A009996 Numbers with digits in nonincreasing order.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 21, 22, 30, 31, 32, 33, 40, 41, 42, 43, 44, 50, 51, 52, 53, 54, 55, 60, 61, 62, 63, 64, 65, 66, 70, 71, 72, 73, 74, 75, 76, 77, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 110, 111, 200, 210, 211
Offset: 1
Examples
As 10000 = C(10+6,10) - 6 + C(7+6,1+6) + C(5+5,1+5) + C(4+4,1+4) + C(3+3,1+3) + C(1+2,1+2) + C(0+1,1+1), C(0+0,1+0), a(10000) = 7543100.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
- D. Applegate, M. LeBrun, and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
- David A. Corneth, Table of n, a(n) for n = 1..20000, Jun 03 2014
- Eric Weisstein's World of Mathematics, Digit
Programs
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Mathematica
Select[Range[0,211], GreaterEqual@@IntegerDigits[#]&] (* Ray Chandler, Oct 25 2011 *)
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PARI
is(n)=my(d=digits(n)); for(i=2,#d,if(d[i]>d[i-1],return(0))); 1 \\ Charles R Greathouse IV, Jan 02 2014
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PARI
\\ This program is optimized for fast calculation of a(n) for large n. a(n)={my(q,m=10,i,r=0);n--;while(binomial(m+1,10)<=n+m-9,m++);n-=binomial(m,10);n+=m-9;q=m-9;i=q;while(n>0,m=i;while(binomial(m+1,i)<=n,m++);r=10*r+m+1-i;n-=binomial(m,i);i--;);z=q-#digits(r);r*=10^z;r} \\ David A. Corneth, Jun 01 2014 -
PARI
\\recursive--feed an element a(n)>0 and it gives a(n+1). nxt(n)={my(r,d=digits(n),y,t); if(d[#d]!=9,y=1; while(y-#d-1&&d[y]==9,y++); t=#d; forstep(i=t,y+1,-1,if(d[i-1]!=d[i],t=i-1;break)); if(t!=#d,d[t+1]++; for(i=t+2,#d,d[i]=0),d[y]++; for(i=y+1,#d,d[i]=0));r=d ,d=vector(#d+1); d[1]=1;for(i=2,#d,d[i]=0); r=d); sum(i=1,#r,10^(#r-i)*r[i])} \\ David A. Corneth, Jun 01 2014 -
Python
from itertools import count, islice, combinations_with_replacement as mc def agen(): # generator of terms yield 0 for d in count(1): ni = (int("".join(m)) for m in mc("9876543210", d) if m[0]!="0") yield from sorted(ni) print(list(islice(agen(), 70))) # Michael S. Branicky, Jun 24 2022
Formula
Binomial(n+k,k) = (n+k)!/(n!*k!). d(i) is the i-th digit of a(n). q is the number of digits of a(n). Find the highest m such that C(10 + m, 10) - m + 1 <= n. a(n) has m+1 digits. Set n = n - C(10+m,10). Find the highest d(m+1), then d(m), then ..., then d(1) each iteration such that C(d(m+1)+m+1,1+m+1)<=n. Then set n = n-C(d(m+1)+m+1,m+2). If n = 0 then stop. All remaining digits are 0.
Extensions
Corrected by Rick L. Shepherd, Jun 06 2002
Comments