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.

Showing 1-2 of 2 results.

A131865 Partial sums of powers of 16.

Original entry on oeis.org

1, 17, 273, 4369, 69905, 1118481, 17895697, 286331153, 4581298449, 73300775185, 1172812402961, 18764998447377, 300239975158033, 4803839602528529, 76861433640456465, 1229782938247303441, 19676527011956855057, 314824432191309680913, 5037190915060954894609
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 22 2007

Keywords

Comments

16 = 2^4 is the growth measure for the Jacobsthal spiral (compare with phi^4 for the Fibonacci spiral). - Paul Barry, Mar 07 2008
Second quadrisection of A115451. - Paul Curtz, May 21 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=16, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014899. Also, the sequence is related to A014931 by A014931(n+1) = (n+1)*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Nov 07 2012
a(n) is the total number of holes in a certain box fractal (start with 16 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015
Except for 1 and 17, all terms are Brazilian repunits numbers in base 16, and so belong to A125134. All terms >= 273 are composite because a(n) = ((4^(n+1) + 1) * (4^(n+1) - 1))/15. - Bernard Schott, Jun 06 2017
The sequence in binary is 1, 10001, 100010001, 1000100010001, 10001000100010001, ... cf. Plouffe link, A330135. - Frank Ellermann, Mar 05 2020

Examples

			a(3) = 1 + 16 + 256 + 4096 = 4369 = in binary: 1000100010001.
a(4) = (16^5 - 1)/15 = (4^5 + 1) * (4^5 - 1)/15 = 1025 * 1023/15 = 205 * 341 = 69905 = 11111_16. - _Bernard Schott_, Jun 06 2017
		

Crossrefs

Programs

Formula

a(n) = if n=0 then 1 else a(n-1) + A001025(n).
for n > 0: A131851(a(n)) = n and abs(A131851(m)) < n for m < a(n).
a(n) = A098704(n+2)/2.
a(n) = (16^(n+1) - 1)/15. - Bernard Schott, Jun 06 2017
a(n) = (A001025(n+1) - 1)/15.
a(n) = 16*a(n-1) + 1. - Paul Curtz, May 20 2008
G.f.: 1 / ( (16*x-1)*(x-1) ). - R. J. Mathar, Feb 06 2011
E.g.f.: exp(x)*(16*exp(15*x) - 1)/15. - Stefano Spezia, Mar 06 2020

A333402 Numbers m such that the largest digit in the decimal expansion of 1/m is 1.

Original entry on oeis.org

1, 9, 10, 90, 99, 100, 900, 909, 990, 999, 1000, 9000, 9009, 9090, 9900, 9990, 9999, 10000, 90000, 90009, 90090, 90900, 90909, 99000, 99900, 99990, 99999, 100000, 900000, 900009, 900090, 900900, 909000, 909090, 990000, 990099, 999000, 999900, 999990, 999999, 1000000
Offset: 1

Views

Author

Bernard Schott, Mar 19 2020

Keywords

Comments

If m is a term, 10*m is also a term.
If m is a term then m has only digits {1}, {9}, {1,0} or {9,0} in its decimal representation, but this is not sufficient to be a term (see examples).
Some subsequences below (not exhaustive, see crossrefs):
m = 10^k, k >= 0, hence m is in A011557 = {1, 10, 100, 1000, 10000, ...};
m = 9*10^k, k >= 0, hence m is in A052268 = {9, 90, 900, 9000, 90000, ...};
m = 10^k-1, k >= 1, hence m is in A002283 = {9, 99, 999, 9999, 99999, ...};
m = 9*(10^k+1), k >= 1, hence m is in 9*A000533 = {99, 909, 9009, 90009, ...};
m = 9+100*(100^k-1)/11, k >= 0, hence m is in 9*A094028 = {9, 909, 90909, 9090909, ...}.

Examples

			As 1/101 = 0.009900990099..., 101 is not a term.
As 1/909 = 0.001100110011..., 909 is a term.
As 1/9099 = 0.000109902187..., 9099 is not a term.
As 1/9999 = 0.000100010001..., 9999 is also a term.
		

Crossrefs

Cf. A333236, A333237 (similar, with 9).
Subsequences: A002283, A011557, A052268.
Subsequences: 9*A000533, 9*A094028, 9*A135577, 9*A261544, 9*A330135.

Programs

  • Mathematica
    Select[Range[10^4], Max @ RealDigits[1/#][[1]] == 1 &] (* Amiram Eldar, Mar 19 2020 *)
  • Python
    from itertools import count, islice
    def A333402_gen(startvalue=1): # generator of terms >= startvalue
        for m in count(max(startvalue,1)):
            k = 1
            while k <= m:
                k *= 10
            rset = {0}
            while True:
                k, r = divmod(k, m)
                if max(str(k)) > '1':
                    break
                else:
                    if r in rset:
                        yield m
                        break
                rset.add(r)
                k = r
                while k <= m:
                    k *= 10
    A333402_list = list(islice(A333402_gen(),30)) # Chai Wah Wu, Feb 17 2022

Formula

A333236(a(n))= 1.

Extensions

More terms from Jinyuan Wang, Mar 19 2020
Showing 1-2 of 2 results.