A237424 Numbers of the form (10^a + 10^b + 1)/3.

1, 4, 7, 34, 37, 67, 334, 337, 367, 667, 3334, 3337, 3367, 3667, 6667, 33334, 33337, 33367, 33667, 36667, 66667, 333334, 333337, 333367, 333667, 336667, 366667, 666667, 3333334, 3333337, 3333367, 3333667, 3336667

Offset: 1

Author

Ahmad J. Masad, Feb 07 2014

Keywords

• nonn

Comments

Has the property that the product of any two (not necessarily distinct) terms has digits in nondecreasing order.

Conjecture: This sequence is in a sense the maximally dense sequence with this nondecreasing products property. That is, it appears that every maximal sequence is either (i) A237424, (ii) a finite set of "extra" terms plus A237424 restricted to b=0 (which is A093137), or (iii) a finite set of "extra" terms plus A237424 restricted to a=b (which is A067275). (There might be one more case, not yet identified.) - David Applegate, Feb 12 2014

See A254143 and link for products a(i)*a(j) in natural order. - Reinhard Zumkeller, Jan 28 2015

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 1035 terms from Robert G. Wilson v)
• Reinhard Zumkeller, First 10000 products of any two terms of A237424

Crossrefs

Cf. A028819, A028820, A028821, A052216, A067275, A093137, A234841, A254143.

Programs

Haskell

a237424 = flip div 3 . (+ 1) . a052216
-- Reinhard Zumkeller, Jan 28 2015

Magma

A052216:=[10^(n-1) + 10^(k-1): k in [1..n], n in [1..100]];
A237424:= func< n | (A052216[n]+1)/3 >;
[A237424(n): n in [1..100]]; // G. C. Greubel, Feb 22 2024

Mathematica

Union@ Flatten@ Table[(10^a + 10^b + 1)/3, {a, 0, 8}, {b, a, 8}] (* Robert G. Wilson v, Jan 26 2015 *)
(10^#[[1]]+10^#[[2]]+1)/3&/@Tuples[Range[0,8],2]//Union (* Harvey P. Dale, May 28 2019 *)

PARI

list(lim)=my(v=List(),a,t); while(1, for(b=0,a, t=(10^a+10^b+1)/3; if(t>lim, return(Set(v))); listput(v, t)); a++) \\ Charles R Greathouse IV, May 13 2015

Python

from math import isqrt
def A237424(n): return (10**(a:=(k:=isqrt(m:=n<<1))+(m>k*(k+1))-1)+10**(n-1-(a*(a+1)>>1))+1)//3 # Chai Wah Wu, Apr 08 2025

SageMath

A052216=flatten([[10^(n-1) + 10^(k-1) for k in range(1,n+1)] for n in range(1,101)])
def A237424(n): return (A052216[n-1]+1)//3
[A237424(n) for n in range(1,101)] # G. C. Greubel, Feb 22 2024

Formula

a(n) = (A052216(n) + 1)/3. - Reinhard Zumkeller, Jan 28 2015

Extensions

Edited by David Applegate, Feb 07 2014