct. - David A. Corneth, Sep 19 2016

Examples

			289 is a term because 289+298+829+892+928+982 = 4218 has distinct decimal digits.
		

Crossrefs

Programs

Extensions

Clarified comment. - Harvey P. Dale, Apr 30 2022

A304657 Number of ways to represent n/100 years in Shinigami time.

Original entry on oeis.org

26, 76, 76, 208, 176, 252, 176, 544, 208, 604, 176, 768, 176, 604, 604, 1376, 176, 768, 176, 1888, 604, 604, 176, 2208, 818, 604, 544, 1888, 176, 2316, 176, 3392, 604, 604, 1460, 2568, 176, 604, 604, 5536, 176, 2316, 176, 1888, 1888, 604, 176, 6080, 818, 3172
Offset: 1

Views

Author

Felix Fröhlich, May 16 2018

Keywords

Comments

"Lifespan" and "Shinigami time", as used in this sequence, are concepts from the Death Note manga and anime series by Tsugumi Ohba and Takeshi Obata. The lifespan is represented in the series as a set of numbers floating above a person's head when that person is looked at through the Shinigami Eyes. For example, the lifespan of Light Yagami, the main character in the series, is represented as 9 3 31 2 6 3 9 in chapter 5 of the manga and episode 3 of the anime. (In the anime, it is unclear whether the 2 and the 6 are supposed to be connected or not. If they are treated as 26, the resulting number of years would be too large, see below). However, the exact meaning of these numbers is never revealed. Tsugumi Ohba, the author of the manga, supposedly said in an interview he created "a complicated math equation for Light's Life span when the numbers first appeared above his head", but later forgot what it was.
The following is my own theory regarding the meaning of the lifespan numbers of Light Yagami. I started by making an assumption: Light was supposed to live for about 70 to 90 years, neither extremely short, nor extremely long for a Japanese around the year 2000 (cf. WHO, p. 167). When one treats the lifespan as displayed in the manga as a sequence of numbers, computes their product and divides the number by 3600, one obtains 9 * 3 * 31 * 2 * 6 * 3 * 9 = 271188 and 271188/3600 = 75.33, which could be interpreted as meaning that Light was supposed to live for 75 years and 4 months. That number lies within the range 70-90 and also is too nice a number to be a coincidence in my opinion. This suggests that Ohba probably came up with the lifespan by calculating 75.33 * 3600 = 271188, factoring that number into 2 * 2 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 31, multiplying some of the primes to obtain 2 * 6 * 9 * 9 * 3 * 3 * 31 and then rearranging those factors.
With the above assumptions it is clear that, due to the commutative property of multiplication, a given lifespan in years can have more than one representation in Shinigami time.
Also, while it is unclear whether the lifespan numbers represent the original or remaining lifespan of an individual, rule no. 42 could be interpreted as meaning that the lifespan of Light that Ryuk sees is the original lifespan that probably never changes as Light ages (cf. Death Note Wiki, Manga Chapter Rules - Rule #42).
It seems the terms of the sequence are also terms of A034776. Why?
0.01 years, equal to roughly 3.5 days or 5256 minutes, is the shortest timespan that can be represented in Shinigami time. - Felix Fröhlich, Apr 04 2021

Examples

			For n = 1, we have 0.01*3600 = 36 and 36 = 2 * 2 * 3 * 3.
Thus, the possible representations are as follows:
2 2 3 3; 2 3 2 3; 2 3 3 2; 3 2 2 3; 3 2 3 2; 3 3 2 2
2 2 9; 2 3 6; 2 6 3; 2 9 2; 3 2 6; 3 3 4; 3 4 3; 3 6 2; 4 3 3; 6 2 3; 6 3 2; 9 2 2
2 18; 3 12; 4 9; 6 6; 9 4; 12 3; 18 2
36
There are 6 representations of length 4, 12 representations of length 3, 7 representations of length 2 and 1 representation of length 1. 6 + 12 + 7 + 1 = 26, so a(1) = 26.
		

References

  • Tsugumi Ohba and Takeshi Obata, Death Note 13: How to Read, Paw Prints, 2008.

Crossrefs

Programs

  • PARI
    step_a(n) = n/100
    step_b(n) = floor(n*3600)
    step_c(n, o=[1])=if(n>1, concat(apply(t->vector(t[2], i, t[1]), Vec(factor(n)~))), o) \\ after M. F. Hasler in A027746
    find_index_a(vec) = my(r=#vec-1); while(1, if(vec[r] < vec[r+1], return(r)); r--; if(r==0, return(-1)))
    find_index_b(r, vec) = my(s=#vec); while(1, if(vec[r] < vec[s], return(s)); s--; if(s==r, return(-1)))
    switch_elements(vec, firstpos, secondpos) = my(g); g=vec[secondpos]; vec[secondpos]=vec[firstpos]; vec[firstpos] = g; vec \\ from David A. Corneth
    reverse_order(vec, r) = my(v=[], w=[]); for(x=1, r, v=concat(v, vec[x])); for(y=r+1, #vec, w=concat(w, vec[y])); w=Vecrev(w); concat(v, w)
    next_permutation(vec) = my(r=find_index_a(vec)); if(r==-1, return(0), my(s=find_index_b(r, vec)); vec=switch_elements(vec, r, s); vec=reverse_order(vec, r)); vec
    multiply_neighboring_elements(vec, pos) = my(x=vec[pos]*vec[pos+1], w=[]); for(k=1, pos-1, w=concat(w, vec[k])); w=concat(w, [x]); for(r=pos+2, #vec, w=concat(w, vec[r])); w
    all_permutations(vec) = my(v=[vecsort(vec)], w=vecsort(vec)); while(1, w=next_permutation(w); if(w==0, return(v), v=concat(v, [0]); v[#v]=w))
    all_sets_from_multiplying_neighboring_elements(vec) = my(v=[]); for(k=1, #vec-1, v=concat(v, [0]); v[#v]=multiply_neighboring_elements(vec, k)); v
    count_eligible_sets(vec) = my(v=all_permutations(vec), w=[], i=#v); while(1, for(x=1, #v, my(asfmne=all_sets_from_multiplying_neighboring_elements(v[x])); if(#asfmne==0, return(i)); for(y=1, #asfmne, w=concat(w, [0]); w[#w]=asfmne[y]); w=vecsort(w, , 8)); i=i+#w; v=w; w=[])
    a(n) = my(f=step_c(step_b(step_a(n)))); count_eligible_sets(f)

Extensions

Entry revised by Felix Fröhlich, Apr 03 2021
Previous Showing 21-27 of 27 results.