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.

A143164 Numbers with digitsum 13, in increasing order.

Original entry on oeis.org

49, 58, 67, 76, 85, 94, 139, 148, 157, 166, 175, 184, 193, 229, 238, 247, 256, 265, 274, 283, 292, 319, 328, 337, 346, 355, 364, 373, 382, 391, 409, 418, 427, 436, 445, 454, 463, 472, 481, 490, 508, 517, 526, 535, 544, 553, 562, 571, 580, 607, 616, 625, 634, 643, 652
Offset: 1

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Author

Wolfdieter Lang, Sep 15 2008

Keywords

Comments

If 13 is considered as an 'unlucky' number: the 'unlucky years'.
A007953(a(n)) = 13; number of repdigits = A242627(13) = 1. - Reinhard Zumkeller, Jul 17 2014

Examples

			2029 is the next 'unlucky year'. Solution to the guardian weekly puzzle.
a(10^ 1) = 166
a(10^ 2) = 1309
a(10^ 3) = 21370
a(10^ 4) = 1100254
a(10^ 5) = 111032122
a(10^ 6) = 30611101000
a(10^ 7) = 40100300100301
a(10^ 8) = 200011001012211010
a(10^ 9) = 10001220000100012002100
a(10^10) = 1100000001010021010000000230 - _David A. Corneth_, Jan 31 2015
		

References

  • The Guardian Weekly, July 25-31, 2008, p.39 puzzles 5., p31.

Crossrefs

Cf. A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052223 (9), A052224 (10), A166311 (11), A235151 (12), A235225(14), A235226 (15), A235227 (16), A166370 (17), A235228 (18), A166459 (19), A235229 (20).

Programs

  • Haskell
    a143164 n = a143164_list !! (n-1)
    a143164_list = filter ((== 13) . a007953) [0..]
    -- Reinhard Zumkeller, Jul 17 2014
    
  • Mathematica
    f[n_] := Rest@ Select[Range@ n, NestWhile[Plus @@ IntegerDigits[#] &, #, # > 14 &] == 13 &]; f@ 652 (* Michael De Vlieger, Feb 03 2015 *)
    Select[Range[700],Total[IntegerDigits[#]]==13&] (* Harvey P. Dale, Oct 11 2017 *)
  • PARI
    \\This algorithm needs a modified binomial.
    C(n,k)=if(n>=k,binomial(n,k),0)
    \\ways to roll s-q with q dice having sides 0 through n - 1.
    b(s,q,n)=if(s<=q*(n-1),s+=q;sum(i=0,q-1,(-1)^i*C(q,i)*C(s-1-n*i,q-1)),0)
    \\main algorithm
    a(n) = {my(q); q = 2; while(b(13, q, 10) < n, q++); q--; s = 13; os = 13; r=0; while(q, if(b(s,q,10) < n, n-=b(s,q,10);s--, r+=(os-s)*10^(q); os = s; q--)); r+= s;r}
    \\inverse
    inv(n)={r = 1; v=digits(n); l=v[#v]; forstep(i = #v-1, 1, -1, for(j=1, v[i], r+=b(l+j, #v-i, 10)); l+=v[i]); r} \\ David A. Corneth, Jan 31 2015
    
  • PARI
    transform(n,b)=my(d=digits(n),nd=#d,v=vector(b,i,[i\10,b-(b+1-i)\10]),k); v[b][2]=d[1]; v
    list(lim)=my(v=List(),d=transform(lim\=1,13)); forvec(u=transform(lim\1,13), if(u[4]Charles R Greathouse IV, May 30 2019

Formula

digitsum(a(n))=13, ordered increasingly.